Modular Forms and L-functions, Math 8207-8208
Modular Forms and L-functions, Math 8207-8208
A course in modern number theory and harmonic analysis
1 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}(xy)^{xy}\,dx\,dy=\int _{0}^{1}x^{x}\,dx}{\displaystyle \int _{0}^{1}\int _{0}^{1}(xy)^{xy}\,dx\,dy=\int _{0}^{1}x^{x}\,dx}
Beweis
{\displaystyle \int _{0}^{1}\int _{0}^{1}(xy)^{xy}\,dx\,dy}{\displaystyle \int _{0}^{1}\int _{0}^{1}(xy)^{xy}\,dx\,dy} ist nach Substitution {\displaystyle x\mapsto {\frac {x}{y}}}{\displaystyle x\mapsto {\frac {x}{y}}} gleich {\displaystyle \int _{0}^{1}\int _{0}^{y}x^{x}\,{\frac {dx}{y}}\,dy}{\displaystyle \int _{0}^{1}\int _{0}^{y}x^{x}\,{\frac {dx}{y}}\,dy}.
Nach dem Vertauschen der Integrationsreihenfolge ist dies {\displaystyle \int _{0}^{1}\int _{x}^{1}x^{x}\,{\frac {dy}{y}}\,dx=\int _{0}^{1}x^{x}\,(-\log x)\,dx}{\displaystyle \int _{0}^{1}\int _{x}^{1}x^{x}\,{\frac {dy}{y}}\,dx=\int _{0}^{1}x^{x}\,(-\log x)\,dx}.
Wegen {\displaystyle \int _{0}^{1}x^{x}\,(1+\log x)\,dx=\left[x^{x}\right]_{0}^{1}=0}{\displaystyle \int _{0}^{1}x^{x}\,(1+\log x)\,dx=\left[x^{x}\right]_{0}^{1}=0} ist {\displaystyle \int _{0}^{1}x^{x}\,dx=\int _{0}^{1}x^{x}\,(-\log x)\,dx}{\displaystyle \int _{0}^{1}x^{x}\,dx=\int _{0}^{1}x^{x}\,(-\log x)\,dx}.
2 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\sqrt {x^{2}+y^{2}}}\,dx\,dy={\frac {1}{3}}\left({\sqrt {2}}+\log \left({\sqrt {2}}+1\right)\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\sqrt {x^{2}+y^{2}}}\,dx\,dy={\frac {1}{3}}\left({\sqrt {2}}+\log \left({\sqrt {2}}+1\right)\right)}
Beweis
Teile das Einheitsquadrat {\displaystyle Q=[0,1]^{2}\,}{\displaystyle Q=[0,1]^{2}\,} in zwei kongruente Dreiecke, jeweils mit dem Ursprung als Spitze.
{\displaystyle Q_{1}=\{(x,y)\in Q\,:\,x\geq y\}\,,\,Q_{2}=\{(x,y)\in Q\,:\,y\geq x\}}{\displaystyle Q_{1}=\{(x,y)\in Q\,:\,x\geq y\}\,,\,Q_{2}=\{(x,y)\in Q\,:\,y\geq x\}}
{\displaystyle I:=\iint _{Q}{\sqrt {x^{2}+y^{2}}}\,dx\,dy=2\iint _{Q_{1}}{\sqrt {x^{2}+y^{2}}}\,dx\,dy}{\displaystyle I:=\iint _{Q}{\sqrt {x^{2}+y^{2}}}\,dx\,dy=2\iint _{Q_{1}}{\sqrt {x^{2}+y^{2}}}\,dx\,dy}
Verwende nun Polarkoordinaten:
{\displaystyle {\begin{bmatrix}x=r\cos \varphi \\y=r\sin \varphi \\\end{bmatrix}}\quad \qquad dx\,dy=r\,dr\,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}}}}{\displaystyle {\begin{bmatrix}x=r\cos \varphi \\y=r\sin \varphi \\\end{bmatrix}}\quad \qquad dx\,dy=r\,dr\,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}}}}
Wegen
{\displaystyle x\geq y\iff r\cos \varphi \geq r\sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}}{\displaystyle x\geq y\iff r\cos \varphi \geq r\sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}}
{\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\cos \varphi }}}{\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\cos \varphi }}}
ist {\displaystyle I=2\int _{0}^{\frac {\pi }{4}}\int _{0}^{\frac {1}{\cos \varphi }}r^{2}\,dr\,d\varphi ={\frac {2}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {d\varphi }{\cos ^{3}\varphi }}}{\displaystyle I=2\int _{0}^{\frac {\pi }{4}}\int _{0}^{\frac {1}{\cos \varphi }}r^{2}\,dr\,d\varphi ={\frac {2}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {d\varphi }{\cos ^{3}\varphi }}}
{\displaystyle ={\frac {1}{3}}\left[{\frac {\sin \varphi }{\cos ^{2}\varphi }}+\log(\sec \varphi +\tan \varphi )\right]_{0}^{\frac {\pi }{4}}={\frac {1}{3}}\left({\sqrt {2}}+\log \left({\sqrt {2}}+1\right)\right)}{\displaystyle ={\frac {1}{3}}\left[{\frac {\sin \varphi }{\cos ^{2}\varphi }}+\log(\sec \varphi +\tan \varphi )\right]_{0}^{\frac {\pi }{4}}={\frac {1}{3}}\left({\sqrt {2}}+\log \left({\sqrt {2}}+1\right)\right)}.
3 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=\log({\sqrt {3}}+1)-{\frac {\log 2}{2}}+{\frac {\sqrt {3}}{4}}-{\frac {\pi }{24}}}{\displaystyle \int _{0}^{1}\int _{0}^{1}\int _{0}^{1}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=\log({\sqrt {3}}+1)-{\frac {\log 2}{2}}+{\frac {\sqrt {3}}{4}}-{\frac {\pi }{24}}}
Beweis
Teile den Einheitswürfel {\displaystyle V=[0,1]^{3}\,}{\displaystyle V=[0,1]^{3}\,} in drei kongruente Pyramiden mit dem Ursprung als Spitze.
{\displaystyle V_{1}=\{(x,y,z)\in V\,:\,x\geq y,z\}\,,\,V_{2}=\{(x,y,z)\in V\,:\,y\geq x,z\}\,,\,V_{3}=\{(x,y,z)\in V\,:\,z\geq x,y\}}{\displaystyle V_{1}=\{(x,y,z)\in V\,:\,x\geq y,z\}\,,\,V_{2}=\{(x,y,z)\in V\,:\,y\geq x,z\}\,,\,V_{3}=\{(x,y,z)\in V\,:\,z\geq x,y\}}
{\displaystyle I:=\iiint _{V}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=3\iiint _{V_{1}}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz}{\displaystyle I:=\iiint _{V}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz=3\iiint _{V_{1}}{\sqrt {x^{2}+y^{2}+z^{2}}}\,dx\,dy\,dz}
Verwende nun Kugelkoordinaten:
{\displaystyle {\begin{bmatrix}x=r\sin \theta \cos \varphi \\y=r\sin \theta \sin \varphi \\z=r\cos \theta \end{bmatrix}}\quad \qquad dx\,dy\,dz=r^{2}\sin \theta \,dr\,d\theta \,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}+z^{2}}}}{\displaystyle {\begin{bmatrix}x=r\sin \theta \cos \varphi \\y=r\sin \theta \sin \varphi \\z=r\cos \theta \end{bmatrix}}\quad \qquad dx\,dy\,dz=r^{2}\sin \theta \,dr\,d\theta \,d\varphi \quad \qquad r={\sqrt {x^{2}+y^{2}+z^{2}}}}
Wegen
{\displaystyle x\geq y\iff r\sin \theta \cos \varphi \geq r\sin \theta \sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}}{\displaystyle x\geq y\iff r\sin \theta \cos \varphi \geq r\sin \theta \sin \varphi \iff \cos \varphi \geq \sin \varphi \iff 0\leq \varphi \leq {\frac {\pi }{4}}}
{\displaystyle x\geq z\iff r\sin \theta \cos \varphi \geq r\cos \theta \iff \tan \theta \geq \sec \varphi \iff \arctan(\sec \varphi )\leq \theta \leq {\frac {\pi }{2}}}{\displaystyle x\geq z\iff r\sin \theta \cos \varphi \geq r\cos \theta \iff \tan \theta \geq \sec \varphi \iff \arctan(\sec \varphi )\leq \theta \leq {\frac {\pi }{2}}}
{\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\sin \theta \,\cos \varphi }}}{\displaystyle x\leq 1\iff 0\leq r\leq {\frac {1}{\sin \theta \,\cos \varphi }}}
ist {\displaystyle I=3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\int _{0}^{\frac {1}{\sin \theta \,\cos \varphi }}r\cdot r^{2}\,\sin \theta \,dr\,d\theta \,d\varphi }{\displaystyle I=3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\int _{0}^{\frac {1}{\sin \theta \,\cos \varphi }}r\cdot r^{2}\,\sin \theta \,dr\,d\theta \,d\varphi }
{\displaystyle =3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\left[{\frac {r^{4}}{4}}\right]_{0}^{\frac {1}{\sin \theta \,\cos \varphi }}\,\sin \theta \,d\theta \,d\varphi ={\frac {3}{4}}\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}\,{\frac {d\varphi }{\cos ^{4}\varphi }}}{\displaystyle =3\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}\left[{\frac {r^{4}}{4}}\right]_{0}^{\frac {1}{\sin \theta \,\cos \varphi }}\,\sin \theta \,d\theta \,d\varphi ={\frac {3}{4}}\int _{0}^{\frac {\pi }{4}}\int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}\,{\frac {d\varphi }{\cos ^{4}\varphi }}}.
Substituiere {\displaystyle \theta ={\text{arccot}}\,s}{\displaystyle \theta ={\text{arccot}}\,s}, dann ist {\displaystyle \int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}=\int _{0}^{\cos \varphi }{\sqrt {1+s^{2}}}\,ds=\left[{\frac {s}{2}}{\sqrt {1+s^{2}}}+{\frac {1}{2}}\,{\text{arsinh}}\,s\right]_{0}^{\cos \varphi }}{\displaystyle \int _{\arctan(\sec \varphi )}^{\frac {\pi }{2}}{\frac {d\theta }{\sin ^{3}\theta }}=\int _{0}^{\cos \varphi }{\sqrt {1+s^{2}}}\,ds=\left[{\frac {s}{2}}{\sqrt {1+s^{2}}}+{\frac {1}{2}}\,{\text{arsinh}}\,s\right]_{0}^{\cos \varphi }}.
Also ist {\displaystyle I={\frac {3}{8}}\int _{0}^{\frac {\pi }{4}}{\frac {\sqrt {1+\cos ^{2}\varphi }}{\cos ^{3}\varphi }}\,d\varphi +{\frac {3}{8}}\int _{0}^{\frac {\pi }{4}}{\frac {{\text{arsinh}}(\cos \varphi )}{\cos ^{4}\varphi }}\,d\varphi }{\displaystyle I={\frac {3}{8}}\int _{0}^{\frac {\pi }{4}}{\frac {\sqrt {1+\cos ^{2}\varphi }}{\cos ^{3}\varphi }}\,d\varphi +{\frac {3}{8}}\int _{0}^{\frac {\pi }{4}}{\frac {{\text{arsinh}}(\cos \varphi )}{\cos ^{4}\varphi }}\,d\varphi },
wobei {\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sqrt {1+\cos ^{2}\varphi }}{\cos ^{3}\varphi }}\,d\varphi =\int _{0}^{1}{\frac {\sqrt {1+{\frac {1}{1+x^{2}}}}}{\frac {1}{{\sqrt {1+x^{2}}}^{3}}}}\,{\frac {dx}{1+x^{2}}}=\int _{0}^{1}{\sqrt {2+x^{2}}}\,dx}{\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sqrt {1+\cos ^{2}\varphi }}{\cos ^{3}\varphi }}\,d\varphi =\int _{0}^{1}{\frac {\sqrt {1+{\frac {1}{1+x^{2}}}}}{\frac {1}{{\sqrt {1+x^{2}}}^{3}}}}\,{\frac {dx}{1+x^{2}}}=\int _{0}^{1}{\sqrt {2+x^{2}}}\,dx}
{\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}+{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}}{\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}+{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}} ist. Und {\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {{\text{arsinh}}(\cos \varphi )}{\cos ^{4}\varphi }}\,d\varphi }{\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {{\text{arsinh}}(\cos \varphi )}{\cos ^{4}\varphi }}\,d\varphi }
{\displaystyle =\left[{\text{arsinh}}(\cos \varphi )\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}{\frac {\sin \varphi }{\sqrt {1+\cos ^{2}\varphi }}}\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)d\varphi }{\displaystyle =\left[{\text{arsinh}}(\cos \varphi )\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)\right]_{0}^{\frac {\pi }{4}}+\int _{0}^{\frac {\pi }{4}}{\frac {\sin \varphi }{\sqrt {1+\cos ^{2}\varphi }}}\left({\frac {1}{3}}\,{\frac {\sin \varphi }{\cos ^{3}\varphi }}+{\frac {2}{3}}{\frac {\sin \varphi }{\cos \varphi }}\right)d\varphi }
{\displaystyle ={\frac {4}{3}}\,{\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {1}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos ^{3}\varphi }}\,d\varphi +{\frac {2}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos \varphi }}\,d\varphi }{\displaystyle ={\frac {4}{3}}\,{\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {1}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos ^{3}\varphi }}\,d\varphi +{\frac {2}{3}}\int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos \varphi }}\,d\varphi },
wobei {\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos ^{3}\varphi }}\,d\varphi =\int _{0}^{1}{\frac {\frac {x^{2}}{1+x^{2}}}{{\sqrt {1+{\frac {1}{1+x^{2}}}}}\,{\frac {1}{{\sqrt {1+x^{2}}}^{3}}}}}\,{\frac {dx}{1+x^{2}}}=\int _{0}^{1}{\frac {x^{2}}{\sqrt {2+x^{2}}}}\,dx}{\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos ^{3}\varphi }}\,d\varphi =\int _{0}^{1}{\frac {\frac {x^{2}}{1+x^{2}}}{{\sqrt {1+{\frac {1}{1+x^{2}}}}}\,{\frac {1}{{\sqrt {1+x^{2}}}^{3}}}}}\,{\frac {dx}{1+x^{2}}}=\int _{0}^{1}{\frac {x^{2}}{\sqrt {2+x^{2}}}}\,dx}
{\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}-{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}}{\displaystyle =\left[{\frac {x}{2}}{\sqrt {2+x^{2}}}-{\text{arsinh}}{\frac {x}{\sqrt {2}}}\right]_{0}^{1}={\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}} ist,
und {\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos \varphi }}\,d\varphi =\int _{0}^{1}{\frac {\frac {x^{2}}{1+x^{2}}}{{\sqrt {1+{\frac {1}{1+x^{2}}}}}\,{\frac {1}{\sqrt {1+x^{2}}}}}}\,{\frac {dx}{1+x^{2}}}}{\displaystyle \int _{0}^{\frac {\pi }{4}}{\frac {\sin ^{2}\varphi }{{\sqrt {1+\cos ^{2}\varphi }}\,\cos \varphi }}\,d\varphi =\int _{0}^{1}{\frac {\frac {x^{2}}{1+x^{2}}}{{\sqrt {1+{\frac {1}{1+x^{2}}}}}\,{\frac {1}{\sqrt {1+x^{2}}}}}}\,{\frac {dx}{1+x^{2}}}}
{\displaystyle =\int _{0}^{1}{\frac {x^{2}}{{\sqrt {2+x^{2}}}\,(1+x^{2})}}\,dx=\left[{\text{arsinh}}{\frac {x}{\sqrt {2}}}-\arctan {\frac {x}{\sqrt {2+x^{2}}}}\right]_{0}^{1}={\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}}{\displaystyle =\int _{0}^{1}{\frac {x^{2}}{{\sqrt {2+x^{2}}}\,(1+x^{2})}}\,dx=\left[{\text{arsinh}}{\frac {x}{\sqrt {2}}}-\arctan {\frac {x}{\sqrt {2+x^{2}}}}\right]_{0}^{1}={\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}}.
Also ist {\displaystyle I={\frac {3}{8}}\left({\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}\right)+{\frac {3}{8}}\left({\frac {4}{3}}\,{\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {1}{3}}\left({\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}\right)+{\frac {2}{3}}\left({\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}\right)\right)}{\displaystyle I={\frac {3}{8}}\left({\frac {\sqrt {3}}{2}}+{\text{arsinh}}{\frac {1}{\sqrt {2}}}\right)+{\frac {3}{8}}\left({\frac {4}{3}}\,{\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {1}{3}}\left({\frac {\sqrt {3}}{2}}-{\text{arsinh}}{\frac {1}{\sqrt {2}}}\right)+{\frac {2}{3}}\left({\text{arsinh}}{\frac {1}{\sqrt {2}}}-{\frac {\pi }{6}}\right)\right)}
{\displaystyle ={\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {\sqrt {3}}{4}}-{\frac {\pi }{24}}}{\displaystyle ={\text{arsinh}}{\frac {1}{\sqrt {2}}}+{\frac {\sqrt {3}}{4}}-{\frac {\pi }{24}}}, wobei {\displaystyle {\text{arsinh}}{\frac {1}{\sqrt {2}}}=\log({\sqrt {3}}+1)-{\frac {\log 2}{2}}}{\displaystyle {\text{arsinh}}{\frac {1}{\sqrt {2}}}=\log({\sqrt {3}}+1)-{\frac {\log 2}{2}}} ist.
4 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}\max \left\{x^{\alpha _{1}-1}\,y^{\beta _{1}-1},x^{\alpha _{2}-1}\,y^{\beta _{2}-1}\right\}\,dx\,dy={\frac {{\frac {\alpha _{1}-\alpha _{2}}{\alpha _{2}}}+{\frac {\beta _{2}-\beta _{1}}{\beta _{1}}}}{\alpha _{1}\beta _{2}-\alpha _{2}\beta _{1}}}\qquad \alpha _{1}>\alpha _{2}>0\,,\,\beta _{2}>\beta _{1}>0}{\displaystyle \int _{0}^{1}\int _{0}^{1}\max \left\{x^{\alpha _{1}-1}\,y^{\beta _{1}-1},x^{\alpha _{2}-1}\,y^{\beta _{2}-1}\right\}\,dx\,dy={\frac {{\frac {\alpha _{1}-\alpha _{2}}{\alpha _{2}}}+{\frac {\beta _{2}-\beta _{1}}{\beta _{1}}}}{\alpha _{1}\beta _{2}-\alpha _{2}\beta _{1}}}\qquad \alpha _{1}>\alpha _{2}>0\,,\,\beta _{2}>\beta _{1}>0}
ohne Beweis
5 Bearbeiten
Ist {\displaystyle V=\left\{(x_{1},...,x_{n})\geq 0\,\left|\,\left({\frac {x_{1}}{a_{1}}}\right)^{b_{1}}+...+\left({\frac {x_{n}}{a_{n}}}\right)^{b_{n}}\leq 1\right.\right\}}{\displaystyle V=\left\{(x_{1},...,x_{n})\geq 0\,\left|\,\left({\frac {x_{1}}{a_{1}}}\right)^{b_{1}}+...+\left({\frac {x_{n}}{a_{n}}}\right)^{b_{n}}\leq 1\right.\right\}}, so gilt
{\displaystyle \int _{V}x_{1}^{c_{1}-1}\cdots x_{n}^{c_{n}-1}\,dx_{1}\cdots dx_{n}={\frac {{a_{1}}^{c_{1}}\cdots {a_{n}}^{c_{n}}}{b_{1}\cdots b_{n}}}\,{\frac {\Gamma \left({\frac {c_{1}}{b_{1}}}\right)\cdots \Gamma \left({\frac {c_{n}}{b_{n}}}\right)}{\Gamma \left({\frac {c_{1}}{b_{1}}}+...+{\frac {c_{n}}{b_{n}}}+1\right)}}}{\displaystyle \int _{V}x_{1}^{c_{1}-1}\cdots x_{n}^{c_{n}-1}\,dx_{1}\cdots dx_{n}={\frac {{a_{1}}^{c_{1}}\cdots {a_{n}}^{c_{n}}}{b_{1}\cdots b_{n}}}\,{\frac {\Gamma \left({\frac {c_{1}}{b_{1}}}\right)\cdots \Gamma \left({\frac {c_{n}}{b_{n}}}\right)}{\Gamma \left({\frac {c_{1}}{b_{1}}}+...+{\frac {c_{n}}{b_{n}}}+1\right)}}}
ohne Beweis
6 Bearbeiten
{\displaystyle \int _{0}^{1}\cdots \int _{0}^{1}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}\,dx_{1}\cdots dx_{n}={\frac {{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}\qquad \alpha _{1},...,\alpha _{n}>0}{\displaystyle \int _{0}^{1}\cdots \int _{0}^{1}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}\,dx_{1}\cdots dx_{n}={\frac {{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}\qquad \alpha _{1},...,\alpha _{n}>0}
Beweis
Es sei {\displaystyle V=[0,1]^{n}\,}{\displaystyle V=[0,1]^{n}\,} der {\displaystyle n\,}n\,-dimensionale Einheitswürfel und {\displaystyle V_{j}=\left\{(x_{1},...,x_{n})\in V\,{\Big |}\,x_{i}^{\alpha _{i}}<x_{j}^{\alpha _{j}}\quad \forall i\neq j\right\}}{\displaystyle V_{j}=\left\{(x_{1},...,x_{n})\in V\,{\Big |}\,x_{i}^{\alpha _{i}}<x_{j}^{\alpha _{j}}\quad \forall i\neq j\right\}}.
