Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,log)

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1+x)-\log 2}{1+x^{2}}}\,dx=-{\frac {\pi }{8}}\log 2}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1+x)-\log 2}{1-x^{2}}}\,dx=-{\frac {\pi ^{2}}{24}}}

ohne Beweis

 

 

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{\displaystyle \int _{1}^{\infty }{\frac {\log(1+x)-\log 2}{1+x^{2}}}\,dx=G-{\frac {\pi }{8}}\log 2}

ohne Beweis

 

 

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{\displaystyle \int _{1}^{\infty }{\frac {\log(1+x)-\log 2}{1-x^{2}}}\,dx=-{\frac {\pi ^{2}}{12}}}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log x}{1+x^{2}}}\,dx=-G}

Beweis

Aus {\displaystyle {\frac {\log x}{1+x^{2}}}=\sum _{n=0}^{\infty }(-1)^{n}\,x^{2n}\,\log x}

folgt {\displaystyle \int _{0}^{1}{\frac {\log x}{1+x^{2}}}dx=\sum _{n=0}^{\infty }(-1)^{n}\int _{0}^{1}x^{2n}\log x\,dx=\sum _{n=0}^{\infty }(-1)^{n}\,{\frac {-1}{(2n+1)^{2}}}=-G}.

 

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{\displaystyle \int _{0}^{1}{\frac {\log x}{1-x^{2}}}\,dx=-{\frac {\pi ^{2}}{8}}}

Beweis

Aus {\displaystyle {\frac {\log x}{1-x^{2}}}=\sum _{n=0}^{\infty }x^{2n}\,\log x}

folgt {\displaystyle \int _{0}^{1}{\frac {\log x}{1-x^{2}}}dx=\sum _{n=0}^{\infty }\int _{0}^{1}x^{2n}\log x\,dx=\sum _{n=0}^{\infty }{\frac {-1}{(2n+1)^{2}}}=-{\frac {\pi ^{2}}{8}}}.

 

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{\displaystyle \int _{0}^{\infty }{\frac {\log(1+x+x^{2})}{1+x^{2}}}\,dx={\frac {\pi }{3}}\log(2+{\sqrt {3}})+{\frac {4}{3}}\,G}

Beweis

In der Formel {\displaystyle \int _{0}^{\infty }{\frac {\log(1+2\sin \alpha \cdot x+x^{2})}{1+x^{2}}}\,dx=\pi \log \left(2\cos {\frac {\alpha }{2}}\right)+\alpha \log \left(\tan {\frac {\alpha }{2}}\right)+2\sum _{k=0}^{\infty }{\frac {\sin(2k+1)\alpha }{(2k+1)^{2}}}} setze {\displaystyle \alpha ={\frac {\pi }{6}}}.

Wegen {\displaystyle \pi \log \left(2\cos {\frac {\pi }{12}}\right)={\frac {\pi }{2}}\left(\left(2\cos {\frac {\pi }{12}}\right)^{2}\right)={\frac {\pi }{2}}\log(2+{\sqrt {3}})\,,\;{\frac {\pi }{6}}\log \left(\tan {\frac {\pi }{12}}\right)=-{\frac {\pi }{6}}\log \left(\cot {\frac {\pi }{12}}\right)=-{\frac {\pi }{6}}\log(2+{\sqrt {3}})}

und {\displaystyle 2\sum _{k=0}^{3N}{\frac {\sin(2k+1){\frac {\pi }{6}}}{(2k+1)^{2}}}=\sum _{k=0}^{3N}{\frac {(-1)^{k}}{(2k+1)^{2}}}+3\sum _{k=0}^{N-1}{\frac {(-1)^{k}}{(6k+3)^{2}}}\to G+{\frac {G}{3}}={\frac {4}{3}}\,G}

ist dann {\displaystyle \int _{0}^{\infty }{\frac {\log(1+x+x^{2})}{1+x^{2}}}\,dx={\frac {\pi }{3}}\log(2+{\sqrt {3}})+{\frac {4}{3}}\,G}.

 

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{\displaystyle \int _{0}^{1}\log(-\log x)\,dx=-\gamma }

Beweis

Differenziere {\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}\,e^{-t}\,dt}.

{\displaystyle \Gamma '(z)=\int _{0}^{\infty }t^{z-1}\,\log(t)\,e^{-t}\,dt}.

und setze {\displaystyle z=1\,}.