Für alle Tupel {\displaystyle (x_{1},...,x_{n})\in V_{j}\,}{\displaystyle (x_{1},...,x_{n})\in V_{j}\,} ist dann {\displaystyle x_{j}^{\alpha _{j}}=\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}}{\displaystyle x_{j}^{\alpha _{j}}=\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}} und {\displaystyle x_{i}<x_{j}^{\alpha _{j}/\alpha _{i}}\,\,\forall i\neq j}{\displaystyle x_{i}<x_{j}^{\alpha _{j}/\alpha _{i}}\,\,\forall i\neq j}.
{\displaystyle V\,}V\, ist die abgeschlossene Hülle der disjunkten Vereinigung aller {\displaystyle V_{j}}{\displaystyle V_{j}}'s. Daher ist {\displaystyle \int _{V}=\int _{V_{1}}+...+\int _{V_{n}}}{\displaystyle \int _{V}=\int _{V_{1}}+...+\int _{V_{n}}}.
{\displaystyle \int _{V_{j}}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}dx_{1}\cdots dx_{n}=\int _{V_{j}}x_{j}^{\alpha _{j}}dx_{1}\cdots dx_{n}}{\displaystyle \int _{V_{j}}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}dx_{1}\cdots dx_{n}=\int _{V_{j}}x_{j}^{\alpha _{j}}dx_{1}\cdots dx_{n}}
{\displaystyle =\int _{0}^{1}\,\int _{0}^{x_{j}^{\alpha _{j}/\alpha _{n}}}\cdots {\widehat {\int _{0}^{x_{j}^{\alpha _{j}/\alpha _{j}}}}}\cdots \int _{0}^{x_{j}^{\alpha _{j}/\alpha _{1}}}x_{j}^{\alpha _{j}}\,dx_{1}\cdots {\widehat {dx_{j}}}\cdots dx_{n}\cdot dx_{j}}{\displaystyle =\int _{0}^{1}\,\int _{0}^{x_{j}^{\alpha _{j}/\alpha _{n}}}\cdots {\widehat {\int _{0}^{x_{j}^{\alpha _{j}/\alpha _{j}}}}}\cdots \int _{0}^{x_{j}^{\alpha _{j}/\alpha _{1}}}x_{j}^{\alpha _{j}}\,dx_{1}\cdots {\widehat {dx_{j}}}\cdots dx_{n}\cdot dx_{j}}
{\displaystyle =\int _{0}^{1}x_{j}^{\alpha _{j}}\,\,x_{j}^{\alpha _{j}/\alpha _{1}}\cdots {\widehat {x_{j}^{\alpha _{j}/\alpha _{j}}}}\cdots x_{j}^{\alpha _{j}/\alpha _{n}}\,dx_{j}=\int _{0}^{1}x_{j}^{\alpha _{j}\left(1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}\right)}\,dx_{j}}{\displaystyle =\int _{0}^{1}x_{j}^{\alpha _{j}}\,\,x_{j}^{\alpha _{j}/\alpha _{1}}\cdots {\widehat {x_{j}^{\alpha _{j}/\alpha _{j}}}}\cdots x_{j}^{\alpha _{j}/\alpha _{n}}\,dx_{j}=\int _{0}^{1}x_{j}^{\alpha _{j}\left(1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}\right)}\,dx_{j}}
{\displaystyle ={\frac {1}{\alpha _{j}\left(1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}\right)+1}}={\frac {\frac {1}{\alpha _{j}}}{1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}+{\frac {1}{\alpha _{j}}}}}={\frac {\frac {1}{\alpha _{j}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}}{\displaystyle ={\frac {1}{\alpha _{j}\left(1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}\right)+1}}={\frac {\frac {1}{\alpha _{j}}}{1+{\frac {1}{\alpha _{1}}}+...+{\widehat {\frac {1}{\alpha _{j}}}}+...+{\frac {1}{\alpha _{n}}}+{\frac {1}{\alpha _{j}}}}}={\frac {\frac {1}{\alpha _{j}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}}.
Also ist {\displaystyle \int _{0}^{1}\cdots \int _{0}^{1}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}\,dx_{1}\cdots dx_{n}=\int _{V_{1}}+...+\int _{V_{n}}={\frac {{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}}{\displaystyle \int _{0}^{1}\cdots \int _{0}^{1}\max \left\{x_{1}^{\alpha _{1}},...,x_{n}^{\alpha _{n}}\right\}\,dx_{1}\cdots dx_{n}=\int _{V_{1}}+...+\int _{V_{n}}={\frac {{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}{1+{\frac {1}{\alpha _{1}}}+...+{\frac {1}{\alpha _{n}}}}}}.
7 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(-\log xy)^{s-2}}{1-xy}}\,(1-x)\,dx\,dy=\Gamma (s)\left(\zeta (s)-{\frac {1}{s-1}}\right)\qquad {\text{Re}}(s)>0}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {(-\log xy)^{s-2}}{1-xy}}\,(1-x)\,dx\,dy=\Gamma (s)\left(\zeta (s)-{\frac {1}{s-1}}\right)\qquad {\text{Re}}(s)>0}
Beweis (Hadjicostas Formel)
{\displaystyle I:=\int _{0}^{1}\int _{0}^{1}{\frac {(-\log xy)^{s-2}}{1-xy}}\,(1-x)\,dy\,dx}{\displaystyle I:=\int _{0}^{1}\int _{0}^{1}{\frac {(-\log xy)^{s-2}}{1-xy}}\,(1-x)\,dy\,dx}
ist nach Substitution {\displaystyle u=xy\,}{\displaystyle u=xy\,} gleich {\displaystyle \int _{0}^{1}\int _{0}^{x}{\frac {(-\log u)^{s-2}}{1-u}}\,(1-x)\,{\frac {du}{x}}\,dx}{\displaystyle \int _{0}^{1}\int _{0}^{x}{\frac {(-\log u)^{s-2}}{1-u}}\,(1-x)\,{\frac {du}{x}}\,dx}.
Vertauscht man die Integrationsreihenfolge, so ist {\displaystyle I=\int _{0}^{1}\int _{u}^{1}\left({\frac {1}{x}}-1\right)dx\,{\frac {(-\log u)^{s-2}}{1-u}}\,du}{\displaystyle I=\int _{0}^{1}\int _{u}^{1}\left({\frac {1}{x}}-1\right)dx\,{\frac {(-\log u)^{s-2}}{1-u}}\,du}
{\displaystyle =\int _{0}^{1}\left(-\log u-(1-u)\right){\frac {(-\log u)^{s-2}}{1-u}}\,du=\int _{0}^{1}\left({\frac {(-\log u)^{s-1}}{1-u}}-(-\log u)^{s-2}\right)du}{\displaystyle =\int _{0}^{1}\left(-\log u-(1-u)\right){\frac {(-\log u)^{s-2}}{1-u}}\,du=\int _{0}^{1}\left({\frac {(-\log u)^{s-1}}{1-u}}-(-\log u)^{s-2}\right)du}.
Substituiert man {\displaystyle u=e^{-t}\,}{\displaystyle u=e^{-t}\,}, so ist {\displaystyle I=\int _{0}^{\infty }\left({\frac {t^{s-1}}{e^{t}-1}}-t^{s-2}\,e^{-t}\right)dt}{\displaystyle I=\int _{0}^{\infty }\left({\frac {t^{s-1}}{e^{t}-1}}-t^{s-2}\,e^{-t}\right)dt}.
Für {\displaystyle {\text{Re}}(s)>1\,}{\displaystyle {\text{Re}}(s)>1\,} ist das {\displaystyle \Gamma (s)\zeta (s)-\Gamma (s-1)=\Gamma (s)\left(\zeta (s)-{\frac {1}{s-1}}\right)}{\displaystyle \Gamma (s)\zeta (s)-\Gamma (s-1)=\Gamma (s)\left(\zeta (s)-{\frac {1}{s-1}}\right)}.
Wegen der analytischen Fortsetzbarkeit gilt dies auch für {\displaystyle {\text{Re}}(s)>0\,}{\displaystyle {\text{Re}}(s)>0\,},
wobei der Fall {\displaystyle s=1\,}{\displaystyle s=1\,} als Grenzwert {\displaystyle \gamma \,}{\displaystyle \gamma \,} zu interpretieren ist.
8 Bearbeiten
{\displaystyle \int _{0}^{\pi }\int _{0}^{\pi }\int _{0}^{\pi }{\frac {dx\,dy\,dz}{1-\cos x\,\cos y\,\cos z}}={\frac {1}{4}}\left[\Gamma \left({\frac {1}{4}}\right)\right]^{4}}{\displaystyle \int _{0}^{\pi }\int _{0}^{\pi }\int _{0}^{\pi }{\frac {dx\,dy\,dz}{1-\cos x\,\cos y\,\cos z}}={\frac {1}{4}}\left[\Gamma \left({\frac {1}{4}}\right)\right]^{4}}
Beweis (Watson Integral)
{\displaystyle I:=\int _{0}^{\pi }\int _{0}^{\pi }\int _{0}^{\pi }{\frac {dx\,dy\,dz}{1-\cos x\,\cos y\,\cos z}}}{\displaystyle I:=\int _{0}^{\pi }\int _{0}^{\pi }\int _{0}^{\pi }{\frac {dx\,dy\,dz}{1-\cos x\,\cos y\,\cos z}}} ist nach den Substitutionen {\displaystyle \left[{\begin{matrix}x\mapsto 2\arctan x\\y\mapsto 2\arctan y\\z\mapsto 2\arctan z\end{matrix}}\right]}{\displaystyle \left[{\begin{matrix}x\mapsto 2\arctan x\\y\mapsto 2\arctan y\\z\mapsto 2\arctan z\end{matrix}}\right]} gleich
{\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {{\frac {2dx}{1+x^{2}}}\cdot {\frac {2dy}{1+y^{2}}}\cdot {\frac {2dz}{1+z^{2}}}}{1-{\frac {1-x^{2}}{1+x^{2}}}\cdot {\frac {1-y^{2}}{1+y^{2}}}\cdot {\frac {1-z^{2}}{1+z^{2}}}}}=8\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {dx\,dy\,dz}{(1+x^{2})(1+y^{2})(1+z^{2})-(1-x^{2})(1-y^{2})(1-z^{2})}}}{\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {{\frac {2dx}{1+x^{2}}}\cdot {\frac {2dy}{1+y^{2}}}\cdot {\frac {2dz}{1+z^{2}}}}{1-{\frac {1-x^{2}}{1+x^{2}}}\cdot {\frac {1-y^{2}}{1+y^{2}}}\cdot {\frac {1-z^{2}}{1+z^{2}}}}}=8\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {dx\,dy\,dz}{(1+x^{2})(1+y^{2})(1+z^{2})-(1-x^{2})(1-y^{2})(1-z^{2})}}}
{\displaystyle =4\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {dx\,dy\,dz}{x^{2}+y^{2}+z^{2}+x^{2}y^{2}z^{2}}}}{\displaystyle =4\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty }{\frac {dx\,dy\,dz}{x^{2}+y^{2}+z^{2}+x^{2}y^{2}z^{2}}}}.
Wechselt man von kartesischen Koordinaten zu Kugelkoordinaten {\displaystyle \left[{\begin{matrix}x=r\,\sin \theta \,\cos \varphi \\y=r\,\sin \theta \,\sin \varphi \\z=r\,\cos \theta \qquad \;\end{matrix}}\right]}{\displaystyle \left[{\begin{matrix}x=r\,\sin \theta \,\cos \varphi \\y=r\,\sin \theta \,\sin \varphi \\z=r\,\cos \theta \qquad \;\end{matrix}}\right]}, so ist
{\displaystyle I=4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {r^{2}\,\sin \theta \;dr\,d\theta \,d\varphi }{r^{2}+r^{2}\,\sin ^{2}\theta \,\cos ^{2}\varphi \cdot r^{2}\,\sin ^{2}\theta \,\sin ^{2}\varphi \cdot r^{2}\,\cos ^{2}\theta }}}{\displaystyle I=4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {r^{2}\,\sin \theta \;dr\,d\theta \,d\varphi }{r^{2}+r^{2}\,\sin ^{2}\theta \,\cos ^{2}\varphi \cdot r^{2}\,\sin ^{2}\theta \,\sin ^{2}\varphi \cdot r^{2}\,\cos ^{2}\theta }}}
{\displaystyle =4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {dr}{1+{\Big (}r\,\sin \theta \,{\sqrt {\cos \theta }}\,{\sqrt {\cos \varphi \,\sin \varphi }}{\Big )}^{4}}}\,\sin \theta \;d\theta \,d\varphi }{\displaystyle =4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {dr}{1+{\Big (}r\,\sin \theta \,{\sqrt {\cos \theta }}\,{\sqrt {\cos \varphi \,\sin \varphi }}{\Big )}^{4}}}\,\sin \theta \;d\theta \,d\varphi }
Das ist nach Substitution {\displaystyle t=r\,\sin \theta \,{\sqrt {\cos \theta }}\,{\sqrt {\sin \varphi \,\cos \varphi }}}{\displaystyle t=r\,\sin \theta \,{\sqrt {\cos \theta }}\,{\sqrt {\sin \varphi \,\cos \varphi }}} gleich {\displaystyle 4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {dt\,d\theta \,d\varphi }{(1+t^{4})\,{\sqrt {\cos \theta }}\,{\sqrt {\sin \varphi \,\cos \varphi }}}}}{\displaystyle 4\int _{0}^{\frac {\pi }{2}}\int _{0}^{\frac {\pi }{2}}\int _{0}^{\infty }{\frac {dt\,d\theta \,d\varphi }{(1+t^{4})\,{\sqrt {\cos \theta }}\,{\sqrt {\sin \varphi \,\cos \varphi }}}}}
{\displaystyle =4\int _{0}^{\infty }{\frac {dt}{1+t^{4}}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {\cos \theta }}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {\sin \varphi \,\cos \varphi }}}=4\cdot {\frac {{\sqrt {2}}\,\pi }{4}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2\,{\sqrt {2\pi }}}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2{\sqrt {\pi }}}}={\frac {1}{4}}\,\left[\Gamma \left({\frac {1}{4}}\right)\right]^{4}}{\displaystyle =4\int _{0}^{\infty }{\frac {dt}{1+t^{4}}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\theta }{\sqrt {\cos \theta }}}\cdot \int _{0}^{\frac {\pi }{2}}{\frac {d\varphi }{\sqrt {\sin \varphi \,\cos \varphi }}}=4\cdot {\frac {{\sqrt {2}}\,\pi }{4}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2\,{\sqrt {2\pi }}}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{2{\sqrt {\pi }}}}={\frac {1}{4}}\,\left[\Gamma \left({\frac {1}{4}}\right)\right]^{4}}.
9 Bearbeiten
{\displaystyle \int _{0}^{1}\cdots \int _{0}^{1}\prod _{i=1}^{n}\left(t_{i}^{\alpha -1}\,(1-t_{i})^{\beta -1}\right)\prod _{1\leq i<j\leq n}|t_{i}-t_{j}|^{2\gamma }\,dt_{1}\cdots dt_{n}=\prod _{j=0}^{n-1}{\frac {\Gamma (\alpha +j\gamma )\,\Gamma (\beta +j\gamma )\,\Gamma (1+(j+1)\gamma )}{\Gamma (\alpha +\beta +(n+j-1)\gamma )\,\Gamma (1+\gamma )}}}{\displaystyle \int _{0}^{1}\cdots \int _{0}^{1}\prod _{i=1}^{n}\left(t_{i}^{\alpha -1}\,(1-t_{i})^{\beta -1}\right)\prod _{1\leq i<j\leq n}|t_{i}-t_{j}|^{2\gamma }\,dt_{1}\cdots dt_{n}=\prod _{j=0}^{n-1}{\frac {\Gamma (\alpha +j\gamma )\,\Gamma (\beta +j\gamma )\,\Gamma (1+(j+1)\gamma )}{\Gamma (\alpha +\beta +(n+j-1)\gamma )\,\Gamma (1+\gamma )}}}
ohne Beweis (Selberg Integral)
10.1 Bearbeiten
{\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy={\frac {2\pi }{\sqrt {3}}}\,e^{-3\lambda }}{\displaystyle \int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy={\frac {2\pi }{\sqrt {3}}}\,e^{-3\lambda }}
Beweis (Formel von Liouville)
Setze {\displaystyle R(\lambda )=\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}{\displaystyle R(\lambda )=\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}.