{\displaystyle -\gamma =\Gamma '(1)=\int _{0}^{\infty }\log(t)\,e^{-t}\,dt}.

Dies ist nach Substitution {\displaystyle t\mapsto -\log x} gleich {\displaystyle \int _{0}^{1}\log(-\log x)\,dx}.

 

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{\displaystyle \int _{0}^{1}\left({\frac {2\log x}{x^{2}-4x+8}}-{\frac {3\log x}{x^{2}+2x+2}}\right)dx=G}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{\infty }{\frac {\log(x+1)}{\log ^{2}x+\pi ^{2}}}\,{\frac {dx}{x^{2}}}=\gamma }

Beweis

Betrachte die Formel {\displaystyle \int _{0}^{\infty }{\frac {1}{x+1-u}}\cdot {\frac {1}{\log ^{2}x+\pi ^{2}}}\,dx={\frac {1}{u}}+{\frac {1}{\log(1-u)}}}.

Wegen {\displaystyle \int _{0}^{1}{\frac {1}{x+1-u}}\,du=\left[-\log(x+1-u)\right]_{0}^{1}=\log \left(1+{\frac {1}{x}}\right)}

und {\displaystyle \int _{0}^{1}\left[{\frac {1}{u}}+{\frac {1}{\log(1-u)}}\right]du=\gamma } ist {\displaystyle \int _{0}^{\infty }{\frac {\log \left(1+{\frac {1}{x}}\right)}{\log ^{2}x+\pi ^{2}}}\,dx=\gamma }.

Nach der Substitution {\displaystyle x\mapsto {\frac {1}{x}}} erhält man das gesuchte Integral.

 

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{\displaystyle \int _{0}^{1}{\frac {\log \left(1+x^{2+{\sqrt {3}}}\,\right)}{1+x}}\,dx={\frac {\pi ^{2}}{12}}\cdot \left(1-{\sqrt {3}}\,\right)+\log(2)\cdot \log \left(1+{\sqrt {3}}\,\right)}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log \left(1+x^{4+{\sqrt {15}}}\,\right)}{1+x}}\,dx={\frac {\pi ^{2}}{12}}\cdot \left(2-{\sqrt {15}}\,\right)+\log \left({\frac {1+{\sqrt {5}}}{2}}\right)\cdot \log \left(2+{\sqrt {3}}\,\right)+\log(2)\cdot \log \left({\sqrt {3}}+{\sqrt {5}}\,\right)}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log \left(1+x^{6+{\sqrt {35}}}\,\right)}{1+x}}\,dx={\frac {\pi ^{2}}{12}}\cdot \left(3-{\sqrt {35}}\,\right)+\log \left({\frac {1+{\sqrt {5}}}{2}}\right)\cdot \log \left(8+3{\sqrt {7}}\,\right)+\log(2)\cdot \log \left({\sqrt {5}}+{\sqrt {7}}\,\right)}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}\log(1-x)\,\log(x)\,\log(1+x)\,dx=-6+{\frac {5\pi ^{2}}{12}}-(\log 2)^{2}+4\cdot \log 2-{\frac {\pi ^{2}}{2}}\log 2+{\frac {21}{8}}\zeta (3)}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{1+x}}\,dx=-{\frac {\zeta (3)}{8}}}

Beweis

Wegen {\displaystyle {\frac {\log(1+x)}{1+x}}=\sum _{k=1}^{\infty }(-1)^{k-1}\,H_{k}\ x^{k}} ist

{\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{1+x}}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,H_{k}\,\int _{0}^{1}x^{k}\,\log x\,\,dx=\sum _{k=1}^{\infty }(-1)^{k}\,{\frac {H_{k}}{(k+1)^{2}}}=-{\frac {\zeta (3)}{8}}}.

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{x}}\,dx=-{\frac {3}{4}}\,\zeta (3)}

Beweis

{\displaystyle \log(1+x)=\sum _{k=1}^{\infty }(-1)^{k-1}\,{\frac {x^{k}}{k}}}

{\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{x}}\,dx=\sum _{k=1}^{\infty }(-1)^{k-1}\,{\frac {1}{k}}\,\int _{0}^{1}x^{k-1}\,\log x\,dx=\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k^{3}}}=-{\frac {3}{4}}\,\zeta (3)}

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1+x)\,\log x}{1-x}}\,dx=-{\frac {\pi ^{2}}{4}}\log 2+\zeta (3)}