Differenziere nach {\displaystyle \lambda \,}{\displaystyle \lambda \,} :
{\displaystyle R'(\lambda )=\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot {\frac {3\lambda ^{2}}{xy}}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}{\displaystyle R'(\lambda )=\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(x+y+{\frac {\lambda ^{3}}{xy}}\right)}\cdot {\frac {3\lambda ^{2}}{xy}}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}
Substituiere {\displaystyle z={\frac {\lambda ^{3}}{xy}}\,\Rightarrow \,x={\frac {\lambda ^{3}}{yz}}\,\Rightarrow \,dx=-{\frac {\lambda ^{3}}{yz^{2}}}\,dz}{\displaystyle z={\frac {\lambda ^{3}}{xy}}\,\Rightarrow \,x={\frac {\lambda ^{3}}{yz}}\,\Rightarrow \,dx=-{\frac {\lambda ^{3}}{yz^{2}}}\,dz} :
{\displaystyle R'(\lambda )=\int _{0}^{\infty }\int _{\infty }^{0}e^{-\left({\frac {\lambda ^{3}}{yz}}+y+z\right)}\cdot {\frac {3z}{\lambda }}\cdot {\frac {\lambda ^{-2}}{y^{{\frac {1}{3}}-1}\cdot z^{{\frac {1}{3}}-1}}}\cdot y^{{\frac {2}{3}}-1}\cdot {\frac {-\lambda ^{3}}{yz^{2}}}\,dz\,dy}{\displaystyle R'(\lambda )=\int _{0}^{\infty }\int _{\infty }^{0}e^{-\left({\frac {\lambda ^{3}}{yz}}+y+z\right)}\cdot {\frac {3z}{\lambda }}\cdot {\frac {\lambda ^{-2}}{y^{{\frac {1}{3}}-1}\cdot z^{{\frac {1}{3}}-1}}}\cdot y^{{\frac {2}{3}}-1}\cdot {\frac {-\lambda ^{3}}{yz^{2}}}\,dz\,dy}
{\displaystyle =3\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(y+z+{\frac {\lambda ^{3}}{yz}}\right)}\cdot y^{{\frac {1}{3}}-1}\cdot z^{{\frac {2}{3}}-1}\,dz\,dy=3\cdot R(\lambda )}{\displaystyle =3\int _{0}^{\infty }\int _{0}^{\infty }e^{-\left(y+z+{\frac {\lambda ^{3}}{yz}}\right)}\cdot y^{{\frac {1}{3}}-1}\cdot z^{{\frac {2}{3}}-1}\,dz\,dy=3\cdot R(\lambda )}
{\displaystyle \Rightarrow \,R(\lambda )=R(0)\cdot e^{-3\lambda }}{\displaystyle \Rightarrow \,R(\lambda )=R(0)\cdot e^{-3\lambda }} mit {\displaystyle R(0)=\int _{0}^{\infty }\int _{0}^{\infty }e^{-(x+y)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}{\displaystyle R(0)=\int _{0}^{\infty }\int _{0}^{\infty }e^{-(x+y)}\cdot x^{{\frac {1}{3}}-1}\cdot y^{{\frac {2}{3}}-1}\,dx\,dy}
{\displaystyle =\int _{0}^{\infty }x^{{\frac {1}{3}}-1}\,e^{-x}\,dx\cdot \int _{0}^{\infty }y^{{\frac {2}{3}}-1}\,e^{-y}\,dy=\Gamma \left({\frac {1}{3}}\right)\cdot \Gamma \left({\frac {2}{3}}\right)={\frac {\pi }{\sin {\frac {\pi }{3}}}}={\frac {2\pi }{\sqrt {3}}}}{\displaystyle =\int _{0}^{\infty }x^{{\frac {1}{3}}-1}\,e^{-x}\,dx\cdot \int _{0}^{\infty }y^{{\frac {2}{3}}-1}\,e^{-y}\,dy=\Gamma \left({\frac {1}{3}}\right)\cdot \Gamma \left({\frac {2}{3}}\right)={\frac {\pi }{\sin {\frac {\pi }{3}}}}={\frac {2\pi }{\sqrt {3}}}}
10.2 Bearbeiten
{\displaystyle \int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}={\frac {1}{\sqrt {n}}}\,{\sqrt {2\pi }}^{\,n-1}\,e^{-nz}}\int_0^\infty \cdots \int_0^\infty e^{-\left(x_1+x_2+...+x_{n-1}+\frac{z^n}{x_1\, x_2\cdots x_{n-1}}\right)}\, x_1^{\frac{1}{n}-1}\, x_2^{\frac{2}{n}-1} \cdots x_{n-1}^{\frac{n-1}{n}-1}\, dx_1\, dx_2 \cdots dx_{n-1}=\frac{1}{\sqrt{n}}\, \sqrt{2\pi}^{\, n-1}\, e^{-nz}
Beweis (Formel von Liouville)
Setze {\displaystyle F(z)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}{\displaystyle F(z)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}.
{\displaystyle F'(z)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,{\frac {-n\,z^{n-1}}{x_{1}\,x_{2}\cdots x_{n-1}}}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}{\displaystyle F'(z)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(x_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,{\frac {-n\,z^{n-1}}{x_{1}\,x_{2}\cdots x_{n-1}}}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}
{\displaystyle =n\,\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left({\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}+x_{2}+...+x_{n-1}+x_{1}\right)}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\cdot {\frac {x_{1}^{\frac {1}{n}}}{z}}\cdot {\frac {-z^{n}}{x_{1}^{2}\,x_{2}\cdots x_{n-1}}}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}{\displaystyle =n\,\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left({\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}+x_{2}+...+x_{n-1}+x_{1}\right)}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\cdot {\frac {x_{1}^{\frac {1}{n}}}{z}}\cdot {\frac {-z^{n}}{x_{1}^{2}\,x_{2}\cdots x_{n-1}}}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}
Nach Substitution {\displaystyle u_{1}={\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}}{\displaystyle u_{1}={\frac {z^{n}}{x_{1}\,x_{2}\cdots x_{n-1}}}} ist {\displaystyle du_{1}={\frac {-z^{n}}{x_{1}^{2}\,x_{2}\cdots x_{n-1}}}\,dx_{1}}{\displaystyle du_{1}={\frac {-z^{n}}{x_{1}^{2}\,x_{2}\cdots x_{n-1}}}\,dx_{1}} und wegen {\displaystyle x_{1}={\frac {z^{n}}{u_{1}\cdot x_{2}\cdots x_{n-1}}}}{\displaystyle x_{1}={\frac {z^{n}}{u_{1}\cdot x_{2}\cdots x_{n-1}}}} ist {\displaystyle x_{1}^{\frac {1}{n}}={\frac {z}{u_{1}^{\frac {1}{n}}\,x_{2}^{\frac {1}{n}}\cdots x_{n-1}^{\frac {1}{n}}}}}{\displaystyle x_{1}^{\frac {1}{n}}={\frac {z}{u_{1}^{\frac {1}{n}}\,x_{2}^{\frac {1}{n}}\cdots x_{n-1}^{\frac {1}{n}}}}}.
{\displaystyle F'(z)=-n\,\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(u_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{u_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{2}^{{\frac {1}{n}}-1}\cdots x_{n-1}^{{\frac {n-2}{n}}-1}\,u_{1}^{{\frac {n-1}{n}}-1}\,du_{1}\,dx_{2}\cdots dx_{n-1}}{\displaystyle F'(z)=-n\,\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-\left(u_{1}+x_{2}+...+x_{n-1}+{\frac {z^{n}}{u_{1}\,x_{2}\cdots x_{n-1}}}\right)}\,x_{2}^{{\frac {1}{n}}-1}\cdots x_{n-1}^{{\frac {n-2}{n}}-1}\,u_{1}^{{\frac {n-1}{n}}-1}\,du_{1}\,dx_{2}\cdots dx_{n-1}}
Also ist {\displaystyle F'(z)=-n\,F(z)}{\displaystyle F'(z)=-n\,F(z)}, woraus {\displaystyle F(z)=C\cdot e^{-nz}}{\displaystyle F(z)=C\cdot e^{-nz}} folgt.
Und {\displaystyle C=F(0)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-(x_{1}+x_{2}+...+x_{n-1})}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}{\displaystyle C=F(0)=\int _{0}^{\infty }\cdots \int _{0}^{\infty }e^{-(x_{1}+x_{2}+...+x_{n-1})}\,x_{1}^{{\frac {1}{n}}-1}\,x_{2}^{{\frac {2}{n}}-1}\cdots x_{n-1}^{{\frac {n-1}{n}}-1}\,dx_{1}\,dx_{2}\cdots dx_{n-1}}
{\displaystyle =\int _{0}^{\infty }x_{1}^{{\frac {1}{n}}-1}\,e^{-x_{1}}\,dx_{1}\cdot \int _{0}^{\infty }x_{2}^{{\frac {2}{n}}-1}\,e^{-x_{2}}\,dx_{2}\cdots \int _{0}^{\infty }x_{n-1}^{{\frac {n-1}{n}}-1}\,e^{-x_{n-1}}\,dx_{n-1}}{\displaystyle =\int _{0}^{\infty }x_{1}^{{\frac {1}{n}}-1}\,e^{-x_{1}}\,dx_{1}\cdot \int _{0}^{\infty }x_{2}^{{\frac {2}{n}}-1}\,e^{-x_{2}}\,dx_{2}\cdots \int _{0}^{\infty }x_{n-1}^{{\frac {n-1}{n}}-1}\,e^{-x_{n-1}}\,dx_{n-1}}
{\displaystyle =\Gamma \left({\frac {1}{n}}\right)\cdot \Gamma \left({\frac {2}{n}}\right)\cdots \Gamma \left({\frac {n-1}{n}}\right)={\frac {1}{\sqrt {n}}}\,{\sqrt {2\pi }}^{\,n-1}}{\displaystyle =\Gamma \left({\frac {1}{n}}\right)\cdot \Gamma \left({\frac {2}{n}}\right)\cdots \Gamma \left({\frac {n-1}{n}}\right)={\frac {1}{\sqrt {n}}}\,{\sqrt {2\pi }}^{\,n-1}}.
11 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{(1-xy)\,{\sqrt {x\,(1-y)}}}}\,dx\,dy=8G}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{(1-xy)\,{\sqrt {x\,(1-y)}}}}\,dx\,dy=8G}
Beweis (Formel von Zudilin)
Substituiere {\displaystyle u=xy}{\displaystyle u=xy}:
{\displaystyle I=\int _{0}^{1}\int _{0}^{y}{\frac {1}{(1-u)\,{\sqrt {{\frac {u}{y}}\,(1-y)}}}}\,{\frac {du}{y}}\,dy=\int _{0}^{1}\int _{0}^{y}{\frac {du}{(1-u)\,{\sqrt {u}}}}\,{\frac {dy}{{\sqrt {y}}\,{\sqrt {1-y}}}}}{\displaystyle I=\int _{0}^{1}\int _{0}^{y}{\frac {1}{(1-u)\,{\sqrt {{\frac {u}{y}}\,(1-y)}}}}\,{\frac {du}{y}}\,dy=\int _{0}^{1}\int _{0}^{y}{\frac {du}{(1-u)\,{\sqrt {u}}}}\,{\frac {dy}{{\sqrt {y}}\,{\sqrt {1-y}}}}}
Vertausche die Integrationsreihenfolge:
{\displaystyle I=\int _{0}^{1}\int _{u}^{1}{\frac {dy}{{\sqrt {y}}{\sqrt {1-y}}}}\,{\frac {du}{(1-u)\,{\sqrt {u}}}}=\int _{0}^{1}2\arccos({\sqrt {u}}\,)\,{\frac {du}{(1-u)\,{\sqrt {u}}}}}{\displaystyle I=\int _{0}^{1}\int _{u}^{1}{\frac {dy}{{\sqrt {y}}{\sqrt {1-y}}}}\,{\frac {du}{(1-u)\,{\sqrt {u}}}}=\int _{0}^{1}2\arccos({\sqrt {u}}\,)\,{\frac {du}{(1-u)\,{\sqrt {u}}}}}
Substituiere {\displaystyle u=w^{2}}{\displaystyle u=w^{2}}:
{\displaystyle I=\int _{0}^{1}2\arccos w\,{\frac {2w\,dw}{(1-w^{2})\,w}}=4\int _{0}^{1}{\frac {\arccos w}{1-w^{2}}}\,dw}{\displaystyle I=\int _{0}^{1}2\arccos w\,{\frac {2w\,dw}{(1-w^{2})\,w}}=4\int _{0}^{1}{\frac {\arccos w}{1-w^{2}}}\,dw}
Substituiere {\displaystyle w=\cos t}{\displaystyle w=\cos t}:
{\displaystyle I=4\int _{\frac {\pi }{2}}^{0}{\frac {t}{1-\cos ^{2}t}}\,(-\sin t)\,dt=4\int _{0}^{\frac {\pi }{2}}{\frac {t}{\sin t}}\,dt}{\displaystyle I=4\int _{\frac {\pi }{2}}^{0}{\frac {t}{1-\cos ^{2}t}}\,(-\sin t)\,dt=4\int _{0}^{\frac {\pi }{2}}{\frac {t}{\sin t}}\,dt}
Substituiere {\displaystyle t=2\arctan s}{\displaystyle t=2\arctan s}:
{\displaystyle I=8\int _{0}^{1}{\frac {\arctan s}{\frac {2s}{1+s^{2}}}}\,{\frac {2ds}{1+s^{2}}}=8\int _{0}^{1}{\frac {\arctan s}{s}}\,ds=8G}{\displaystyle I=8\int _{0}^{1}{\frac {\arctan s}{\frac {2s}{1+s^{2}}}}\,{\frac {2ds}{1+s^{2}}}=8\int _{0}^{1}{\frac {\arctan s}{s}}\,ds=8G}
12 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{(x+y)\,{\sqrt {(1-x)(1-y)}}}}\,dx\,dy=4G}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{(x+y)\,{\sqrt {(1-x)(1-y)}}}}\,dx\,dy=4G}
Beweis
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{4\,(x+y)\,{\sqrt {(1-x)(1-y)}}}}\,dx\,dy}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{4\,(x+y)\,{\sqrt {(1-x)(1-y)}}}}\,dx\,dy}
ist nach der Substitution {\displaystyle x=1-u^{2}\,,\,y=1-v^{2}}{\displaystyle x=1-u^{2}\,,\,y=1-v^{2}} gleich
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {2u\cdot 2v}{4\,(2-u^{2}-v^{2})\,{\sqrt {u^{2}v^{2}}}}}\,du\,dv=\int _{0}^{1}\int _{0}^{1}{\frac {1}{2-u^{2}-v^{2}}}\,du\,dv=G}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {2u\cdot 2v}{4\,(2-u^{2}-v^{2})\,{\sqrt {u^{2}v^{2}}}}}\,du\,dv=\int _{0}^{1}\int _{0}^{1}{\frac {1}{2-u^{2}-v^{2}}}\,du\,dv=G}.
13 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{\sqrt {1+x^{2}+y^{2}}}}\,dx\,dy=\log \left(2+{\sqrt {3}}\right)-{\frac {\pi }{6}}}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{\sqrt {1+x^{2}+y^{2}}}}\,dx\,dy=\log \left(2+{\sqrt {3}}\right)-{\frac {\pi }{6}}}
ohne Beweis
14 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=G}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=G}
Beweis
{\displaystyle I:=\int _{0}^{1}\int _{0}^{1}{\frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=2\cdot \int _{0}^{1}\int _{0}^{y}{\frac {1}{2-x^{2}-y^{2}}}\,dx\,dy}{\displaystyle I:=\int _{0}^{1}\int _{0}^{1}{\frac {1}{2-x^{2}-y^{2}}}\,dx\,dy=2\cdot \int _{0}^{1}\int _{0}^{y}{\frac {1}{2-x^{2}-y^{2}}}\,dx\,dy}
Verwende Polarkoordinaten {\displaystyle x=r\cos \varphi \,,\,y=r\sin \varphi }{\displaystyle x=r\cos \varphi \,,\,y=r\sin \varphi }:
{\displaystyle I=\int _{0}^{\frac {\pi }{4}}\int _{0}^{\sec \varphi }{\frac {2r}{2-r^{2}}}\,dr\,d\varphi =\int _{0}^{\frac {\pi }{4}}\left[-\log(2-r^{2})\right]_{0}^{\sec \varphi }d\varphi =\int _{0}^{\frac {\pi }{4}}\left(\log 2-\log \left(2-\sec ^{2}\varphi \right)\right)d\varphi }{\displaystyle I=\int _{0}^{\frac {\pi }{4}}\int _{0}^{\sec \varphi }{\frac {2r}{2-r^{2}}}\,dr\,d\varphi =\int _{0}^{\frac {\pi }{4}}\left[-\log(2-r^{2})\right]_{0}^{\sec \varphi }d\varphi =\int _{0}^{\frac {\pi }{4}}\left(\log 2-\log \left(2-\sec ^{2}\varphi \right)\right)d\varphi }
{\displaystyle ={\frac {\pi }{4}}\log 2-\int _{0}^{\frac {\pi }{4}}\left[\log \left(2\cos ^{2}\varphi -1\right)-\log \left(\cos ^{2}\varphi \right)\right]d\varphi }{\displaystyle ={\frac {\pi }{4}}\log 2-\int _{0}^{\frac {\pi }{4}}\left[\log \left(2\cos ^{2}\varphi -1\right)-\log \left(\cos ^{2}\varphi \right)\right]d\varphi }, wobei {\displaystyle 2\cos ^{2}\varphi -1=\cos 2\varphi }{\displaystyle 2\cos ^{2}\varphi -1=\cos 2\varphi } ist.
Also ist {\displaystyle I={\frac {\pi }{4}}\log 2-\underbrace {\int _{0}^{\frac {\pi }{2}}\log(\cos \varphi )\,{\frac {d\varphi }{2}}} _{=-{\frac {\pi }{4}}\log 2}+\int _{0}^{\frac {\pi }{2}}\log \left(\cos ^{2}{\frac {\varphi }{2}}\right){\frac {d\varphi }{2}}={\frac {\pi }{2}}\log 2+\int _{0}^{\frac {\pi }{2}}\log \left(\cos {\frac {\varphi }{2}}\right)d\varphi =G}{\displaystyle I={\frac {\pi }{4}}\log 2-\underbrace {\int _{0}^{\frac {\pi }{2}}\log(\cos \varphi )\,{\frac {d\varphi }{2}}} _{=-{\frac {\pi }{4}}\log 2}+\int _{0}^{\frac {\pi }{2}}\log \left(\cos ^{2}{\frac {\varphi }{2}}\right){\frac {d\varphi }{2}}={\frac {\pi }{2}}\log 2+\int _{0}^{\frac {\pi }{2}}\log \left(\cos {\frac {\varphi }{2}}\right)d\varphi =G}.
15 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1-x^{2}}{(1+x^{2}y^{2})\,\log ^{2}(xy)}}\,dx\,dy={\frac {2G}{\pi }}-{\frac {1}{2}}-\log \left(2\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1-x^{2}}{(1+x^{2}y^{2})\,\log ^{2}(xy)}}\,dx\,dy={\frac {2G}{\pi }}-{\frac {1}{2}}-\log \left(2\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}
ohne Beweis
16 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)}
Beweis
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\,\int _{0}^{1}\int _{0}^{1}x^{k+\alpha -1}\,y^{k+\beta -1}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {1}{k+\alpha }}\cdot {\frac {1}{k+\beta }}}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\,\int _{0}^{1}\int _{0}^{1}x^{k+\alpha -1}\,y^{k+\beta -1}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\,{\frac {1}{k+\alpha }}\cdot {\frac {1}{k+\beta }}}
{\displaystyle ={\frac {1}{\alpha -\beta }}\left(\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+\beta }}-\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+\alpha }}\right)}{\displaystyle ={\frac {1}{\alpha -\beta }}\left(\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+\beta }}-\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k+\alpha }}\right)} {\displaystyle ={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)}{\displaystyle ={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)}
17 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\log {\frac {\Gamma \left({\frac {\alpha }{2}}\right)\,\Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)}{\Gamma \left({\frac {\beta }{2}}\right)\,\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}}}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\log {\frac {\Gamma \left({\frac {\alpha }{2}}\right)\,\Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)}{\Gamma \left({\frac {\beta }{2}}\right)\,\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}}}
Beweis
Aus der Formel {\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\psi \left({\frac {\beta }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {\alpha }{2}}\right)\right]\right)} folgt
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}\,(xy)^{\gamma }}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)-\psi \left({\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)-\psi \left({\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)\right]\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}\,(xy)^{\gamma }}{1+xy}}\,dx\,dy={\frac {1}{2}}\,{\frac {1}{\alpha -\beta }}\left(\left[\psi \left({\frac {1}{2}}+{\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)-\psi \left({\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)\right]-\left[\psi \left({\frac {1}{2}}+{\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)-\psi \left({\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)\right]\right)}
Integriere nach {\displaystyle \gamma \,}{\displaystyle \gamma \,}:
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}\,(xy)^{\gamma }}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\left(\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)-\log \Gamma \left({\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)\right]-\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)-\log \Gamma \left({\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)\right]\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}\,(xy)^{\gamma }}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\left(\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)-\log \Gamma \left({\frac {\beta }{2}}+{\frac {\gamma }{2}}\right)\right]-\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)-\log \Gamma \left({\frac {\alpha }{2}}+{\frac {\gamma }{2}}\right)\right]\right)}
Setze {\displaystyle \gamma =0\,}{\displaystyle \gamma =0\,}:
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\left(\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\log \Gamma \left({\frac {\beta }{2}}\right)\right]-\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\log \Gamma \left({\frac {\alpha }{2}}\right)\right]\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x^{\alpha -1}\,y^{\beta -1}}{(1+xy)\,\log(xy)}}\,dx\,dy={\frac {1}{\alpha -\beta }}\left(\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\beta }{2}}\right)-\log \Gamma \left({\frac {\beta }{2}}\right)\right]-\left[\log \Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)-\log \Gamma \left({\frac {\alpha }{2}}\right)\right]\right)}
18 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x}{(1+x^{2}y^{2})\,\log(xy)}}\,dx\,dy=\log \left({\sqrt {\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {x}{(1+x^{2}y^{2})\,\log(xy)}}\,dx\,dy=\log \left({\sqrt {\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}
ohne Beweis
19 Bearbeiten
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1-x^{2}}{(1+x^{2}y^{2})\,\log ^{2}(xy)}}\,dx\,dy={\frac {2G}{\pi }}-{\frac {1}{2}}-\log \left(2\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1-x^{2}}{(1+x^{2}y^{2})\,\log ^{2}(xy)}}\,dx\,dy={\frac {2G}{\pi }}-{\frac {1}{2}}-\log \left(2\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}
Beweis
{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\cdot (xy)^{\alpha -1}\,dx\,dy=\int _{0}^{1}\int _{0}^{1}x^{\alpha -1}\,y^{\alpha -1}\,dx\,dy-\int _{0}^{1}\int _{0}^{1}x^{\alpha }\,y^{\alpha -1}\,dx\,dy}{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\cdot (xy)^{\alpha -1}\,dx\,dy=\int _{0}^{1}\int _{0}^{1}x^{\alpha -1}\,y^{\alpha -1}\,dx\,dy-\int _{0}^{1}\int _{0}^{1}x^{\alpha }\,y^{\alpha -1}\,dx\,dy}
{\displaystyle ={\frac {1}{\alpha ^{2}}}-{\frac {1}{\alpha +1}}\cdot {\frac {1}{\alpha }}={\frac {1}{\alpha ^{2}}}-{\frac {1}{\alpha }}+{\frac {1}{\alpha +1}}}{\displaystyle ={\frac {1}{\alpha ^{2}}}-{\frac {1}{\alpha +1}}\cdot {\frac {1}{\alpha }}={\frac {1}{\alpha ^{2}}}-{\frac {1}{\alpha }}+{\frac {1}{\alpha +1}}}
Integriere zweimal hintereinander nach {\displaystyle \alpha :}{\displaystyle \alpha :}
{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\,{\frac {(xy)^{\alpha -1}}{\log(xy)}}\,dx\,dy=-{\frac {1}{\alpha }}-\log(\alpha )+\log(\alpha +1)}{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\,{\frac {(xy)^{\alpha -1}}{\log(xy)}}\,dx\,dy=-{\frac {1}{\alpha }}-\log(\alpha )+\log(\alpha +1)}
{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\,{\frac {(xy)^{\alpha -1}}{\log ^{2}(xy)}}\,dx\,dy=(\alpha +1)\cdot \log \left(1+{\frac {1}{\alpha }}\right)-1}{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x)\,{\frac {(xy)^{\alpha -1}}{\log ^{2}(xy)}}\,dx\,dy=(\alpha +1)\cdot \log \left(1+{\frac {1}{\alpha }}\right)-1}
(Die Integrationskonstanten sind null, da beide Terme auf der rechten Seite gegen null gehen für {\displaystyle \alpha \to \infty }{\displaystyle \alpha \to \infty }.)
Substituiere {\displaystyle x\mapsto x^{2}\,,\,y\mapsto y^{2}\,:}{\displaystyle x\mapsto x^{2}\,,\,y\mapsto y^{2}\,:}
{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x^{2})\,{\frac {(xy)^{2\alpha -1}}{\log ^{2}(xy)}}\,dx\,dy=(\alpha +1)\cdot \log \left(1+{\frac {1}{\alpha }}\right)-1}{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x^{2})\,{\frac {(xy)^{2\alpha -1}}{\log ^{2}(xy)}}\,dx\,dy=(\alpha +1)\cdot \log \left(1+{\frac {1}{\alpha }}\right)-1}
Substituiere {\displaystyle \alpha ={\frac {2k+1}{2}}\,:}{\displaystyle \alpha ={\frac {2k+1}{2}}\,:}
{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x^{2})\,{\frac {(xy)^{2k}}{\log ^{2}(xy)}}\,dx\,dy={\frac {2k+3}{2}}\cdot \log \left(1+{\frac {2}{2k+1}}\right)-1}{\displaystyle \int _{0}^{1}\int _{0}^{1}(1-x^{2})\,{\frac {(xy)^{2k}}{\log ^{2}(xy)}}\,dx\,dy={\frac {2k+3}{2}}\cdot \log \left(1+{\frac {2}{2k+1}}\right)-1}
Generiere eine geometrische Reihe:
{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1-x^{2}}{(1+x^{2}y^{2})\,\log ^{2}(xy)}}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\left[{\frac {2k+3}{2}}\cdot \log \left(1+{\frac {2}{2k+1}}\right)-1\right]}{\displaystyle \int _{0}^{1}\int _{0}^{1}{\frac {1-x^{2}}{(1+x^{2}y^{2})\,\log ^{2}(xy)}}\,dx\,dy=\sum _{k=0}^{\infty }(-1)^{k}\left[{\frac {2k+3}{2}}\cdot \log \left(1+{\frac {2}{2k+1}}\right)-1\right]}
{\displaystyle =\sum _{k=1}^{\infty }(-1)^{k-1}\left[{\frac {2k+1}{2}}\cdot \log \left({\frac {2k+1}{2k-1}}\right)-1\right]}{\displaystyle =\sum _{k=1}^{\infty }(-1)^{k-1}\left[{\frac {2k+1}{2}}\cdot \log \left({\frac {2k+1}{2k-1}}\right)-1\right]}
{\displaystyle =\underbrace {\sum _{k=1}^{\infty }(-1)^{k-1}\left[2k\cdot {\frac {1}{2}}\log \left({\frac {2k+1}{2k-1}}\right)-1\right]} _{-{\frac {1}{2}}+{\frac {2G}{\pi }}}+{\frac {1}{2}}\underbrace {\sum _{k=1}^{\infty }(-1)^{k-1}\,\log \left({\frac {2k+1}{2k-1}}\right)} _{-2\log \left(2\cdot {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}}{\displaystyle =\underbrace {\sum _{k=1}^{\infty }(-1)^{k-1}\left[2k\cdot {\frac {1}{2}}\log \left({\frac {2k+1}{2k-1}}\right)-1\right]} _{-{\frac {1}{2}}+{\frac {2G}{\pi }}}+{\frac {1}{2}}\underbrace {\sum _{k=1}^{\infty }(-1)^{k-1}\,\log \left({\frac {2k+1}{2k-1}}\right)} _{-2\log \left(2\cdot {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}}
Nun müssen noch die unterschweiften Ausdrücke nachgerechnet werden.
{\displaystyle F(x):=\sum _{k=1}^{\infty }(-1)^{k-1}\left[2k\cdot {\frac {1}{2}}\log \left({\frac {1+{\frac {x}{2k}}}{1-{\frac {x}{2k}}}}\right)-x\right]}{\displaystyle F(x):=\sum _{k=1}^{\infty }(-1)^{k-1}\left[2k\cdot {\frac {1}{2}}\log \left({\frac {1+{\frac {x}{2k}}}{1-{\frac {x}{2k}}}}\right)-x\right]}
{\displaystyle =\sum _{k=1}^{\infty }(-1)^{k-1}\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\,{\frac {x^{2n+1}}{(2k)^{2n}}}=\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{(2k)^{2n}}}\,x^{2n+1}=\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\,{\frac {\eta (2n)}{2^{2n}}}\,x^{2n+1}}{\displaystyle =\sum _{k=1}^{\infty }(-1)^{k-1}\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\,{\frac {x^{2n+1}}{(2k)^{2n}}}=\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{(2k)^{2n}}}\,x^{2n+1}=\sum _{n=1}^{\infty }{\frac {1}{2n+1}}\,{\frac {\eta (2n)}{2^{2n}}}\,x^{2n+1}}
{\displaystyle F'(x)=\sum _{n=1}^{\infty }\eta (2n)\,\left({\frac {x}{2}}\right)^{2n}=-{\frac {1}{2}}+{\frac {1}{4}}\cdot {\frac {\pi x}{\sin {\frac {\pi x}{2}}}}}{\displaystyle F'(x)=\sum _{n=1}^{\infty }\eta (2n)\,\left({\frac {x}{2}}\right)^{2n}=-{\frac {1}{2}}+{\frac {1}{4}}\cdot {\frac {\pi x}{\sin {\frac {\pi x}{2}}}}}
Wegen {\displaystyle F(0)=0}{\displaystyle F(0)=0} ist {\displaystyle F(1)=\int _{0}^{1}F'(x)\,dx=-{\frac {1}{2}}+{\frac {1}{4}}\int _{0}^{1}{\frac {\pi x}{\sin {\frac {\pi x}{2}}}}\,dx=-{\frac {1}{2}}+{\frac {2G}{\pi }}}{\displaystyle F(1)=\int _{0}^{1}F'(x)\,dx=-{\frac {1}{2}}+{\frac {1}{4}}\int _{0}^{1}{\frac {\pi x}{\sin {\frac {\pi x}{2}}}}\,dx=-{\frac {1}{2}}+{\frac {2G}{\pi }}}.
{\displaystyle \sum _{k=1}^{\infty }(-1)^{k-1}\,\log \left({\frac {2k+1}{2k-1}}\right)=\sum _{k=0}^{\infty }(-1)^{k}\,\log \left({\frac {2k+3}{2k+1}}\right)}{\displaystyle \sum _{k=1}^{\infty }(-1)^{k-1}\,\log \left({\frac {2k+1}{2k-1}}\right)=\sum _{k=0}^{\infty }(-1)^{k}\,\log \left({\frac {2k+3}{2k+1}}\right)}
{\displaystyle =\sum _{k=0}^{\infty }\left(\log {\frac {4k+3}{4k+1}}-\log {\frac {4k+5}{4k+3}}\right)=\sum _{k=0}^{\infty }\log {\frac {(4k+3)(4k+3)}{(4k+1)(4k+5)}}}{\displaystyle =\sum _{k=0}^{\infty }\left(\log {\frac {4k+3}{4k+1}}-\log {\frac {4k+5}{4k+3}}\right)=\sum _{k=0}^{\infty }\log {\frac {(4k+3)(4k+3)}{(4k+1)(4k+5)}}}
{\displaystyle =\log \prod _{k=0}^{\infty }{\frac {\left(k+{\frac {3}{4}}\right)\left(k+{\frac {3}{4}}\right)}{\left(k+{\frac {1}{4}}\right)\left(k+{\frac {5}{4}}\right)}}=\log {\frac {\Gamma \left({\frac {1}{4}}\right)\,\Gamma \left({\frac {5}{4}}\right)}{\Gamma \left({\frac {3}{4}}\right)\,\Gamma \left({\frac {3}{4}}\right)}}=\log \left({\frac {1}{4}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{\Gamma ^{2}\left({\frac {3}{4}}\right)}}\right)}{\displaystyle =\log \prod _{k=0}^{\infty }{\frac {\left(k+{\frac {3}{4}}\right)\left(k+{\frac {3}{4}}\right)}{\left(k+{\frac {1}{4}}\right)\left(k+{\frac {5}{4}}\right)}}=\log {\frac {\Gamma \left({\frac {1}{4}}\right)\,\Gamma \left({\frac {5}{4}}\right)}{\Gamma \left({\frac {3}{4}}\right)\,\Gamma \left({\frac {3}{4}}\right)}}=\log \left({\frac {1}{4}}\cdot {\frac {\Gamma ^{2}\left({\frac {1}{4}}\right)}{\Gamma ^{2}\left({\frac {3}{4}}\right)}}\right)}
{\displaystyle =2\log \left({\frac {1}{2}}\cdot {\frac {\Gamma \left({\frac {1}{4}}\right)}{\Gamma \left({\frac {3}{4}}\right)}}\right)=-2\log \left(2\cdot {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}{\displaystyle =2\log \left({\frac {1}{2}}\cdot {\frac {\Gamma \left({\frac {1}{4}}\right)}{\Gamma \left({\frac {3}{4}}\right)}}\right)=-2\log \left(2\cdot {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}
20 Bearbeiten
Ist {\displaystyle V:=\left\{(x_{1},...,x_{n})\geq 0\,{\Big |}\,x_{1}+...+x_{n}=1\right\}}{\displaystyle V:=\left\{(x_{1},...,x_{n})\geq 0\,{\Big |}\,x_{1}+...+x_{n}=1\right\}} und ist {\displaystyle 0<a_{1}<...<a_{n}}{\displaystyle 0<a_{1}<...<a_{n}}, so gilt
{\displaystyle \int \limits _{V}{\frac {1}{a_{1}x_{1}+...+a_{n}x_{n}}}\,dx_{1}\cdots dx_{n}={\frac {1}{(n-2)!}}\sum _{k=1}^{n}{\frac {\log a_{k}}{a_{k}}}\,\prod _{j=1 \atop j\neq k}^{n}{\frac {a_{k}}{a_{k}-a_{j}}}}{\displaystyle \int \limits _{V}{\frac {1}{a_{1}x_{1}+...+a_{n}x_{n}}}\,dx_{1}\cdots dx_{n}={\frac {1}{(n-2)!}}\sum _{k=1}^{n}{\frac {\log a_{k}}{a_{k}}}\,\prod _{j=1 \atop j\neq k}^{n}{\frac {a_{k}}{a_{k}-a_{j}}}}.
ohne Beweis
\href{https://de.m.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Allgemeine_Integralformeln}{Formelsammlung Mathematik: Bestimmte Integrale: Allgemeine Integralformeln}
1.1 Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)dx=\int _{-\infty }^{\infty }f(x)\,dx\qquad b>0}{\displaystyle \int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)dx=\int _{-\infty }^{\infty }f(x)\,dx\qquad b>0}
Beweis (Formel nach Cauchy)
{\displaystyle I:=\int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)dx=\int _{-\infty }^{0}f\left(x-{\frac {b}{x}}\right)dx+\int _{0}^{\infty }f\left(x-{\frac {b}{x}}\right)dx}{\displaystyle I:=\int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)dx=\int _{-\infty }^{0}f\left(x-{\frac {b}{x}}\right)dx+\int _{0}^{\infty }f\left(x-{\frac {b}{x}}\right)dx}
Substituiert man bei beiden Integralen auf der rechten Seite {\displaystyle x\mapsto -{\frac {b}{x}}}{\displaystyle x\mapsto -{\frac {b}{x}}}, so ist
{\displaystyle I=\int _{0}^{\infty }f\left(-{\frac {b}{x}}+x\right){\frac {b}{x^{2}}}\,dx+\int _{-\infty }^{0}f\left(-{\frac {b}{x}}+x\right){\frac {b}{x^{2}}}\,dx=\int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right){\frac {b}{x^{2}}}\,dx}{\displaystyle I=\int _{0}^{\infty }f\left(-{\frac {b}{x}}+x\right){\frac {b}{x^{2}}}\,dx+\int _{-\infty }^{0}f\left(-{\frac {b}{x}}+x\right){\frac {b}{x^{2}}}\,dx=\int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right){\frac {b}{x^{2}}}\,dx}.
Daher ist
{\displaystyle 2I=\int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)\left(1+{\frac {b}{x^{2}}}\right)dx=\int _{-\infty }^{0}f\left(x-{\frac {b}{x}}\right)\left(1+{\frac {b}{x^{2}}}\right)dx+\int _{0}^{\infty }f\left(x-{\frac {b}{x}}\right)\left(1+{\frac {b}{x^{2}}}\right)dx}{\displaystyle 2I=\int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)\left(1+{\frac {b}{x^{2}}}\right)dx=\int _{-\infty }^{0}f\left(x-{\frac {b}{x}}\right)\left(1+{\frac {b}{x^{2}}}\right)dx+\int _{0}^{\infty }f\left(x-{\frac {b}{x}}\right)\left(1+{\frac {b}{x^{2}}}\right)dx}.
Substituiert man nun bei beiden Integralen auf der rechten Seite {\displaystyle y=x-{\frac {b}{x}}\;\;\left[dy=\left(1+{\frac {b}{x^{2}}}\right)dx\right]}{\displaystyle y=x-{\frac {b}{x}}\;\;\left[dy=\left(1+{\frac {b}{x^{2}}}\right)dx\right]}, so ist
{\displaystyle 2I=\int _{-\infty }^{\infty }f(y)dy+\int _{-\infty }^{\infty }f(y)dy}{\displaystyle 2I=\int _{-\infty }^{\infty }f(y)dy+\int _{-\infty }^{\infty }f(y)dy}, woraus die Behauptung folgt.
1.2 Bearbeiten
Ist {\displaystyle \phi (x):=x-\sum _{k=0}^{n}{\frac {a_{k}}{x+b_{k}}}}{\displaystyle \phi (x):=x-\sum _{k=0}^{n}{\frac {a_{k}}{x+b_{k}}}} mit {\displaystyle a_{0},a_{1},...,a_{n}>0\,}{\displaystyle a_{0},a_{1},...,a_{n}>0\,} und {\displaystyle b_{0},b_{1},...,b_{n}\in \mathbb {R} }{\displaystyle b_{0},b_{1},...,b_{n}\in \mathbb {R} }, so gilt {\displaystyle \int _{-\infty }^{\infty }f(\phi (x))\,dx=\int _{-\infty }^{\infty }f(x)\,dx}{\displaystyle \int _{-\infty }^{\infty }f(\phi (x))\,dx=\int _{-\infty }^{\infty }f(x)\,dx}.
ohne Beweis (Glassers Formel)
2 Bearbeiten
Für ein {\displaystyle 0<\delta <1\,}{\displaystyle 0<\delta <1\,} sei {\displaystyle H(\delta ):=\left\{z\in \mathbb {C} \,|\,{\text{Re}}(z)\geq -\delta \right\}}{\displaystyle H(\delta ):=\left\{z\in \mathbb {C} \,|\,{\text{Re}}(z)\geq -\delta \right\}}.
{\displaystyle f:\,H(\delta )\to \mathbb {C} }{\displaystyle f:\,H(\delta )\to \mathbb {C} } sei eine analytische Funktion, zu der es Konstanten {\displaystyle C,P\in \mathbb {R} }{\displaystyle C,P\in \mathbb {R} } und {\displaystyle A<\pi \,}{\displaystyle A<\pi \,} gibt,
so dass {\displaystyle |f(z)|\leq C\cdot e^{P\cdot {\text{Re}}(z)+A\cdot |{\text{Im}}(z)|}}{\displaystyle |f(z)|\leq C\cdot e^{P\cdot {\text{Re}}(z)+A\cdot |{\text{Im}}(z)|}} gilt {\displaystyle \forall z\in H(\delta )}{\displaystyle \forall z\in H(\delta )}.
Wenn man für {\displaystyle x>0\,}x>0\, und ein {\displaystyle 0<c<\delta \qquad F(x):={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }{\frac {\pi }{\sin \pi z}}\,f(-z)\,x^{-z}\,dz}{\displaystyle 0<c<\delta \qquad F(x):={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }{\frac {\pi }{\sin \pi z}}\,f(-z)\,x^{-z}\,dz} setzt, so ist
{\displaystyle F(x)=\sum _{k=0}^{\infty }f(k)\,(-x)^{k}}{\displaystyle F(x)=\sum _{k=0}^{\infty }f(k)\,(-x)^{k}} für {\displaystyle 0<x<e^{-P}\,}{\displaystyle 0<x<e^{-P}\,} und {\displaystyle \int _{0}^{\infty }F(t)\,t^{s-1}\,dt={\frac {\pi }{\sin \pi s}}\,f(-s)}{\displaystyle \int _{0}^{\infty }F(t)\,t^{s-1}\,dt={\frac {\pi }{\sin \pi s}}\,f(-s)} für {\displaystyle 0<{\text{Re}}(s)<\delta \,}{\displaystyle 0<{\text{Re}}(s)<\delta \,}.
ohne Beweis (Hardys Version von Ramanujan Master Theorem)
3 Bearbeiten
Es sei {\displaystyle f\,}f\, eine rationale Funktion ohne Polstellen auf der positiven reellen Achse und mit höchstens einer einfachen Polstelle im Ursprung.