Beweis

{\displaystyle I:=\int _{0}^{1}{\frac {\log(1+x)\,\log x}{1-x}}\,dx=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}\int _{0}^{1}{\frac {x^{n}\,\log x}{1-x}}\,dx},

wobei {\displaystyle \int _{0}^{1}{\frac {x^{n}\,\log x}{1-x}}\,dx=\sum _{k=0}^{\infty }\int _{0}^{1}x^{n+k}\,\log x\,dx=\sum _{k=0}^{\infty }{\frac {-1}{(n+k+1)^{2}}}=\sum _{k=1}^{n}{\frac {1}{k^{2}}}-{\frac {\pi ^{2}}{6}}} ist.

{\displaystyle I=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}\left(H_{n,2}-{\frac {\pi ^{2}}{6}}\right)=\underbrace {\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{n}}\,H_{n,2}} _{=-{\frac {\pi ^{2}}{12}}\log 2+\zeta (3)}-{\frac {\pi ^{2}}{6}}\log 2=-{\frac {\pi ^{2}}{4}}\log 2+\zeta (3)}

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log x}{1+x}}\,dx=-{\frac {\pi ^{2}}{4}}\log 2+{\frac {13}{8}}\,\zeta (3)}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log x}{x}}\,dx=\zeta (3)}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log x}{1-x}}\,dx=\zeta (3)}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log(1+x)}{x}}\,dx=-{\frac {5}{8}}\,\zeta (3)}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1-x)\,\log(1+x)}{1+x}}\,dx={\frac {(\log 2)^{2}}{8}}-{\frac {\pi ^{2}}{12}}\log 2+{\frac {\zeta (3)}{8}}}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\log ^{2n}x}{1+x^{2}}}\,dx={\frac {1}{2}}\,|E_{2n}|\,\left({\frac {\pi }{2}}\right)^{2n+1}\qquad n\in \mathbb {Z} ^{\geq 0}}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{\infty }{\frac {\log ^{n-1}x}{1+x^{2}}}\,dx=|E_{n-1}|\,\left({\frac {\pi }{2}}\right)^{n}\qquad n\in \mathbb {Z} ^{\geq 1}}

Beweis

In der Formel {\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha -1}}{1+x^{\beta }}}\,dx={\frac {\pi }{\beta }}\csc \left({\frac {\alpha \pi }{\beta }}\right)} setze {\displaystyle \beta =2\,} und verschiebe {\displaystyle \alpha \,} um {\displaystyle 1\,} nach rechts.

{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha }}{1+x^{2}}}\,dx={\frac {\pi }{2}}\,\csc \left({\frac {(\alpha +1)\pi }{2}}\right)={\frac {\pi }{2}}\sec \left({\frac {\alpha \pi }{2}}\right)={\frac {\pi }{2}}\sum _{k=0}^{\infty }|E_{k}|\,{\frac {1}{k!}}\,\left({\frac {\alpha \pi }{2}}\right)^{k}=\sum _{k=0}^{\infty }|E_{k}|\,\left({\frac {\pi }{2}}\right)^{k+1}\,{\frac {\alpha ^{k}}{k!}}}

Differenziere {\displaystyle n-1\,} mal nach {\displaystyle \alpha \,}

{\displaystyle \int _{0}^{\infty }{\frac {x^{\alpha }\,\log ^{n-1}x}{1+x^{2}}}\,dx=\sum _{k=n-1}^{\infty }|E_{k}|\,\left({\frac {\pi }{2}}\right)^{k+1}\,{\frac {\alpha ^{k-n+1}}{(k-n+1)!}}}

Und setze {\displaystyle \alpha =0\,}

{\displaystyle \int _{0}^{\infty }{\frac {\log ^{n-1}x}{1+x^{2}}}\,dx=|E_{n-1}|\,\left({\frac {\pi }{2}}\right)^{n}}

 

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{\displaystyle \int _{0}^{\infty }{\frac {\log ^{n-1}x}{1-x^{2}}}\,dx={\frac {2^{n}(1-2^{n})|B_{n}|}{n}}\,\left({\frac {\pi }{2}}\right)^{n}\qquad n\in \mathbb {Z} ^{\geq 1}}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{\infty }{\frac {dx}{(x+1)^{n}\,(\log ^{2}x+\pi ^{2})}}=C_{n}} Fontana-Zahlen genügen der Rekursion: {\displaystyle \quad C_{0}=-1,\quad \sum _{k=0}^{n-1}{\frac {C_{k}}{n-k}}=0}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{\infty }{\frac {1}{x+1-u}}\cdot {\frac {1}{\log ^{2}x+\pi ^{2}}}\,dx={\frac {1}{u}}+{\frac {1}{\log(1-u)}}\qquad u\in \mathbb {C} ^{\times }\setminus \mathbb {R} ^{\geq 1}}

Beweis

Die Funktion {\displaystyle f(z)={\frac {1}{z+1-u}}\cdot {\frac {1}{\log(-z)}}} ist auf {\displaystyle \mathbb {C} \setminus \mathbb {R} ^{\geq 0}} meromorph.

Die Polstellen liegen bei {\displaystyle -1\,} und {\displaystyle u-1\,}. Dabei ist {\displaystyle {\text{res}}(f,-1)={\frac {1}{u}}} und {\displaystyle {\text{res}}(f,u-1)={\frac {1}{\log(1-u)}}}.

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=2\pi i\,\left({\frac {1}{u}}+{\frac {1}{\log(1-u)}}\right)}.

Aus {\displaystyle L(C_{R})\sim 2\pi R\,} und {\displaystyle M(C_{R})=\max _{z\in C_{R}}|f(z)|\sim {\frac {1}{R\,\log R}}} folgt {\displaystyle \left|\int _{C_{R}}f\,dz\right|\leq L(C_{R})\,M(C_{R})\sim {\frac {2\pi }{\log R}}}.

Daher geht {\displaystyle \int _{C_{R}}fdz} gegen null für {\displaystyle R\to \infty \,}.

Für {\displaystyle z\,}, nahe bei 0, ist {\displaystyle \log(-z)\,} groß, und somit {\displaystyle f(z)={\frac {1}{z+1-u}}\cdot {\frac {1}{\log(-z)}}} klein.

Daher geht {\displaystyle \int _{c_{\varepsilon }}f\,dz} für {\displaystyle \varepsilon \to 0+\,} auch gegen null.

Im Grenzübergang {\displaystyle R\to \infty ,\varepsilon \to 0+\,} ergibt sich:

{\displaystyle \int _{0}^{\infty }{\frac {1}{x+1-u}}\left({\frac {1}{\log(-x-i0^{+})}}-{\frac {1}{\log(-x+i0^{+})}}\right)dx=2\pi i\left({\frac {1}{u}}+{\frac {1}{\log(1-u)}}\right)}.

Dabei ist {\displaystyle {\frac {1}{\log x-i\pi }}-{\frac {1}{\log x+i\pi }}={\frac {2\pi i}{\log ^{2}x+\pi ^{2}}}}, und somit gilt

{\displaystyle \int _{0}^{\infty }{\frac {1}{x+1-u}}\cdot {\frac {1}{\log ^{2}x+\pi ^{2}}}\,dx={\frac {1}{u}}+{\frac {1}{\log(1-u)}}}.

 

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{\displaystyle \int _{0}^{1}{\frac {\left(\log \,{\frac {1}{x}}\right)^{z-1}}{1+x}}\,dx=\eta (z)\,\Gamma (z)\qquad {\text{Re}}(z)>0}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {\left(\log \,{\frac {1}{x}}\right)^{z-1}}{1-x}}\,dx=\zeta (z)\,\Gamma (z)\qquad {\text{Re}}(z)>1}

ohne Beweis

 

 

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{\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}-x^{-\alpha }}{(x+1)\,\log x}}\,dx=\log \left(\tan {\frac {\alpha \pi }{2}}\right)\qquad 0<\mathrm {Re} (\alpha )<1}

Beweis

{\displaystyle \int _{0}^{1}{\frac {x^{-\alpha }}{1+x}}\,dx} ist nach Substitution {\displaystyle x\mapsto {\frac {1}{x}}} gleich {\displaystyle \int _{1}^{\infty }{\frac {x^{\alpha -1}}{1+x}}\,dx}.

Also ist {\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}+x^{-\alpha }}{1+x}}\,dx=\int _{0}^{1}{\frac {x^{\alpha -1}}{1+x}}\,dx+\int _{1}^{\infty }{\frac {x^{\alpha -1}}{1+x}}\,dx}   {\displaystyle =\int _{0}^{\infty }{\frac {x^{\alpha -1}}{1+x}}\,dx={\frac {\pi }{\sin \alpha \pi }}}.