Ist dann der Nennergrad um mindestens 2 größer als der Zählergrad, so gilt:
{\displaystyle \int _{0}^{\infty }f(x)\,x^{\alpha }\,dx={\frac {\pi }{\sin \alpha \pi }}\,\sum {\text{res}}{\Big (}f(-z)\,z^{\alpha }{\Big )}\qquad 0<\alpha <1}{\displaystyle \int _{0}^{\infty }f(x)\,x^{\alpha }\,dx={\frac {\pi }{\sin \alpha \pi }}\,\sum {\text{res}}{\Big (}f(-z)\,z^{\alpha }{\Big )}\qquad 0<\alpha <1}
ohne Beweis
4 Bearbeiten
Ist {\displaystyle f:\mathbb {R} \to \mathbb {R} }{\displaystyle f:\mathbb {R} \to \mathbb {R} } integrierbar und {\displaystyle \pi \,}\pi \,-periodisch, so gilt
{\displaystyle \int _{-\infty }^{\infty }f(x)\,{\frac {\sin x}{x}}\,dx=\int _{0}^{\pi }f(x)\,dx}{\displaystyle \int _{-\infty }^{\infty }f(x)\,{\frac {\sin x}{x}}\,dx=\int _{0}^{\pi }f(x)\,dx} und {\displaystyle {\text{p.V.}}\int _{-\infty }^{\infty }f(x)\,{\frac {\tan x}{x}}\,dx=\int _{0}^{\pi }f(x)\,dx}{\displaystyle {\text{p.V.}}\int _{-\infty }^{\infty }f(x)\,{\frac {\tan x}{x}}\,dx=\int _{0}^{\pi }f(x)\,dx}
Beweis (Formel von Lobatschewski)
{\displaystyle \int _{-\infty }^{\infty }f(x)\,{\frac {\sin x}{x}}\,dx=\sum _{k\in \mathbb {Z} }\int _{k\pi }^{(k+1)\pi }f(x)\,{\frac {\sin x}{x}}\,dx=\sum _{k\in \mathbb {Z} }\int _{0}^{\pi }f(x)\,{\frac {(-1)^{k}\,\sin x}{x+k\pi }}\,dx}{\displaystyle \int _{-\infty }^{\infty }f(x)\,{\frac {\sin x}{x}}\,dx=\sum _{k\in \mathbb {Z} }\int _{k\pi }^{(k+1)\pi }f(x)\,{\frac {\sin x}{x}}\,dx=\sum _{k\in \mathbb {Z} }\int _{0}^{\pi }f(x)\,{\frac {(-1)^{k}\,\sin x}{x+k\pi }}\,dx}
{\displaystyle =\int _{0}^{\pi }f(x)\underbrace {\sum _{k\in \mathbb {Z} }{\frac {(-1)^{k}}{x+k\pi }}} _{\csc x}\,\sin x\,dx=\int _{0}^{\pi }f(x)\,dx}{\displaystyle =\int _{0}^{\pi }f(x)\underbrace {\sum _{k\in \mathbb {Z} }{\frac {(-1)^{k}}{x+k\pi }}} _{\csc x}\,\sin x\,dx=\int _{0}^{\pi }f(x)\,dx}.
Analog ist
{\displaystyle \sum _{k\in \mathbb {Z} }\int _{\left(k-{\frac {1}{2}}\right)\pi +\varepsilon }^{\left(k+{\frac {1}{2}}\right)\pi -\varepsilon }f(x)\,{\frac {\tan x}{x}}\,dx=\sum _{k\in \mathbb {Z} }\int _{{\frac {\pi }{2}}+\varepsilon }^{{\frac {\pi }{2}}-\varepsilon }f(x)\,{\frac {\tan x}{x+k\pi }}\,dx=\int _{{\frac {\pi }{2}}+\varepsilon }^{{\frac {\pi }{2}}-\varepsilon }f(x)\,\underbrace {\sum _{k\in \mathbb {Z} }{\frac {1}{x+k\pi }}} _{\cot x}\,\tan x\,dx}{\displaystyle \sum _{k\in \mathbb {Z} }\int _{\left(k-{\frac {1}{2}}\right)\pi +\varepsilon }^{\left(k+{\frac {1}{2}}\right)\pi -\varepsilon }f(x)\,{\frac {\tan x}{x}}\,dx=\sum _{k\in \mathbb {Z} }\int _{{\frac {\pi }{2}}+\varepsilon }^{{\frac {\pi }{2}}-\varepsilon }f(x)\,{\frac {\tan x}{x+k\pi }}\,dx=\int _{{\frac {\pi }{2}}+\varepsilon }^{{\frac {\pi }{2}}-\varepsilon }f(x)\,\underbrace {\sum _{k\in \mathbb {Z} }{\frac {1}{x+k\pi }}} _{\cot x}\,\tan x\,dx}
{\displaystyle =\int _{{\frac {\pi }{2}}+\varepsilon }^{{\frac {\pi }{2}}-\varepsilon }f(x)\,dx\to \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}f(x)\,dx=\int _{0}^{\pi }f(x)\,dx}{\displaystyle =\int _{{\frac {\pi }{2}}+\varepsilon }^{{\frac {\pi }{2}}-\varepsilon }f(x)\,dx\to \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}f(x)\,dx=\int _{0}^{\pi }f(x)\,dx} für {\displaystyle \varepsilon \to 0+}\varepsilon \to 0+.
5 Bearbeiten
Für {\displaystyle |R|>|r|\geq 0}{\displaystyle |R|>|r|\geq 0} und {\displaystyle \phi \in \mathbb {C} }{\displaystyle \phi \in \mathbb {C} }, mit {\displaystyle \left|{\text{Im}}\phi \right|<\log \left|{\frac {R}{r}}\right|}{\displaystyle \left|{\text{Im}}\phi \right|<\log \left|{\frac {R}{r}}\right|} falls {\displaystyle r\neq 0\,}{\displaystyle r\neq 0\,} ist, sei der Poissonsche Integralkern {\displaystyle P_{R}(r,\phi )\,}{\displaystyle P_{R}(r,\phi )\,} definiert als {\displaystyle {\frac {R^{2}-r^{2}}{R^{2}-2Rr\cos \phi +r^{2}}}}{\displaystyle {\frac {R^{2}-r^{2}}{R^{2}-2Rr\cos \phi +r^{2}}}}.
Ist {\displaystyle f:{\overline {B_{R}(0)}}\to \mathbb {C} \;,\;z\mapsto \sum _{k=0}^{\infty }a_{k}z^{k}}{\displaystyle f:{\overline {B_{R}(0)}}\to \mathbb {C} \;,\;z\mapsto \sum _{k=0}^{\infty }a_{k}z^{k}} eine holomorphe Funktion, so gilt {\displaystyle {\frac {1}{2\pi }}\int _{-\pi }^{\pi }P_{R}(r,\phi -\varphi )\,f(Re^{i\varphi })\,d\varphi =f(re^{i\phi })}{\displaystyle {\frac {1}{2\pi }}\int _{-\pi }^{\pi }P_{R}(r,\phi -\varphi )\,f(Re^{i\varphi })\,d\varphi =f(re^{i\phi })}.
Beweis (Poissonsche Integralformel)
Der Kern {\displaystyle P_{R}(r,\phi )\,}{\displaystyle P_{R}(r,\phi )\,} besitzt die komplexe Fourierreihenentwicklung {\displaystyle \sum _{n\in \mathbb {Z} }\left({\frac {r}{R}}\right)^{|n|}e^{in\phi }}{\displaystyle \sum _{n\in \mathbb {Z} }\left({\frac {r}{R}}\right)^{|n|}e^{in\phi }}.
Nun ist {\displaystyle \int _{-\pi }^{\pi }P_{R}(r,\phi -\varphi )\,f(Re^{i\varphi })\,d\varphi =\int _{-\pi }^{\pi }\sum _{n\in \mathbb {Z} }\left({\frac {r}{R}}\right)^{|n|}e^{in(\phi -\varphi )}\sum _{k=0}^{\infty }a_{k}R^{k}e^{ik\varphi }\,\;d\varphi }{\displaystyle \int _{-\pi }^{\pi }P_{R}(r,\phi -\varphi )\,f(Re^{i\varphi })\,d\varphi =\int _{-\pi }^{\pi }\sum _{n\in \mathbb {Z} }\left({\frac {r}{R}}\right)^{|n|}e^{in(\phi -\varphi )}\sum _{k=0}^{\infty }a_{k}R^{k}e^{ik\varphi }\,\;d\varphi }
{\displaystyle =\sum _{k=0}^{\infty }\sum _{n\in \mathbb {Z} }a_{k}R^{k}\left({\frac {r}{R}}\right)^{|n|}e^{in\phi }\underbrace {\int _{-\pi }^{\pi }e^{-in\varphi }\,e^{ik\varphi }\,d\varphi } _{2\pi \delta _{nk}}=2\pi \sum _{k=0}^{\infty }a_{k}r^{k}e^{ik\phi }=2\pi \,f(re^{i\phi })}{\displaystyle =\sum _{k=0}^{\infty }\sum _{n\in \mathbb {Z} }a_{k}R^{k}\left({\frac {r}{R}}\right)^{|n|}e^{in\phi }\underbrace {\int _{-\pi }^{\pi }e^{-in\varphi }\,e^{ik\varphi }\,d\varphi } _{2\pi \delta _{nk}}=2\pi \sum _{k=0}^{\infty }a_{k}r^{k}e^{ik\phi }=2\pi \,f(re^{i\phi })}.
6 Bearbeiten
{\displaystyle \int _{0}^{\infty }f\left(x+{\frac {\alpha }{x}}\right)dx=\int _{0}^{\infty }f\left({\sqrt {x^{2}+4\alpha }}\,\right)dx\qquad \alpha >0}{\displaystyle \int _{0}^{\infty }f\left(x+{\frac {\alpha }{x}}\right)dx=\int _{0}^{\infty }f\left({\sqrt {x^{2}+4\alpha }}\,\right)dx\qquad \alpha >0}
Beweis
{\displaystyle I:=\int _{0}^{\infty }f\left(x+{\frac {\alpha }{x}}\right)dx=\underbrace {\int \limits _{0}^{\sqrt {\alpha }}f\left(u+{\frac {\alpha }{u}}\right)du} _{=:J}+\int \limits _{\sqrt {a}}^{\infty }f\left(t+{\frac {\alpha }{t}}\right)dt}{\displaystyle I:=\int _{0}^{\infty }f\left(x+{\frac {\alpha }{x}}\right)dx=\underbrace {\int \limits _{0}^{\sqrt {\alpha }}f\left(u+{\frac {\alpha }{u}}\right)du} _{=:J}+\int \limits _{\sqrt {a}}^{\infty }f\left(t+{\frac {\alpha }{t}}\right)dt}
Nach Substitution {\displaystyle u={\frac {\alpha }{t}}}{\displaystyle u={\frac {\alpha }{t}}} ist {\displaystyle J=\int \limits _{\infty }^{\sqrt {a}}f\left({\frac {\alpha }{t}}+t\right){\frac {-\alpha }{t^{2}}}\,dt=\int \limits _{\sqrt {a}}^{\infty }f\left(t+{\frac {\alpha }{t}}\right)\,{\frac {\alpha }{t^{2}}}\,dt}{\displaystyle J=\int \limits _{\infty }^{\sqrt {a}}f\left({\frac {\alpha }{t}}+t\right){\frac {-\alpha }{t^{2}}}\,dt=\int \limits _{\sqrt {a}}^{\infty }f\left(t+{\frac {\alpha }{t}}\right)\,{\frac {\alpha }{t^{2}}}\,dt}.
Also ist {\displaystyle I=\int \limits _{\sqrt {a}}^{\infty }f\left(t+{\frac {\alpha }{t}}\right)\left(1+{\frac {\alpha }{t^{2}}}\right)dt=\int \limits _{\sqrt {a}}^{\infty }f\left({\sqrt {\left(t-{\frac {\alpha }{t}}\right)^{2}+4\alpha }}\,\right)\cdot \left(1+{\frac {\alpha }{t^{2}}}\right)dt}{\displaystyle I=\int \limits _{\sqrt {a}}^{\infty }f\left(t+{\frac {\alpha }{t}}\right)\left(1+{\frac {\alpha }{t^{2}}}\right)dt=\int \limits _{\sqrt {a}}^{\infty }f\left({\sqrt {\left(t-{\frac {\alpha }{t}}\right)^{2}+4\alpha }}\,\right)\cdot \left(1+{\frac {\alpha }{t^{2}}}\right)dt}.
Nach Substitution {\displaystyle x=t-{\frac {\alpha }{t}}}{\displaystyle x=t-{\frac {\alpha }{t}}} ist {\displaystyle I=\int _{0}^{\infty }f\left({\sqrt {x^{2}+4\alpha }}\,\right)dx}{\displaystyle I=\int _{0}^{\infty }f\left({\sqrt {x^{2}+4\alpha }}\,\right)dx}.
\href{https://de.m.wikibooks.org/wiki/Formelsammlung_Mathematik:_Bestimmte_Integrale:_Form_R(x,exp)}{Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,exp)}
0.1 Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}
1. Beweis
{\displaystyle I^{2}=\left(\int _{-\infty }^{\infty }e^{-x^{2}}\,dx\right)\,\left(\int _{-\infty }^{\infty }e^{-y^{2}}\,dy\right)=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-(x^{2}+y^{2})}\,dx\,dy}{\displaystyle I^{2}=\left(\int _{-\infty }^{\infty }e^{-x^{2}}\,dx\right)\,\left(\int _{-\infty }^{\infty }e^{-y^{2}}\,dy\right)=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }e^{-(x^{2}+y^{2})}\,dx\,dy}
lässt sich in Polarkoordinaten schreiben als {\displaystyle \int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}\,r\,dr\,d\varphi }{\displaystyle \int _{0}^{2\pi }\int _{0}^{\infty }e^{-r^{2}}\,r\,dr\,d\varphi }.
Und das ist {\displaystyle \int _{0}^{2\pi }\left[-{\frac {e^{-r^{2}}}{2}}\right]_{0}^{\infty }\,d\varphi =\int _{0}^{2\pi }{\frac {1}{2}}\,d\varphi =\pi \Rightarrow I={\sqrt {\pi }}}{\displaystyle \int _{0}^{2\pi }\left[-{\frac {e^{-r^{2}}}{2}}\right]_{0}^{\infty }\,d\varphi =\int _{0}^{2\pi }{\frac {1}{2}}\,d\varphi =\pi \Rightarrow I={\sqrt {\pi }}}.
2. Beweis
Die Fläche, die entsteht wenn {\displaystyle f(x)=e^{-x^{2}}}{\displaystyle f(x)=e^{-x^{2}}} um die z-Achse rotiert, schließt mit der xy-Ebene das gleiche Volumen ein
wie die Fläche, die entsteht, wenn {\displaystyle f^{-1}(x)={\sqrt {-\log x}}}{\displaystyle f^{-1}(x)={\sqrt {-\log x}}} um die x-Achse rotiert, mit der yz-Ebene.
Also {\displaystyle I^{2}=\pi \int _{0}^{1}{\sqrt {-\log x}}^{\,2}\,dx=\pi \Rightarrow I={\sqrt {\pi }}}{\displaystyle I^{2}=\pi \int _{0}^{1}{\sqrt {-\log x}}^{\,2}\,dx=\pi \Rightarrow I={\sqrt {\pi }}}.
3. Beweis
Definiert man {\displaystyle F(x)=\left(\int _{0}^{x}e^{-t^{2}}\,dt\right)^{2}}{\displaystyle F(x)=\left(\int _{0}^{x}e^{-t^{2}}\,dt\right)^{2}} und {\displaystyle G(x)=\int _{0}^{1}{\frac {e^{-x^{2}\,(1+t^{2})}}{1+t^{2}}}\,dt}{\displaystyle G(x)=\int _{0}^{1}{\frac {e^{-x^{2}\,(1+t^{2})}}{1+t^{2}}}\,dt}, so ist {\displaystyle F'(x)=2\,\int _{0}^{x}e^{-t^{2}}\,dt\;e^{-x^{2}}}{\displaystyle F'(x)=2\,\int _{0}^{x}e^{-t^{2}}\,dt\;e^{-x^{2}}}
und {\displaystyle G'(x)=\int _{0}^{1}e^{-x^{2}\,(1+t^{2})}\,(-2x)\,dt=-2\,\int _{0}^{1}e^{-x^{2}\,t^{2}}\,x\,dt\;e^{-x^{2}}=-2\int _{0}^{x}e^{-t^{2}}\,dt\;e^{-x^{2}}}{\displaystyle G'(x)=\int _{0}^{1}e^{-x^{2}\,(1+t^{2})}\,(-2x)\,dt=-2\,\int _{0}^{1}e^{-x^{2}\,t^{2}}\,x\,dt\;e^{-x^{2}}=-2\int _{0}^{x}e^{-t^{2}}\,dt\;e^{-x^{2}}}.
Es ist also {\displaystyle F'(x)+G'(x)=0\,}{\displaystyle F'(x)+G'(x)=0\,}. Folglich muss {\displaystyle F(x)+G(x)\,}{\displaystyle F(x)+G(x)\,} konstant sein.
{\displaystyle F(\infty )+G(\infty )=F(0)+G(0)\Rightarrow \left({\frac {I}{2}}\right)^{2}+0=0+{\frac {\pi }{4}}\Rightarrow I={\sqrt {\pi }}}{\displaystyle F(\infty )+G(\infty )=F(0)+G(0)\Rightarrow \left({\frac {I}{2}}\right)^{2}+0=0+{\frac {\pi }{4}}\Rightarrow I={\sqrt {\pi }}}
4. Beweis
Es sei {\displaystyle a={\sqrt {i\pi }}=(1+i){\sqrt {\frac {\pi }{2}}}}{\displaystyle a={\sqrt {i\pi }}=(1+i){\sqrt {\frac {\pi }{2}}}} und {\displaystyle f(z)={\frac {e^{-z^{2}}}{1+e^{-2az}}}}{\displaystyle f(z)={\frac {e^{-z^{2}}}{1+e^{-2az}}}}. Gaussintegralberechnung.PNG
Wegen {\displaystyle {\text{Re}}(a)>0\,}{\displaystyle {\text{Re}}(a)>0\,} gilt {\displaystyle e^{-2ax}\to \left\{{\begin{matrix}0&,&x\to \infty \\\infty &,&x\to -\infty \end{matrix}}\right.}{\displaystyle e^{-2ax}\to \left\{{\begin{matrix}0&,&x\to \infty \\\infty &,&x\to -\infty \end{matrix}}\right.}.
Ist {\displaystyle 0\leq y\leq {\sqrt {\frac {\pi }{2}}}}{\displaystyle 0\leq y\leq {\sqrt {\frac {\pi }{2}}}}, so geht für {\displaystyle x\to \pm \infty \,}{\displaystyle x\to \pm \infty \,} der Nenner von {\displaystyle f(x+iy)={\frac {e^{-x^{2}}\,e^{y^{2}-2ixy}}{1+e^{-2ax}\,e^{-2iay}}}}{\displaystyle f(x+iy)={\frac {e^{-x^{2}}\,e^{y^{2}-2ixy}}{1+e^{-2ax}\,e^{-2iay}}}} gegen {\displaystyle 1\,}1\, oder {\displaystyle \infty \,}{\displaystyle \infty \,}
und der Zähler geht gegen Null. Also verschwinden die beiden Integrale {\displaystyle \int _{\pm R}^{\pm R+a}\,f\,dz}{\displaystyle \int _{\pm R}^{\pm R+a}\,f\,dz} für {\displaystyle R\to \infty \,}R\to \infty \,.