Integriere nach {\displaystyle \alpha \,}:

{\displaystyle \int _{0}^{1}{\frac {x^{\alpha -1}-x^{-\alpha }}{(1+x)\log x}}\,dx=\log \left(\tan {\frac {\alpha \pi }{2}}\right)+C}

Dass dabei {\displaystyle C=0\,} ist, erkennt man, wenn man {\displaystyle \alpha ={\frac {1}{2}}} setzt.

 

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{\displaystyle \int _{0}^{1}\left(\log {\frac {1}{x}}\right)^{z-1}dx=\Gamma (z)\qquad {\text{Re}}(z)>0}

Beweis

In der Formel {\displaystyle \Gamma (z)=\int _{0}^{\infty }t^{z-1}e^{-t}\,dt} substituiere {\displaystyle t=-\log x\,}:

{\displaystyle \Gamma (z)=\int _{0}^{1}\left(-\log x\right)^{z-1}dx}

 

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{\displaystyle \int _{0}^{1}\left({\frac {\log x}{a+1-x}}-{\frac {\log x}{a+x}}\right)dx={\frac {1}{2}}\left(\log a-\log(a+1)\right)^{2}\qquad \forall a\in \mathbb {C} \setminus [-1,0]}

Beweis

Definiert man {\displaystyle F:\,]0,1]\to \mathbb {C} } als {\displaystyle F(x)={\frac {x\log x}{a+x}}-\log(a+x)},

so ist {\displaystyle F'(x)={\frac {\log x+1}{a+x}}-{\frac {x\log x}{(a+x)^{2}}}-{\frac {1}{a+x}}}

{\displaystyle ={\frac {a\log x+a+x\log x+x-x\log x-(a+x)}{(a+x)^{2}}}={\frac {a\log x}{(a+x)^{2}}}}.

Also ist {\displaystyle a\int _{0}^{1}{\frac {\log x}{(a+x)^{2}}}\,dx=F(1)-F(0+)=\log a-\log(a+1)}.


Definiert man {\displaystyle G:\,]0,1]\to \mathbb {C} } als {\displaystyle G(x)={\frac {x\log x}{a+1-x}}+\log(a+1-x)},

so ist {\displaystyle G'(x)={\frac {\log x+1}{a+1-x}}+{\frac {x\log x}{(a+1-x)^{2}}}-{\frac {1}{a+1-x}}}

{\displaystyle ={\frac {(a+1)\log x+a+1-x\log x-x+x\log x-(a+1-x)}{(a+1-x)^{2}}}={\frac {(a+1)\log x}{(a+1-x)^{2}}}}

Also ist {\displaystyle (a+1)\int _{0}^{1}{\frac {\log x}{(a+1-x)^{2}}}\,dx=G(1)-G(0+)=\log a-\log(a+1)}.


Mit den beiden Integralen erhält man {\displaystyle \int _{0}^{1}\left({\frac {\log x}{(a+x)^{2}}}-{\frac {\log x}{(a+1-x)^{2}}}\right)dx=\left(\log a-\log(a+1)\right)\left({\frac {1}{a}}-{\frac {1}{a+1}}\right)}.

Integriert man beide Seiten nach {\displaystyle a\,}, so ist {\displaystyle \int _{0}^{1}\left(-{\frac {\log x}{a+x}}+{\frac {\log x}{a+1-x}}\right)dx={\frac {1}{2}}\left(\log a-\log(a+1)\right)^{2}+C}.

Dass die Konstante {\displaystyle C=0\,} sein muss, erkennt man wenn man {\displaystyle a\to \infty \,} gehen lässt.

 

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{\displaystyle \int _{0}^{1}{\frac {2\log x}{(a\pi )^{2}+\log ^{2}x}}\,{\frac {x}{1-x^{2}}}\,dx={\frac {1}{2a}}+\psi (a)-\log a\qquad {\text{Re}}(a)>0}

Beweis

{\displaystyle \int _{0}^{1}{\frac {2\log x}{(a\pi )^{2}+\log ^{2}x}}\,{\frac {x}{1-x^{2}}}\,dx=\int _{0}^{1}}   {\displaystyle \int _{0}^{\infty }2\sin(t\log x)\,e^{-a\pi t}\,dt\,}   {\displaystyle \,{\frac {x}{1-x^{2}}}\,dx}