Wegen {\displaystyle f(z)-f(z+a)=e^{-z^{2}}}{\displaystyle f(z)-f(z+a)=e^{-z^{2}}} gilt nun {\displaystyle \int _{-\infty }^{\infty }e^{-z^{2}}\,dz=\lim _{R\to \infty }\oint f\,dz=2\pi i\,{\text{res}}\left(f,{\frac {a}{2}}\right)={\sqrt {\pi }}}{\displaystyle \int _{-\infty }^{\infty }e^{-z^{2}}\,dz=\lim _{R\to \infty }\oint f\,dz=2\pi i\,{\text{res}}\left(f,{\frac {a}{2}}\right)={\sqrt {\pi }}}.
0.2 Bearbeiten
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{x\,e^{x}}}\right)\,dx=\gamma }{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{x\,e^{x}}}\right)\,dx=\gamma }
Beweis
Für {\displaystyle {\text{Re}}(z)>1\,}{\displaystyle {\text{Re}}(z)>1\,} gilt {\displaystyle \int _{0}^{\infty }\left({\frac {x^{z-1}}{e^{x}-1}}-{\frac {x^{z-1}}{x\,e^{x}}}\right)\,dx=\Gamma (z)\,\left(\zeta (z)-{\frac {1}{z-1}}\right)\to \gamma }{\displaystyle \int _{0}^{\infty }\left({\frac {x^{z-1}}{e^{x}-1}}-{\frac {x^{z-1}}{x\,e^{x}}}\right)\,dx=\Gamma (z)\,\left(\zeta (z)-{\frac {1}{z-1}}\right)\to \gamma } für {\displaystyle z\to 1\,}{\displaystyle z\to 1\,}.
1.1 Bearbeiten
{\displaystyle \int _{0}^{\infty }x^{z-1}e^{-x}\,dx=\Gamma (z)\qquad {\text{Re}}(z)>0}{\displaystyle \int _{0}^{\infty }x^{z-1}e^{-x}\,dx=\Gamma (z)\qquad {\text{Re}}(z)>0}
ohne Beweis
1.2 Bearbeiten
{\displaystyle \int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx={\frac {\sqrt {\pi }}{2}}\,e^{-2\alpha }\qquad \alpha >0}{\displaystyle \int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx={\frac {\sqrt {\pi }}{2}}\,e^{-2\alpha }\qquad \alpha >0}
1. Beweis
{\displaystyle 2I=\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx=\int _{-\infty }^{\infty }\exp \left(-\left(x-{\frac {\alpha }{x}}\right)^{2}-2\alpha \right)\,dx=\int _{-\infty }^{\infty }e^{-\left(x-{\frac {\alpha }{x}}\right)^{2}}dx\cdot e^{-2\alpha }}{\displaystyle 2I=\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx=\int _{-\infty }^{\infty }\exp \left(-\left(x-{\frac {\alpha }{x}}\right)^{2}-2\alpha \right)\,dx=\int _{-\infty }^{\infty }e^{-\left(x-{\frac {\alpha }{x}}\right)^{2}}dx\cdot e^{-2\alpha }}.
Nach der Formel {\displaystyle \int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)dx=\int _{-\infty }^{\infty }f(x)dx}{\displaystyle \int _{-\infty }^{\infty }f\left(x-{\frac {b}{x}}\right)dx=\int _{-\infty }^{\infty }f(x)dx}, gilt im Fall {\displaystyle f(x)=e^{-x^{2}}}{\displaystyle f(x)=e^{-x^{2}}},
{\displaystyle \int _{-\infty }^{\infty }e^{-\left(x-{\frac {\alpha }{x}}\right)^{2}}dx=\int _{-\infty }^{\infty }e^{-x^{2}}dx}{\displaystyle \int _{-\infty }^{\infty }e^{-\left(x-{\frac {\alpha }{x}}\right)^{2}}dx=\int _{-\infty }^{\infty }e^{-x^{2}}dx}, und das ist {\displaystyle {\sqrt {\pi }}}\sqrt{\pi}.
2. Beweis
{\displaystyle I'(\alpha )=\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,{\frac {-2\alpha }{x^{2}}}\,dx}{\displaystyle I'(\alpha )=\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,{\frac {-2\alpha }{x^{2}}}\,dx} ist nach Substitution {\displaystyle x\mapsto {\frac {\alpha }{x}}}{\displaystyle x\mapsto {\frac {\alpha }{x}}} gleich {\displaystyle -2\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx}{\displaystyle -2\int _{0}^{\infty }\exp \left(-x^{2}-{\frac {\alpha ^{2}}{x^{2}}}\right)\,dx}.
Die Differenzialgleichung {\displaystyle I'(\alpha )=-2\,I(\alpha )}{\displaystyle I'(\alpha )=-2\,I(\alpha )} wird gelöst durch {\displaystyle I(\alpha )=C\,e^{-2\alpha }}{\displaystyle I(\alpha )=C\,e^{-2\alpha }}, wobei {\displaystyle C=I(0)=\int _{0}^{\infty }e^{-x^{2}}\,dx={\frac {\sqrt {\pi }}{2}}}{\displaystyle C=I(0)=\int _{0}^{\infty }e^{-x^{2}}\,dx={\frac {\sqrt {\pi }}{2}}} ist.
1.3 Bearbeiten
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{x\,e^{x}}}-{\frac {e^{-x(z-1)}}{e^{x}-1}}\right)dx=\psi (z)\qquad {\text{Re}}(z)>0}{\displaystyle \int _{0}^{\infty }\left({\frac {1}{x\,e^{x}}}-{\frac {e^{-x(z-1)}}{e^{x}-1}}\right)dx=\psi (z)\qquad {\text{Re}}(z)>0}
Beweis
In der Formel {\displaystyle \int _{0}^{1}{\frac {1-x^{z-1}}{1-x}}\,dx=\psi (z)+\gamma }{\displaystyle \int _{0}^{1}{\frac {1-x^{z-1}}{1-x}}\,dx=\psi (z)+\gamma } wird das Integral
nach Substitution {\displaystyle x\mapsto e^{-x}}{\displaystyle x\mapsto e^{-x}} zu {\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {e^{-x(z-1)}}{e^{x}-1}}\right)\,dx}{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {e^{-x(z-1)}}{e^{x}-1}}\right)\,dx},
und {\displaystyle \gamma \,}{\displaystyle \gamma \,} lässt sich schreiben als {\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{x\,e^{x}}}\right)\,dx}{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {1}{x\,e^{x}}}\right)\,dx}.
1.4 Bearbeiten
{\displaystyle \int _{0}^{\infty }\left({\frac {z-1}{x\,e^{x}}}-{\frac {1-e^{-x(z-1)}}{x\,(e^{x}-1)}}\right)\,dx=\log \Gamma (z)\qquad {\text{Re}}(z)>0}{\displaystyle \int _{0}^{\infty }\left({\frac {z-1}{x\,e^{x}}}-{\frac {1-e^{-x(z-1)}}{x\,(e^{x}-1)}}\right)\,dx=\log \Gamma (z)\qquad {\text{Re}}(z)>0}
Beweis (Formel nach Malmstén)
Integriere die Formel {\displaystyle \int _{0}^{\infty }\left({\frac {1}{x\,e^{x}}}-{\frac {e^{-x(z'-1)}}{e^{x}-1}}\right)dx=\psi (z')}{\displaystyle \int _{0}^{\infty }\left({\frac {1}{x\,e^{x}}}-{\frac {e^{-x(z'-1)}}{e^{x}-1}}\right)dx=\psi (z')} nach z' von 1 bis z.
1.5 Bearbeiten
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}}}-{\frac {1}{(1+x)^{z}}}\right)\,{\frac {dx}{x}}=\psi (z)\qquad {\text{Re}}(z)>0}{\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}}}-{\frac {1}{(1+x)^{z}}}\right)\,{\frac {dx}{x}}=\psi (z)\qquad {\text{Re}}(z)>0}
Beweis (Formel nach Cauchy)
{\displaystyle \forall \varepsilon >0}{\displaystyle \forall \varepsilon >0} ist {\displaystyle \int _{0}^{\infty }\left({\frac {x^{\varepsilon -1}}{e^{x}}}-{\frac {x^{\varepsilon -1}}{(1+x)^{z}}}\right)dx=\Gamma (\varepsilon )-{\frac {\Gamma (z-\varepsilon )\,\Gamma (\varepsilon )}{\Gamma (z)}}}{\displaystyle \int _{0}^{\infty }\left({\frac {x^{\varepsilon -1}}{e^{x}}}-{\frac {x^{\varepsilon -1}}{(1+x)^{z}}}\right)dx=\Gamma (\varepsilon )-{\frac {\Gamma (z-\varepsilon )\,\Gamma (\varepsilon )}{\Gamma (z)}}}
{\displaystyle ={\frac {\Gamma (z)\,\Gamma (\varepsilon )-\Gamma (z-\varepsilon )\,\Gamma (\varepsilon )}{\Gamma (z)}}={\frac {\Gamma (1+\varepsilon )}{\Gamma (z)}}\cdot {\frac {\Gamma (z)-\Gamma (z-\varepsilon )}{\varepsilon }}\,{\xrightarrow {\,\,\varepsilon \to 0\,\,}}\,{\frac {1}{\Gamma (z)}}\cdot \Gamma '(z)=\psi (z)}{\displaystyle ={\frac {\Gamma (z)\,\Gamma (\varepsilon )-\Gamma (z-\varepsilon )\,\Gamma (\varepsilon )}{\Gamma (z)}}={\frac {\Gamma (1+\varepsilon )}{\Gamma (z)}}\cdot {\frac {\Gamma (z)-\Gamma (z-\varepsilon )}{\varepsilon }}\,{\xrightarrow {\,\,\varepsilon \to 0\,\,}}\,{\frac {1}{\Gamma (z)}}\cdot \Gamma '(z)=\psi (z)}.
1.6 Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}-1}}\,dx=\Gamma (z)\,\zeta (z)\qquad {\text{Re}}(z)>1}{\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}-1}}\,dx=\Gamma (z)\,\zeta (z)\qquad {\text{Re}}(z)>1}
Beweis
Wegen {\displaystyle {\frac {1}{e^{x}-1}}={\frac {e^{-x}}{1-e^{-x}}}=\sum _{k=1}^{\infty }e^{-kz}}{\displaystyle {\frac {1}{e^{x}-1}}={\frac {e^{-x}}{1-e^{-x}}}=\sum _{k=1}^{\infty }e^{-kz}} ist {\displaystyle {\frac {x^{z-1}}{e^{x}-1}}=\sum _{k=1}^{\infty }x^{z-1}\,e^{-kx}}{\displaystyle {\frac {x^{z-1}}{e^{x}-1}}=\sum _{k=1}^{\infty }x^{z-1}\,e^{-kx}}
und somit {\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}-1}}\,dx=\sum _{k=1}^{\infty }\int _{0}^{\infty }x^{z-1}\,e^{-kx}\,dx=\sum _{k=1}^{\infty }{\frac {\Gamma (z)}{k^{z}}}=\Gamma (z)\,\zeta (z)}{\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}-1}}\,dx=\sum _{k=1}^{\infty }\int _{0}^{\infty }x^{z-1}\,e^{-kx}\,dx=\sum _{k=1}^{\infty }{\frac {\Gamma (z)}{k^{z}}}=\Gamma (z)\,\zeta (z)}.
1.7 Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}+1}}\,dx=\Gamma (z)\,\eta (z)\qquad {\text{Re}}(z)>0}{\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}+1}}\,dx=\Gamma (z)\,\eta (z)\qquad {\text{Re}}(z)>0}
Beweis
Wegen {\displaystyle {\frac {1}{e^{x}+1}}={\frac {e^{-x}}{1+e^{-x}}}={\frac {-(-e^{-x})}{1-(-e^{-x})}}=-\sum _{k=1}^{\infty }(-e^{-x})^{k}=\sum _{k=1}^{\infty }(-1)^{k-1}\,e^{-kx}}{\displaystyle {\frac {1}{e^{x}+1}}={\frac {e^{-x}}{1+e^{-x}}}={\frac {-(-e^{-x})}{1-(-e^{-x})}}=-\sum _{k=1}^{\infty }(-e^{-x})^{k}=\sum _{k=1}^{\infty }(-1)^{k-1}\,e^{-kx}} ist {\displaystyle {\frac {x^{z-1}}{e^{x}+1}}=\sum _{k=1}^{\infty }(-1)^{k-1}\,x^{z-1}\,e^{-kx}}{\displaystyle {\frac {x^{z-1}}{e^{x}+1}}=\sum _{k=1}^{\infty }(-1)^{k-1}\,x^{z-1}\,e^{-kx}}
und somit {\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}+1}}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,\int _{0}^{\infty }x^{z-1}\,e^{-kx}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,{\frac {\Gamma (z)}{k^{z}}}=\Gamma (z)\,\eta (z)}{\displaystyle \int _{0}^{\infty }{\frac {x^{z-1}}{e^{x}+1}}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,\int _{0}^{\infty }x^{z-1}\,e^{-kx}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,{\frac {\Gamma (z)}{k^{z}}}=\Gamma (z)\,\eta (z)}.
1.8 Bearbeiten
{\displaystyle {\frac {1}{\Gamma (z)}}={\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt\qquad C\,}{\displaystyle {\frac {1}{\Gamma (z)}}={\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt\qquad C\,} ist eine Kurve in {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}{\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}, die von {\displaystyle -\infty -i\varepsilon \,}{\displaystyle -\infty -i\varepsilon \,} nach {\displaystyle -\infty +i\varepsilon \,}{\displaystyle -\infty +i\varepsilon \,} läuft.
Für {\displaystyle {\text{Re}}(z)>0\,}{\displaystyle {\text{Re}}(z)>0\,} kann man als Integrationsweg {\displaystyle C\,}{\displaystyle C\,} auch die Gerade {\displaystyle a+i\mathbb {R} }{\displaystyle a+i\mathbb {R} }, mit {\displaystyle a>0\,}{\displaystyle a>0\,}, hernehmen.
Beweis für Re(z)>0 (Hankelsche Integraldarstellung für die reziproke Gammafunktion)
Die Funktion {\displaystyle f(t)=t^{-z}\,e^{t}}{\displaystyle f(t)=t^{-z}\,e^{t}} mit {\displaystyle {\text{Re}}(z)>0\,}{\displaystyle {\text{Re}}(z)>0\,} ist auf {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}{\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}} holomorph.
Integrationsweg9.PNG
Für {\displaystyle \varepsilon \leq t\leq R\,}{\displaystyle \varepsilon \leq t\leq R\,} ist {\displaystyle |f(-R+it)|\,}{\displaystyle |f(-R+it)|\,}
{\displaystyle =\left|(-R+it)^{-z}\right|\cdot \left|e^{-R+it}\right|=\Theta \left(R^{-{\text{Re}}(z)}\right)\cdot e^{-R}}{\displaystyle =\left|(-R+it)^{-z}\right|\cdot \left|e^{-R+it}\right|=\Theta \left(R^{-{\text{Re}}(z)}\right)\cdot e^{-R}}.
Daher verschwinden die Integrale über den Abschnitten {\displaystyle \sigma _{1},\sigma _{2}\,}{\displaystyle \sigma _{1},\sigma _{2}\,} für {\displaystyle R\to \infty \,}R\to \infty \,.
Und es ist
{\displaystyle \left|\int _{-R}^{a}f(t\pm iR)\,dt\right|\leq \int _{-R}^{a}\left|(t\pm iR)^{-z}\right|\cdot \left|e^{t\pm iR}\right|\,dt}{\displaystyle \left|\int _{-R}^{a}f(t\pm iR)\,dt\right|\leq \int _{-R}^{a}\left|(t\pm iR)^{-z}\right|\cdot \left|e^{t\pm iR}\right|\,dt}
{\displaystyle \leq \max _{-R\leq t\leq a}\left|(t\pm iR)^{-z}\right|\cdot \int _{-R}^{a}e^{t}\,dt=\Theta \left(R^{-{\text{Re}}(z)}\right)\cdot e^{a}}{\displaystyle \leq \max _{-R\leq t\leq a}\left|(t\pm iR)^{-z}\right|\cdot \int _{-R}^{a}e^{t}\,dt=\Theta \left(R^{-{\text{Re}}(z)}\right)\cdot e^{a}}.
Daher verschwinden auch die Integrale über den Abschnitten {\displaystyle \tau _{1},\tau _{2}\,}{\displaystyle \tau _{1},\tau _{2}\,} für {\displaystyle R\to \infty \,}R\to \infty \,.
Also ist
{\displaystyle {\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }s^{-z}\,e^{s}\,ds}{\displaystyle {\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt={\frac {1}{2\pi i}}\int _{a-i\infty }^{a+i\infty }s^{-z}\,e^{s}\,ds}
{\displaystyle ={\mathcal {L}}^{-1}\left[s^{-z}\right](1)=\left.{\frac {t^{z-1}}{\Gamma (z)}}\right|_{t=1}={\frac {1}{\Gamma (z)}}}{\displaystyle ={\mathcal {L}}^{-1}\left[s^{-z}\right](1)=\left.{\frac {t^{z-1}}{\Gamma (z)}}\right|_{t=1}={\frac {1}{\Gamma (z)}}}.
Beweis für Re(z)<1
Die Funktion {\displaystyle f(t)=t^{z-1}e^{t}\,}{\displaystyle f(t)=t^{z-1}e^{t}\,} mit {\displaystyle {\text{Re}}(z)>0\,}{\displaystyle {\text{Re}}(z)>0\,} ist auf {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}{\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}} holomorph.
Hankelintegrationsweg.PNG
Das Integral über dem Kreisbogen {\displaystyle K_{\varepsilon }\,}{\displaystyle K_{\varepsilon }\,} verschwindet für {\displaystyle \varepsilon \to 0+\,}{\displaystyle \varepsilon \to 0+\,}, weil wegen {\displaystyle \left|t^{z-1}\right|=\Theta \left(|t|^{{\text{Re}}(z)-1}\right)=o\left({\frac {1}{|t|}}\right)}{\displaystyle \left|t^{z-1}\right|=\Theta \left(|t|^{{\text{Re}}(z)-1}\right)=o\left({\frac {1}{|t|}}\right)} für {\displaystyle |t|\to 0\,}{\displaystyle |t|\to 0\,}
ist {\displaystyle \max _{t\in K_{\varepsilon }}\left|t^{z-1}e^{-t}\right|=o\left({\frac {1}{\varepsilon }}\right)}{\displaystyle \max _{t\in K_{\varepsilon }}\left|t^{z-1}e^{-t}\right|=o\left({\frac {1}{\varepsilon }}\right)}, und daher gilt {\displaystyle \left|\int _{K_{\varepsilon }}f(t)\,dt\right|\leq \pi \varepsilon \cdot o\left({\frac {1}{\varepsilon }}\right)=o(1)}{\displaystyle \left|\int _{K_{\varepsilon }}f(t)\,dt\right|\leq \pi \varepsilon \cdot o\left({\frac {1}{\varepsilon }}\right)=o(1)}.
Für die horizontalen Integrationswege gilt:
{\displaystyle \int _{-\infty }^{0}f(t\pm i\varepsilon )\,dt=\int _{-\infty }^{0}(t\pm i\varepsilon )^{z-1}\,e^{t\pm i\varepsilon }\,dt=e^{\pm i\pi (z-1)}\int _{-\infty }^{0}(-t\mp i\varepsilon )^{z-1}\,e^{t\pm i\varepsilon }\,dt}{\displaystyle \int _{-\infty }^{0}f(t\pm i\varepsilon )\,dt=\int _{-\infty }^{0}(t\pm i\varepsilon )^{z-1}\,e^{t\pm i\varepsilon }\,dt=e^{\pm i\pi (z-1)}\int _{-\infty }^{0}(-t\mp i\varepsilon )^{z-1}\,e^{t\pm i\varepsilon }\,dt}
{\displaystyle =-e^{\pm i\pi z}\int _{0}^{\infty }(t\mp i\varepsilon )^{z-1}\,e^{-t\pm i\varepsilon }\,dt\to -e^{\pm i\pi z}\,\Gamma (z)}{\displaystyle =-e^{\pm i\pi z}\int _{0}^{\infty }(t\mp i\varepsilon )^{z-1}\,e^{-t\pm i\varepsilon }\,dt\to -e^{\pm i\pi z}\,\Gamma (z)} für {\displaystyle \varepsilon \to 0+\,}{\displaystyle \varepsilon \to 0+\,}.