{\displaystyle =\int _{0}^{\infty }\int _{0}^{1}2\sin(t\log x)\,{\frac {x}{1-x^{2}}}\,dx\,e^{-a\pi t}\,dt\qquad {\text{//subst:}}\,\,x=e^{-u}}

{\displaystyle =\int _{0}^{\infty }\int _{0}^{\infty }{\frac {2\sin(tu)}{1-e^{2u}}}\,du\,e^{-a\pi t}\,dt=\int _{0}^{\infty }}   {\displaystyle \left({\frac {1}{t}}-{\frac {\pi }{2}}\coth {\frac {\pi t}{2}}\right)}   {\displaystyle e^{-a\pi t}\,dt\qquad {\text{//subst:}}\,\,\omega =\pi t}

{\displaystyle =\int _{0}^{\infty }\left({\frac {1}{\omega }}-{\frac {1}{e^{\omega }-1}}-{\frac {1}{2}}\right)e^{-a\omega }\,d\omega =}   {\displaystyle \psi (a+1)-\log a-{\frac {1}{2a}}}

 

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{\displaystyle \int _{0}^{\infty }{\frac {\log(1+2\sin \alpha \,\,x+x^{2})}{1+x^{2}}}\,dx=\pi \log \left(2\cos {\frac {\alpha }{2}}\right)+\alpha \log \left(\tan {\frac {\alpha }{2}}\right)+2\sum _{k=0}^{\infty }{\frac {\sin(2k+1)\alpha }{(2k+1)^{2}}}\qquad 0<\alpha <\pi }

Beweis

Setzt man {\displaystyle F(z):=\int _{0}^{\infty }{\frac {\log(1+2\sin z\;\,x+x^{2})}{1+x^{2}}}\,dx},

so ist {\displaystyle F(0)=\int _{0}^{\infty }{\frac {\log(1+x^{2})}{1+x^{2}}}\,dx=\int _{0}^{\frac {\pi }{2}}\log(1+\tan ^{2}x)\,dx=-2\int _{0}^{\frac {\pi }{2}}\log(\cos x)\,dx=\pi \log 2}

und {\displaystyle F'(z)=\int _{0}^{\infty }{\frac {1}{1+x^{2}}}\,{\frac {2x\cos z}{1+2\sin z\;\,x+x^{2}}}\,dx={\frac {\cos z}{\sin z}}\int _{0}^{\infty }{\frac {1}{1+x^{2}}}\,{\frac {2x\sin z}{1+2\sin z\;\,x+x^{2}}}\,dx}

{\displaystyle ={\frac {\cos z}{\sin z}}\int _{0}^{\infty }\left({\frac {1}{1+x^{2}}}-{\frac {1}{1+2\sin z\;\,x+x^{2}}}\right)dx={\frac {\pi }{2}}\cdot {\frac {\cos z}{\sin z}}-{\frac {\cos z}{\sin z}}\left[{\frac {1}{\cos z}}\,\arctan {\frac {x+\sin z}{\cos z}}\right]_{0}^{\infty }}

{\displaystyle ={\frac {\pi }{2}}\cdot {\frac {\cos z}{\sin z}}-{\frac {\pi }{2}}\cdot {\frac {1}{\sin z}}+{\frac {z}{\sin z}}=-{\frac {\pi }{2}}\tan {\frac {z}{2}}+{\frac {z}{\sin z}}}.

Nun ist {\displaystyle F(\alpha )-\pi \log 2=\int _{0}^{\alpha }F'(z)\,dz=\pi \log \left(\cos {\frac {\alpha }{2}}\right)+\int _{0}^{\alpha }{\frac {z}{\sin z}}\,dz}

und somit ist {\displaystyle F(\alpha )-\pi \log \left(2\cos {\frac {\alpha }{2}}\right)=\int _{0}^{\alpha }{\frac {z}{\sin z}}\,dz=\left[z\log \left(\tan {\frac {z}{2}}\right)\right]_{0}^{\alpha }-\int _{0}^{\alpha }\log \left(\tan {\frac {z}{2}}\right)dz}

{\displaystyle =\alpha \log \left(\tan {\frac {\alpha }{2}}\right)+2\int _{0}^{\alpha }\sum _{k=0}^{\infty }{\frac {\cos(2k+1)z}{2k+1}}\,dz}.