Daher ist {\displaystyle \int _{C}t^{z-1}\,e^{t}\,dt=\left(e^{i\pi z}-e^{-i\pi z}\right)\Gamma (z)=2i\sin \pi z\,\Gamma (z)={\frac {2\pi i}{\Gamma (z)\,\Gamma (1-z)}}\,\Gamma (z)={\frac {2\pi i}{\Gamma (1-z)}}}{\displaystyle \int _{C}t^{z-1}\,e^{t}\,dt=\left(e^{i\pi z}-e^{-i\pi z}\right)\Gamma (z)=2i\sin \pi z\,\Gamma (z)={\frac {2\pi i}{\Gamma (z)\,\Gamma (1-z)}}\,\Gamma (z)={\frac {2\pi i}{\Gamma (1-z)}}}.
Ersetzt man {\displaystyle z\,}z\, durch {\displaystyle 1-z\,}{\displaystyle 1-z\,}, so ist {\displaystyle {\frac {1}{\Gamma (z)}}={\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt}{\displaystyle {\frac {1}{\Gamma (z)}}={\frac {1}{2\pi i}}\int _{C}t^{-z}e^{t}\,dt} für {\displaystyle {\text{Re}}(z)<1\,}{\displaystyle {\text{Re}}(z)<1\,}.
1.9 Bearbeiten
{\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}e^{{\frac {x}{2}}\left(t-{\frac {1}{t}}\right)}\,{\frac {dt}{t^{\nu +1}}}\qquad {\text{Re}}(x)>0\qquad C\,}{\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}e^{{\frac {x}{2}}\left(t-{\frac {1}{t}}\right)}\,{\frac {dt}{t^{\nu +1}}}\qquad {\text{Re}}(x)>0\qquad C\,} ist eine Kurve in {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}{\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}}, die von {\displaystyle -\infty -i\varepsilon \,}{\displaystyle -\infty -i\varepsilon \,} nach {\displaystyle -\infty +i\varepsilon \,}{\displaystyle -\infty +i\varepsilon \,} läuft.
Beweis für x>0 (Hankelsche Integraldarstellung für die Besselfunktion)
{\displaystyle J_{\nu }(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!\,\Gamma (n+\nu +1)}}\left({\frac {x}{2}}\right)^{\nu +2n}}{\displaystyle J_{\nu }(x)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!\,\Gamma (n+\nu +1)}}\left({\frac {x}{2}}\right)^{\nu +2n}}
Ersetze {\displaystyle {\frac {1}{\Gamma (n+\nu +1)}}}{\displaystyle {\frac {1}{\Gamma (n+\nu +1)}}} durch die Hankelsche Integraldarstellung {\displaystyle {\frac {1}{2\pi i}}\int _{C}t^{-n-\nu -1}e^{t}\,dt}{\displaystyle {\frac {1}{2\pi i}}\int _{C}t^{-n-\nu -1}e^{t}\,dt}.
{\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\,{\frac {\left({\frac {x}{2}}\right)^{2n}}{t^{n}}}\,\left({\frac {x}{2}}\right)^{\nu }\,{\frac {e^{t}}{t^{\nu +1}}}\,dt={\frac {1}{2\pi i}}\int _{C}e^{t-{\frac {x^{2}}{4t}}}\,\left({\frac {x}{2}}\right)^{\nu }{\frac {dt}{t^{\nu +1}}}}{\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\,{\frac {\left({\frac {x}{2}}\right)^{2n}}{t^{n}}}\,\left({\frac {x}{2}}\right)^{\nu }\,{\frac {e^{t}}{t^{\nu +1}}}\,dt={\frac {1}{2\pi i}}\int _{C}e^{t-{\frac {x^{2}}{4t}}}\,\left({\frac {x}{2}}\right)^{\nu }{\frac {dt}{t^{\nu +1}}}}
Nach Substitution {\displaystyle t\mapsto {\frac {x}{2}}\cdot t}{\displaystyle t\mapsto {\frac {x}{2}}\cdot t} ändert sich am Integrationsweg {\displaystyle C\,}{\displaystyle C\,} nichts, und es ist {\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}e^{{\frac {x}{2}}\left(t-{\frac {1}{t}}\right)}\,{\frac {dt}{t^{\nu +1}}}}{\displaystyle J_{\nu }(x)={\frac {1}{2\pi i}}\int _{C}e^{{\frac {x}{2}}\left(t-{\frac {1}{t}}\right)}\,{\frac {dt}{t^{\nu +1}}}}.
1.10 Bearbeiten
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)e^{-zx}\,dx=\psi (z+1)-\log z\qquad {\text{Re}}(z)>0}{\displaystyle \int _{0}^{\infty }\left({\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)e^{-zx}\,dx=\psi (z+1)-\log z\qquad {\text{Re}}(z)>0}
Beweis
Dies folgt unmittelbar aus den Formeln {\displaystyle \psi (z+1)=\int _{0}^{\infty }\left({\frac {e^{-x}}{x}}-{\frac {e^{-zx}}{e^{x}-1}}\right)dx}{\displaystyle \psi (z+1)=\int _{0}^{\infty }\left({\frac {e^{-x}}{x}}-{\frac {e^{-zx}}{e^{x}-1}}\right)dx} und {\displaystyle \log z=\int _{0}^{\infty }\left({\frac {e^{-x}}{x}}-{\frac {e^{-zx}}{x}}\right)dx}{\displaystyle \log z=\int _{0}^{\infty }\left({\frac {e^{-x}}{x}}-{\frac {e^{-zx}}{x}}\right)dx}.
1.11 Bearbeiten
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right)\,{\frac {e^{-zx}}{x}}\,dx=\log \left({\frac {z!\,e^{z}}{z^{z}\,{\sqrt {2\pi z}}}}\right)\qquad {\text{Re}}(z)>0}{\displaystyle \int _{0}^{\infty }\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right)\,{\frac {e^{-zx}}{x}}\,dx=\log \left({\frac {z!\,e^{z}}{z^{z}\,{\sqrt {2\pi z}}}}\right)\qquad {\text{Re}}(z)>0}
Beweis (Erste Binetsche Formel)
In der Formel {\displaystyle \psi (z)+{\frac {1}{z}}=\log z+\int _{0}^{\infty }e^{-zx}\left({\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)dx}{\displaystyle \psi (z)+{\frac {1}{z}}=\log z+\int _{0}^{\infty }e^{-zx}\left({\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)dx} ersetze {\displaystyle {\frac {1}{z}}}{\displaystyle {\frac {1}{z}}} durch {\displaystyle \int _{0}^{\infty }e^{-zx}\,dx}{\displaystyle \int _{0}^{\infty }e^{-zx}\,dx}:
{\displaystyle \psi (z)=\log z-{\frac {1}{2z}}+\int _{0}^{\infty }e^{-zx}\left(-{\frac {1}{2}}+{\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)dx}{\displaystyle \psi (z)=\log z-{\frac {1}{2z}}+\int _{0}^{\infty }e^{-zx}\left(-{\frac {1}{2}}+{\frac {1}{x}}-{\frac {1}{e^{x}-1}}\right)dx}, wobei {\displaystyle \lim _{x\to 0}{\frac {{\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}}{x}}={\frac {1}{12}}}{\displaystyle \lim _{x\to 0}{\frac {{\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}}{x}}={\frac {1}{12}}} ist.
Integriert man beide Seiten unbestimmt nach {\displaystyle z\,}z\,, so ist
{\displaystyle \log \Gamma (z)=\left(z-{\frac {1}{2}}\right)\log z-z+\int _{0}^{\infty }e^{-zx}\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right){\frac {dx}{x}}+C}{\displaystyle \log \Gamma (z)=\left(z-{\frac {1}{2}}\right)\log z-z+\int _{0}^{\infty }e^{-zx}\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right){\frac {dx}{x}}+C}.
Daraus folgt {\displaystyle \int _{0}^{\infty }e^{-zx}\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right){\frac {dx}{x}}+C=\log \left({\frac {(z-1)!\,e^{z}}{z^{z-{\frac {1}{2}}}}}\right)}{\displaystyle \int _{0}^{\infty }e^{-zx}\left({\frac {1}{2}}-{\frac {1}{x}}+{\frac {1}{e^{x}-1}}\right){\frac {dx}{x}}+C=\log \left({\frac {(z-1)!\,e^{z}}{z^{z-{\frac {1}{2}}}}}\right)}.
Nachdem für {\displaystyle z\to \infty \,}{\displaystyle z\to \infty \,} das Integral verschwindet, ist {\displaystyle C=\lim _{z\to \infty }\log \left({\frac {z!\,e^{z}}{z^{z}\,{\sqrt {z}}}}\right)=\log {\sqrt {2\pi }}}{\displaystyle C=\lim _{z\to \infty }\log \left({\frac {z!\,e^{z}}{z^{z}\,{\sqrt {z}}}}\right)=\log {\sqrt {2\pi }}}.
1.12 Bearbeiten
{\displaystyle J_{\nu }(z)={\frac {1}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }\int _{-1}^{1}e^{izx}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx\qquad {\text{Re}}(\nu )>-{\frac {1}{2}}}{\displaystyle J_{\nu }(z)={\frac {1}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }\int _{-1}^{1}e^{izx}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx\qquad {\text{Re}}(\nu )>-{\frac {1}{2}}}
Beweis (Poissonsche Darstellung der Besselfunktion)
Setze {\displaystyle I:=\int _{-1}^{1}e^{izx}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}{\displaystyle I:=\int _{-1}^{1}e^{izx}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}.
Verwende die Reihenentwicklung {\displaystyle e^{izx}=\sum _{n=0}^{\infty }{\frac {(iz)^{n}}{n!}}\,x^{n}}{\displaystyle e^{izx}=\sum _{n=0}^{\infty }{\frac {(iz)^{n}}{n!}}\,x^{n}}:
{\displaystyle I=\sum _{n=0}^{\infty }{\frac {(iz)^{n}}{n!}}\int _{-1}^{1}x^{n}(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}{\displaystyle I=\sum _{n=0}^{\infty }{\frac {(iz)^{n}}{n!}}\int _{-1}^{1}x^{n}(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}
Letzter Integrand ist für gerade {\displaystyle n\,}n\, gerade und für ungerade {\displaystyle n\,}n\, ungerade.
{\displaystyle I=\sum _{n=0}^{\infty }{\frac {(iz)^{2n}}{(2n)!}}\cdot 2\int _{0}^{1}x^{2n}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}{\displaystyle I=\sum _{n=0}^{\infty }{\frac {(iz)^{2n}}{(2n)!}}\cdot 2\int _{0}^{1}x^{2n}\,(1-x^{2})^{\nu -{\frac {1}{2}}}\,dx}
Nach Substitution {\displaystyle x={\sqrt {t}}}{\displaystyle x={\sqrt {t}}} ist {\displaystyle I=\sum _{n=0}^{\infty }{\frac {(-1)^{n}\,z^{2n}}{(2n)!}}\,\int _{0}^{1}t^{n-{\frac {1}{2}}}\,(1-t)^{\nu -{\frac {1}{2}}}\,dt}{\displaystyle I=\sum _{n=0}^{\infty }{\frac {(-1)^{n}\,z^{2n}}{(2n)!}}\,\int _{0}^{1}t^{n-{\frac {1}{2}}}\,(1-t)^{\nu -{\frac {1}{2}}}\,dt}.
Dabei ist {\displaystyle \int _{0}^{1}t^{n-{\frac {1}{2}}}\,(1-t)^{\nu -{\frac {1}{2}}}\,dt=B\left(n+{\frac {1}{2}},\nu +{\frac {1}{2}}\right)}{\displaystyle \int _{0}^{1}t^{n-{\frac {1}{2}}}\,(1-t)^{\nu -{\frac {1}{2}}}\,dt=B\left(n+{\frac {1}{2}},\nu +{\frac {1}{2}}\right)} {\displaystyle ={\frac {\Gamma \left(n+{\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}{\Gamma (\nu +n+1)}}}{\displaystyle ={\frac {\Gamma \left(n+{\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}{\Gamma (\nu +n+1)}}},
wobei nach Legendrescher Verdopplungsformel {\displaystyle \Gamma \left(n+{\frac {1}{2}}\right)={\frac {\Gamma \left({\frac {1}{2}}\right)}{2^{2n}}}\,{\frac {(2n)!}{n!}}}{\displaystyle \Gamma \left(n+{\frac {1}{2}}\right)={\frac {\Gamma \left({\frac {1}{2}}\right)}{2^{2n}}}\,{\frac {(2n)!}{n!}}} ist.
Also ist {\displaystyle I=\sum _{n=0}^{\infty }(-1)^{n}\,\left({\frac {z}{2}}\right)^{2n}\,{\frac {\Gamma \left({\frac {1}{2}}\right)\,\Gamma \left(\nu +{\frac {1}{2}}\right)}{n!\,\Gamma (\nu +n+1)}}}{\displaystyle I=\sum _{n=0}^{\infty }(-1)^{n}\,\left({\frac {z}{2}}\right)^{2n}\,{\frac {\Gamma \left({\frac {1}{2}}\right)\,\Gamma \left(\nu +{\frac {1}{2}}\right)}{n!\,\Gamma (\nu +n+1)}}},
und damit ist {\displaystyle {\frac {1}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }I=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!\,\Gamma (\nu +n+1)}}\left({\frac {z}{2}}\right)^{\nu +2n}=J_{\nu }(z)}{\displaystyle {\frac {1}{\Gamma \left({\frac {1}{2}}\right)\Gamma \left(\nu +{\frac {1}{2}}\right)}}\left({\frac {z}{2}}\right)^{\nu }I=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!\,\Gamma (\nu +n+1)}}\left({\frac {z}{2}}\right)^{\nu +2n}=J_{\nu }(z)}.
2.1 Bearbeiten
{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {a}{a^{2}+x^{2}}}\,dx=\sin(as)\,{\text{Ci}}(as)-\cos(as)\,\left({\text{Si}}(as)-{\frac {\pi }{2}}\right)}{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {a}{a^{2}+x^{2}}}\,dx=\sin(as)\,{\text{Ci}}(as)-\cos(as)\,\left({\text{Si}}(as)-{\frac {\pi }{2}}\right)}
Beweis
{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {a}{a^{2}+x^{2}}}\,dx=\int _{0}^{\infty }e^{-sx}\int _{0}^{\infty }\sin(at)\,e^{-xt}\,dt\,dx=\int _{0}^{\infty }\int _{0}^{\infty }\sin(at)\,e^{-(s+t)x}\,dx\,dt}{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {a}{a^{2}+x^{2}}}\,dx=\int _{0}^{\infty }e^{-sx}\int _{0}^{\infty }\sin(at)\,e^{-xt}\,dt\,dx=\int _{0}^{\infty }\int _{0}^{\infty }\sin(at)\,e^{-(s+t)x}\,dx\,dt}
{\displaystyle =\int _{0}^{\infty }{\frac {\sin(at)}{s+t}}\,dt=\int _{s}^{\infty }{\frac {\sin \left(a(t-s)\right)}{t}}\,dt=\int _{s}^{\infty }{\frac {\sin(at)\,\cos(as)-\cos(at)\,\sin(as)}{t}}\,dt}{\displaystyle =\int _{0}^{\infty }{\frac {\sin(at)}{s+t}}\,dt=\int _{s}^{\infty }{\frac {\sin \left(a(t-s)\right)}{t}}\,dt=\int _{s}^{\infty }{\frac {\sin(at)\,\cos(as)-\cos(at)\,\sin(as)}{t}}\,dt}
{\displaystyle =\cos(as)\int _{s}^{\infty }{\frac {\sin(at)}{t}}\,dt-\sin(as)\int _{s}^{\infty }{\frac {\cos(at)}{t}}\,dt=\cos(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)+\sin(as)\,{\text{Ci}}(as)}{\displaystyle =\cos(as)\int _{s}^{\infty }{\frac {\sin(at)}{t}}\,dt-\sin(as)\int _{s}^{\infty }{\frac {\cos(at)}{t}}\,dt=\cos(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)+\sin(as)\,{\text{Ci}}(as)}
2.2 Bearbeiten
{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {x}{a^{2}+x^{2}}}\,dx=\sin(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)-\cos(as)\,{\text{Ci}}(as)}{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {x}{a^{2}+x^{2}}}\,dx=\sin(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)-\cos(as)\,{\text{Ci}}(as)}
Beweis
{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {x}{a^{2}+x^{2}}}\,dx=\int _{0}^{\infty }e^{-sx}\int _{0}^{\infty }\cos(at)\,e^{-xt}\,dt\,dx=\int _{0}^{\infty }\int _{0}^{\infty }\cos(at)\,e^{-(s+t)x}\,dx\,dt}{\displaystyle \int _{0}^{\infty }e^{-sx}\,{\frac {x}{a^{2}+x^{2}}}\,dx=\int _{0}^{\infty }e^{-sx}\int _{0}^{\infty }\cos(at)\,e^{-xt}\,dt\,dx=\int _{0}^{\infty }\int _{0}^{\infty }\cos(at)\,e^{-(s+t)x}\,dx\,dt}
{\displaystyle =\int _{0}^{\infty }{\frac {\cos(at)}{s+t}}\,dt=\int _{s}^{\infty }{\frac {\cos \left(a(t-s)\right)}{t}}\,dt=\int _{s}^{\infty }{\frac {\cos(at)\,\cos(as)+\sin(at)\,\sin(as)}{t}}\,dt}{\displaystyle =\int _{0}^{\infty }{\frac {\cos(at)}{s+t}}\,dt=\int _{s}^{\infty }{\frac {\cos \left(a(t-s)\right)}{t}}\,dt=\int _{s}^{\infty }{\frac {\cos(at)\,\cos(as)+\sin(at)\,\sin(as)}{t}}\,dt}
{\displaystyle =\cos(as)\int _{s}^{\infty }{\frac {\cos(at)}{t}}\,dt+\sin(as)\int _{s}^{\infty }{\frac {\sin(at)}{t}}\,dt=-\cos(as)\,{\text{Ci}}(as)+\sin(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)}{\displaystyle =\cos(as)\int _{s}^{\infty }{\frac {\cos(at)}{t}}\,dt+\sin(as)\int _{s}^{\infty }{\frac {\sin(at)}{t}}\,dt=-\cos(as)\,{\text{Ci}}(as)+\sin(as)\,\left({\frac {\pi }{2}}-{\text{Si}}(as)\right)}
2.3 Bearbeiten
{\displaystyle \int _{0}^{\infty }x^{z-1}e^{-\mu x}\,dx={\frac {\Gamma (z)}{\mu ^{z}}}\qquad {\text{Re}}(z),{\text{Re}}(\mu )>0}{\displaystyle \int _{0}^{\infty }x^{z-1}e^{-\mu x}\,dx={\frac {\Gamma (z)}{\mu ^{z}}}\qquad {\text{Re}}(z),{\text{Re}}(\mu )>0}
ohne Beweis
2.4 Bearbeiten
{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-i\mu x}\,dx={\frac {\Gamma (z)}{(i\mu )^{z}}}\qquad 0<\operatorname {Re} (z)<1\,,\,\mu >0}{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-i\mu x}\,dx={\frac {\Gamma (z)}{(i\mu )^{z}}}\qquad 0<\operatorname {Re} (z)<1\,,\,\mu >0}
ohne Beweis
2.5 Bearbeiten
{\displaystyle \int _{-\infty }^{\infty }{\frac {a^{2}}{(e^{x}-ax-b)^{2}+(a\pi )^{2}}}\,dx={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}\qquad a>0\,,\,b\in \mathbb {R} }{\displaystyle \int _{-\infty }^{\infty }{\frac {a^{2}}{(e^{x}-ax-b)^{2}+(a\pi )^{2}}}\,dx={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}\qquad a>0\,,\,b\in \mathbb {R} }
Beweis
Betrachte die Funktion {\displaystyle f(z)={\frac {1}{a\log(-z)+b-z}}\cdot {\frac {1}{z}}}{\displaystyle f(z)={\frac {1}{a\log(-z)+b-z}}\cdot {\frac {1}{z}}} auf dem Gebiet {\displaystyle D:=\mathbb {C} \setminus \mathbb {R} ^{\geq 0}}{\displaystyle D:=\mathbb {C} \setminus \mathbb {R} ^{\geq 0}}.