Daraus folgt {\displaystyle F(\alpha )=\pi \log \left(2\cos {\frac {\alpha }{2}}\right)+\alpha \log \left(\tan {\frac {\alpha }{2}}\right)+2\sum _{k=0}^{\infty }{\frac {\sin(2k+1)\alpha }{(2k+1)^{2}}}}.

 

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{\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)\qquad 0<\alpha <1}

Beweis

{\displaystyle {\frac {\sin \alpha \pi }{1+2\cos \alpha \pi \cdot x+x^{2}}}=\sum _{n=1}^{\infty }(-x)^{n-1}\,\sin n\pi \alpha \qquad \quad |x|<1}

{\displaystyle \int _{0}^{1}{\frac {\sin \alpha \pi }{1+2\cos \alpha \pi \cdot x+x^{2}}}\,\log \log \left({\frac {1}{x}}\right)dx=\sum _{n=1}^{\infty }(-1)^{n-1}\int _{0}^{1}x^{n-1}\,\log \log \left({\frac {1}{x}}\right)dx\,\;\sin n\pi \alpha }

{\displaystyle =\sum _{n=1}^{\infty }(-1)^{n}\,{\frac {\gamma +\log n}{n}}\,\sin n\pi \alpha =\gamma \cdot \sum _{n=1}^{\infty }{\frac {(-1)^{n}}{n}}\,\sin n\pi \alpha +\sum _{n=1}^{\infty }(-1)^{n}\,{\frac {\log n}{n}}\,\sin n\pi \alpha }

{\displaystyle -{\frac {\alpha \pi }{2}}\,\gamma +{\frac {\alpha \pi }{2}}\left(\gamma +\log 2\pi \right)+{\frac {\pi }{2}}\log {\frac {\Gamma \left({\frac {1}{2}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{2}}-{\frac {\alpha }{2}}\right)}}={\frac {\pi }{2}}\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)}

 

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{\displaystyle \int _{0}^{1}{\frac {\log(1-x)}{x}}\,{\frac {2z}{\log ^{2}x+(2\pi z)^{2}}}\,dx=-\log \left({\frac {z!\,e^{z}}{z^{z}\,{\sqrt {2\pi z}}}}\right)\qquad {\text{Re}}(z)>0}

Beweis

 

 

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{\displaystyle \int _{a}^{b}{\frac {\log x}{(x+a)(x+b)}}\,dx={\frac {\log(ab)}{2\,(b-a)}}\,\log \left({\frac {(a+b)^{2}}{4ab}}\right)}

Beweis

Nach der Substitution {\displaystyle x\mapsto {\frac {ab}{x}}} wird das Integral zu {\displaystyle I=\int _{a}^{b}{\frac {\log(ab)-\log x}{(b+x)(a+x)}}\,dx}

Also ist {\displaystyle 2I=\int _{a}^{b}{\frac {\log(ab)}{(x+a)(x+b)}}\,dx={\frac {\log(ab)}{b-a}}\int _{a}^{b}\left({\frac {1}{x+a}}-{\frac {1}{x+b}}\right)\,dx}

{\displaystyle ={\frac {\log(ab)}{b-a}}\,{\Big [}\log(x+a)-\log(x+b){\Big ]}_{a}^{b}={\frac {\log(ab)}{b-a}}\log \left({\frac {(a+b)^{2}}{4ab}}\right)}.

 

2.2Bearbeiten

{\displaystyle \int _{0}^{\infty }{\frac {\log x}{(x+a)(x+b)}}\,dx={\frac {\log ^{2}(a)-\log ^{2}(b)}{2\,(a-b)}}}

Beweis

Nach der Substitution {\displaystyle x\mapsto {\frac {ab}{x}}} wird das Integral zu {\displaystyle I=\int _{0}^{\infty }{\frac {\log(ab)-\log x}{(b+x)(a+x)}}\,dx}.

Also ist {\displaystyle 2I=\int _{0}^{\infty }{\frac {\log(ab)}{(x+a)(x+b)}}\,dx={\frac {\log(ab)}{a-b}}\int _{0}^{\infty }\left({\frac {1}{x+b}}-{\frac {1}{x+a}}\right)\,dx}

{\displaystyle ={\frac {\log(a)+\log(b)}{a-b}}\,\underbrace {{\Big [}\log(x+b)-\log(x+a){\Big ]}_{0}^{\infty }} _{\log(a)-\log(b)}={\frac {\log ^{2}(a)-\log ^{2}(b)}{a-b}}}.