Eins durch Log.PNG
{\displaystyle \forall z\in D}{\displaystyle \forall z\in D} gibt es genau ein {\displaystyle r>0\,}{\displaystyle r>0\,} und genau ein {\displaystyle -{\frac {\pi }{2}}<\varphi <{\frac {\pi }{2}}}{\displaystyle -{\frac {\pi }{2}}<\varphi <{\frac {\pi }{2}}}, so dass {\displaystyle -z=re^{i\varphi }\,}{\displaystyle -z=re^{i\varphi }\,} ist.
Beim Nenner {\displaystyle a\log(-z)+b-z=a\log \left(re^{i\varphi }\right)+b+re^{i\varphi }}{\displaystyle a\log(-z)+b-z=a\log \left(re^{i\varphi }\right)+b+re^{i\varphi }}
{\displaystyle =a\log r+ia\varphi +b+r\cos \varphi +ir\sin \varphi =(a\log r+r\cos \varphi +b)+i(a\varphi +r\sin \varphi )\,}{\displaystyle =a\log r+ia\varphi +b+r\cos \varphi +ir\sin \varphi =(a\log r+r\cos \varphi +b)+i(a\varphi +r\sin \varphi )\,}
hat der Imaginärteil das selbe Vorzeichen wie {\displaystyle \varphi \,}\varphi \, und der Realteil steigt streng monoton in {\displaystyle r\,}r\,.
Daher ist {\displaystyle z=-a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)}{\displaystyle z=-a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)} die einzige Polstelle von {\displaystyle f\,}f\,.
Diese erhält man, wenn man {\displaystyle \varphi =0\,}{\displaystyle \varphi =0\,} und {\displaystyle r=a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)}{\displaystyle r=a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)} setzt.
Nun ist {\displaystyle {\text{res}}\left(f,-a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)\right)={\frac {1}{a}}\cdot {\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}{\displaystyle {\text{res}}\left(f,-a\cdot W\left({\frac {1}{a}}\,e^{-b/a}\right)\right)={\frac {1}{a}}\cdot {\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.
Also gilt nach dem Residuensatz {\displaystyle \int _{\gamma _{R,\varepsilon }}f\,dz+\int _{C_{R}}f\,dz+\int _{\delta _{R,\varepsilon }}f\,dz+\int _{c_{\varepsilon }}f\,dz={\frac {1}{a}}\cdot {\frac {2\pi i}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}{\displaystyle \int _{\gamma _{R,\varepsilon }}f\,dz+\int _{C_{R}}f\,dz+\int _{\delta _{R,\varepsilon }}f\,dz+\int _{c_{\varepsilon }}f\,dz={\frac {1}{a}}\cdot {\frac {2\pi i}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.
Aus {\displaystyle L(C_{R})\sim 2\pi R\,}{\displaystyle L(C_{R})\sim 2\pi R\,} und {\displaystyle M(C_{R})=\max _{z\in C_{R}}|f(z)|\sim {\frac {1}{R^{2}}}}{\displaystyle M(C_{R})=\max _{z\in C_{R}}|f(z)|\sim {\frac {1}{R^{2}}}} folgt {\displaystyle \left|\int _{C_{R}}f\,dz\right|\leq L(C_{R})\,M(C_{R})\sim {\frac {2\pi }{R}}}{\displaystyle \left|\int _{C_{R}}f\,dz\right|\leq L(C_{R})\,M(C_{R})\sim {\frac {2\pi }{R}}}.
Daher geht {\displaystyle \int _{C_{R}}f\,dz}{\displaystyle \int _{C_{R}}f\,dz} gegen null für {\displaystyle R\to \infty \,}R\to \infty \,.
Und aus {\displaystyle L(c_{\varepsilon })=\pi \varepsilon \,}{\displaystyle L(c_{\varepsilon })=\pi \varepsilon \,} und {\displaystyle M(c_{\varepsilon })=\max _{z\in c_{\varepsilon }}|f(z)|\sim {\frac {-1}{a\,\log \varepsilon }}\cdot {\frac {1}{\varepsilon }}}{\displaystyle M(c_{\varepsilon })=\max _{z\in c_{\varepsilon }}|f(z)|\sim {\frac {-1}{a\,\log \varepsilon }}\cdot {\frac {1}{\varepsilon }}} folgt {\displaystyle \left|\int _{c_{\varepsilon }}f\,dz\right|\leq L(c_{\varepsilon })\,M(c_{\varepsilon })\sim {\frac {-\pi }{a\,\log \varepsilon }}}{\displaystyle \left|\int _{c_{\varepsilon }}f\,dz\right|\leq L(c_{\varepsilon })\,M(c_{\varepsilon })\sim {\frac {-\pi }{a\,\log \varepsilon }}}.
Daher geht {\displaystyle \int _{c_{\varepsilon }}f\,dz}{\displaystyle \int _{c_{\varepsilon }}f\,dz} auch gegen null für {\displaystyle \varepsilon \to 0+\,}{\displaystyle \varepsilon \to 0+\,}.
Im Grenzübergang {\displaystyle R\to \infty \,,\,\varepsilon \to 0+}{\displaystyle R\to \infty \,,\,\varepsilon \to 0+} ergibt sich
{\displaystyle \int _{0}^{\infty }\left({\frac {1}{a\log(-x-i0^{+})+b-x}}-{\frac {1}{a\log(-x+i0^{+})+b-x}}\right){\frac {dx}{x}}={\frac {1}{a}}\cdot {\frac {2\pi i}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}{\displaystyle \int _{0}^{\infty }\left({\frac {1}{a\log(-x-i0^{+})+b-x}}-{\frac {1}{a\log(-x+i0^{+})+b-x}}\right){\frac {dx}{x}}={\frac {1}{a}}\cdot {\frac {2\pi i}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.
Dabei ist {\displaystyle {\frac {1}{a(\log x-i\pi )+b-x}}-{\frac {1}{a(\log x+i\pi )+b-x}}={\frac {2\pi i\cdot a}{(a\log x+b-x)^{2}-(i\pi a)^{2}}}}{\displaystyle {\frac {1}{a(\log x-i\pi )+b-x}}-{\frac {1}{a(\log x+i\pi )+b-x}}={\frac {2\pi i\cdot a}{(a\log x+b-x)^{2}-(i\pi a)^{2}}}},
und somit gilt {\displaystyle \int _{0}^{\infty }{\frac {a^{2}}{(a\log x+b-x)^{2}+(a\pi )^{2}}}\cdot {\frac {dx}{x}}={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}{\displaystyle \int _{0}^{\infty }{\frac {a^{2}}{(a\log x+b-x)^{2}+(a\pi )^{2}}}\cdot {\frac {dx}{x}}={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.
Substituiert man {\displaystyle x\mapsto e^{x}}{\displaystyle x\mapsto e^{x}}, so ist {\displaystyle \int _{-\infty }^{\infty }{\frac {a^{2}}{(ax+b-e^{x})^{2}+(a\pi )^{2}}}\,dx={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}{\displaystyle \int _{-\infty }^{\infty }{\frac {a^{2}}{(ax+b-e^{x})^{2}+(a\pi )^{2}}}\,dx={\frac {1}{1+W\left({\frac {1}{a}}\,e^{-b/a}\right)}}}.
4.1 Bearbeiten
{\displaystyle \int _{0}^{\infty }{\frac {e^{kx}-e^{\lambda x}}{e^{\mu x}-e^{\nu x}}}\,dx={\frac {1}{\mu -\nu }}\left(\psi \left({\frac {\mu -\lambda }{\mu -\nu }}\right)-\psi \left({\frac {\mu -k}{\mu -\nu }}\right)\right)}{\displaystyle \int _{0}^{\infty }{\frac {e^{kx}-e^{\lambda x}}{e^{\mu x}-e^{\nu x}}}\,dx={\frac {1}{\mu -\nu }}\left(\psi \left({\frac {\mu -\lambda }{\mu -\nu }}\right)-\psi \left({\frac {\mu -k}{\mu -\nu }}\right)\right)}
Beweis
Aus der Gaußschen Formel {\displaystyle \psi (z)+\gamma =\int _{0}^{1}{\frac {1-u^{z-1}}{1-u}}\,du}{\displaystyle \psi (z)+\gamma =\int _{0}^{1}{\frac {1-u^{z-1}}{1-u}}\,du}
folgt {\displaystyle \psi (1-\beta )-\psi (1-\alpha )=\int _{0}^{1}{\frac {u^{-\alpha }-u^{-\beta }}{1-u}}\,du=\int _{0}^{\infty }{\frac {e^{\alpha x}-e^{\beta x}}{e^{x}-1}}\,dx\quad \left({\text{nach Substitution}}\,u=e^{-x}\right)}{\displaystyle \psi (1-\beta )-\psi (1-\alpha )=\int _{0}^{1}{\frac {u^{-\alpha }-u^{-\beta }}{1-u}}\,du=\int _{0}^{\infty }{\frac {e^{\alpha x}-e^{\beta x}}{e^{x}-1}}\,dx\quad \left({\text{nach Substitution}}\,u=e^{-x}\right)}.
Nun ist {\displaystyle \int _{0}^{\infty }{\frac {e^{kx}-e^{\lambda x}}{e^{\mu x}-e^{\nu x}}}\,dx=\int _{0}^{\infty }{\frac {e^{(k-\nu )x}-e^{(\lambda -\nu )x}}{e^{(\mu -\nu )x}-1}}\,dx={\frac {1}{\mu -\nu }}\int _{0}^{\infty }{\frac {e^{{\frac {k-\nu }{\mu -\nu }}\,x}-e^{{\frac {\lambda -\nu }{\mu -\nu }}\,x}}{e^{x}-1}}\,dx}{\displaystyle \int _{0}^{\infty }{\frac {e^{kx}-e^{\lambda x}}{e^{\mu x}-e^{\nu x}}}\,dx=\int _{0}^{\infty }{\frac {e^{(k-\nu )x}-e^{(\lambda -\nu )x}}{e^{(\mu -\nu )x}-1}}\,dx={\frac {1}{\mu -\nu }}\int _{0}^{\infty }{\frac {e^{{\frac {k-\nu }{\mu -\nu }}\,x}-e^{{\frac {\lambda -\nu }{\mu -\nu }}\,x}}{e^{x}-1}}\,dx}
{\displaystyle ={\frac {1}{\mu -\nu }}\left(\psi \left(1-{\frac {\lambda -\nu }{\mu -\nu }}\right)-\psi \left(1-{\frac {k-\nu }{\mu -\nu }}\right)\right)={\frac {1}{\mu -\nu }}\left(\psi \left({\frac {\mu -\lambda }{\mu -\nu }}\right)-\psi \left({\frac {\mu -k}{\mu -\nu }}\right)\right)}{\displaystyle ={\frac {1}{\mu -\nu }}\left(\psi \left(1-{\frac {\lambda -\nu }{\mu -\nu }}\right)-\psi \left(1-{\frac {k-\nu }{\mu -\nu }}\right)\right)={\frac {1}{\mu -\nu }}\left(\psi \left({\frac {\mu -\lambda }{\mu -\nu }}\right)-\psi \left({\frac {\mu -k}{\mu -\nu }}\right)\right)}.
Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,log,tan)
0.1 Bearbeiten
{\displaystyle \int _{0}^{1}\log \left(\tan {\frac {\pi x}{2}}\right)\,dx=0}{\displaystyle \int _{0}^{1}\log \left(\tan {\frac {\pi x}{2}}\right)\,dx=0}
ohne Beweis
0.2 Bearbeiten
{\displaystyle \int _{0}^{\pi }\log ^{2}\left(\tan {\frac {x}{2}}\right)dx={\frac {\pi ^{3}}{4}}}{\displaystyle \int _{0}^{\pi }\log ^{2}\left(\tan {\frac {x}{2}}\right)dx={\frac {\pi ^{3}}{4}}}
Beweis
Die Funktion {\displaystyle f(x)=-{\frac {1}{2}}\log \left(\tan {\frac {x}{2}}\right)}{\displaystyle f(x)=-{\frac {1}{2}}\log \left(\tan {\frac {x}{2}}\right)} besitzt die Fourierreihenentwicklung {\displaystyle \sum _{k=0}^{\infty }{\frac {\cos(2k+1)x}{2k+1}}}{\displaystyle \sum _{k=0}^{\infty }{\frac {\cos(2k+1)x}{2k+1}}}.
Nach der Parsevalschen Gleichung {\displaystyle {\frac {1}{\pi }}\int _{-\pi }^{\pi }|f(x)|^{2}\,dx={\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{\infty }{\Big (}a_{k}^{2}+b_{k}^{2}{\Big )}}{\displaystyle {\frac {1}{\pi }}\int _{-\pi }^{\pi }|f(x)|^{2}\,dx={\frac {a_{0}^{2}}{2}}+\sum _{k=1}^{\infty }{\Big (}a_{k}^{2}+b_{k}^{2}{\Big )}}
gilt dann {\displaystyle {\frac {1}{4\pi }}\int _{-\pi }^{\pi }\log ^{2}\left(\tan {\frac {x}{2}}\right)dx=\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{8}}}{\displaystyle {\frac {1}{4\pi }}\int _{-\pi }^{\pi }\log ^{2}\left(\tan {\frac {x}{2}}\right)dx=\sum _{k=0}^{\infty }{\frac {1}{(2k+1)^{2}}}={\frac {\pi ^{2}}{8}}}.
0.3 Bearbeiten
{\displaystyle \int _{0}^{\pi }\log ^{2}\left(\tan {\frac {x}{4}}\right)dx={\frac {\pi ^{3}}{4}}}{\displaystyle \int _{0}^{\pi }\log ^{2}\left(\tan {\frac {x}{4}}\right)dx={\frac {\pi ^{3}}{4}}}
Beweis
Die Funktion {\displaystyle f(x)=\log ^{2}\left(\tan {\frac {x}{2}}\right)}{\displaystyle f(x)=\log ^{2}\left(\tan {\frac {x}{2}}\right)} besitzt die Symmetrie {\displaystyle f(\pi -x)=f(x)\,}{\displaystyle f(\pi -x)=f(x)\,}.
{\displaystyle \int _{0}^{\pi }f\left({\frac {x}{2}}\right)dx=2\int _{0}^{\frac {\pi }{2}}f(x)\,dx}{\displaystyle \int _{0}^{\pi }f\left({\frac {x}{2}}\right)dx=2\int _{0}^{\frac {\pi }{2}}f(x)\,dx} ist daher {\displaystyle \int _{0}^{\pi }f(x)\,dx={\frac {\pi ^{3}}{4}}}{\displaystyle \int _{0}^{\pi }f(x)\,dx={\frac {\pi ^{3}}{4}}}.
0.4 Bearbeiten
{\displaystyle \int _{\pi /4}^{\pi /2}\log \log \tan x\,dx={\frac {\pi }{2}}\,\log \left({\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}{\displaystyle \int _{\pi /4}^{\pi /2}\log \log \tan x\,dx={\frac {\pi }{2}}\,\log \left({\sqrt {2\pi }}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}
1. Beweis (Vardisches Integral)
{\displaystyle I:=\int _{\pi /4}^{\pi /2}\log \log \tan x\,dx}{\displaystyle I:=\int _{\pi /4}^{\pi /2}\log \log \tan x\,dx} ist nach Substitution {\displaystyle x\mapsto \arctan e^{x}}{\displaystyle x\mapsto \arctan e^{x}} gleich {\displaystyle {\frac {1}{2}}\int _{0}^{\infty }{\frac {\log x}{\cosh x}}\,dx}{\displaystyle {\frac {1}{2}}\int _{0}^{\infty }{\frac {\log x}{\cosh x}}\,dx}.
Und das ist {\displaystyle {\frac {\pi }{2}}\,\int _{0}^{\infty }{\frac {\log \pi x}{\cosh \pi x}}\,dx={\frac {\pi }{2}}\log {\sqrt {\pi }}\int _{-\infty }^{\infty }{\frac {dx}{\cosh \pi x}}+{\frac {\pi }{8}}\int _{-\infty }^{\infty }{\frac {\log x^{2}}{\cosh \pi x}}\,dx}{\displaystyle {\frac {\pi }{2}}\,\int _{0}^{\infty }{\frac {\log \pi x}{\cosh \pi x}}\,dx={\frac {\pi }{2}}\log {\sqrt {\pi }}\int _{-\infty }^{\infty }{\frac {dx}{\cosh \pi x}}+{\frac {\pi }{8}}\int _{-\infty }^{\infty }{\frac {\log x^{2}}{\cosh \pi x}}\,dx}.
Dabei ist {\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{\cosh \pi x}}=1}{\displaystyle \int _{-\infty }^{\infty }{\frac {dx}{\cosh \pi x}}=1} und nach der Formel {\displaystyle \int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2}+x^{2})}{\cosh \pi x}}\,dx=4\,\log \left({\sqrt {2}}\;{\frac {\Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)}}\right)}{\displaystyle \int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2}+x^{2})}{\cosh \pi x}}\,dx=4\,\log \left({\sqrt {2}}\;{\frac {\Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)}}\right)} für {\displaystyle \alpha \geq 0}{\displaystyle \alpha \geq 0}
ist {\displaystyle {\frac {\pi }{8}}\,\int _{-\infty }^{\infty }{\frac {\log x^{2}}{\cosh \pi x}}\,dx={\frac {\pi }{2}}\,\log \left({\sqrt {2}}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}{\displaystyle {\frac {\pi }{8}}\,\int _{-\infty }^{\infty }{\frac {\log x^{2}}{\cosh \pi x}}\,dx={\frac {\pi }{2}}\,\log \left({\sqrt {2}}\,\,{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}. Also ist {\displaystyle I={\frac {\pi }{2}}\,\log \left({\sqrt {2\pi }}\;{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}{\displaystyle I={\frac {\pi }{2}}\,\log \left({\sqrt {2\pi }}\;{\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}.
2. Beweis
In der Formel {\displaystyle \int _{0}^{1}{\frac {\log \log \left({\frac {1}{x}}\right)}{1+2\cos \alpha \pi \cdot x+x^{2}}}\,dx={\frac {\pi }{2\sin \alpha \pi }}\left(\alpha \log 2\pi +\log {\frac {\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{2}}-{\frac {\alpha }{2}}\right)}}\right)}{\displaystyle \int _{0}^{1}{\frac {\log \log \left({\frac {1}{x}}\right)}{1+2\cos \alpha \pi \cdot x+x^{2}}}\,dx={\frac {\pi }{2\sin \alpha \pi }}\left(\alpha \log 2\pi +\log {\frac {\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{2}}-{\frac {\alpha }{2}}\right)}}\right)}
setze {\displaystyle \alpha ={\frac {1}{2}}\,:\quad \int _{0}^{1}{\frac {\log \log \left({\frac {1}{x}}\right)}{1+x^{2}}}\,dx={\frac {\pi }{2}}\left(\log {\sqrt {2\pi }}+\log {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}{\displaystyle \alpha ={\frac {1}{2}}\,:\quad \int _{0}^{1}{\frac {\log \log \left({\frac {1}{x}}\right)}{1+x^{2}}}\,dx={\frac {\pi }{2}}\left(\log {\sqrt {2\pi }}+\log {\frac {\Gamma \left({\frac {3}{4}}\right)}{\Gamma \left({\frac {1}{4}}\right)}}\right)}
Durch die Substitution {\displaystyle x\mapsto \cot x}{\displaystyle x\mapsto \cot x} ergibt sich die besagte Gleichung.
