Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,exp,cos)

 

1.1Bearbeiten

{\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,\cos(2ax)\,dx={\sqrt {\pi }}\cdot e^{-a^{2}}\qquad a\in \mathbb {C} }

1. Beweis

Verwende die Reihenentwicklung {\displaystyle \cos(2ax)=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}\,(2ax)^{2k}}.

{\displaystyle I:=\int _{-\infty }^{\infty }e^{-x^{2}}\,\cos(2ax)\,dx=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}\,(2a)^{2k}\,\int _{-\infty }^{\infty }x^{2k}\,e^{-x^{2}}\,dx}

Dabei ist {\displaystyle \int _{-\infty }^{\infty }x^{2k}\,e^{-x^{2}}\,dx=2\int _{0}^{\infty }x^{2k}\,e^{-x^{2}}\,dx=2\int _{0}^{\infty }y^{k}\,e^{-y}\,{\frac {dy}{2{\sqrt {y}}}}=\Gamma \left(k+{\frac {1}{2}}\right)={\frac {\sqrt {\pi }}{2^{2k-1}}}\,{\frac {\Gamma (2k)}{\Gamma (k)}}={\frac {\sqrt {\pi }}{2^{2k}}}\,{\frac {(2k)!}{k!}}}.

{\displaystyle I=\sum _{k=0}^{\infty }{\frac {(-1)^{k}}{(2k)!}}\,(2a)^{2k}\,{\frac {\sqrt {\pi }}{2^{2k}}}\,{\frac {(2k)!}{k!}}={\sqrt {\pi }}\cdot \sum _{k=0}^{\infty }{\frac {(-1)^{k}}{k!}}\,a^{2k}={\sqrt {\pi }}\cdot e^{-a^{2}}}

2. Beweis

Integriert man die holomorphe Funktion {\displaystyle f(z)=e^{-z^{2}}} längs der geschlossenen Kurve in der Zeichnung, so ist {\displaystyle \oint f(z)\,dz=0}.

Die Integrale über den vertikalen Strecken verschwinden für {\displaystyle R\to \infty \,}. Also ist {\displaystyle {\sqrt {\pi }}=\int _{-\infty }^{\infty }f(x)\,dx=\int _{-\infty }^{\infty }f(x+ia)\,dx}

{\displaystyle f(x+ia)=e^{-(x+ia)^{2}}=e^{-x^{2}-i\cdot 2ax+a^{2}}=e^{a^{2}}\cdot e^{-x^{2}}\,{\Big (}\cos(2ax)-i\sin(2ax){\Big )}}

{\displaystyle \Rightarrow \,\int _{-\infty }^{\infty }f(x+ia)\,dx=e^{a^{2}}\cdot {\Bigg [}\int _{-\infty }^{\infty }e^{-x^{2}}\cos(2ax)\,dx-i\underbrace {\int _{-\infty }^{\infty }e^{-x^{2}}\sin(2ax)\,dx} _{=0}{\Bigg ]}}
{\displaystyle \Rightarrow \,\int _{-\infty }^{\infty }e^{-x^{2}}\,\cos(2ax)\,dx={\sqrt {\pi }}\cdot e^{-a^{2}}}

 

3.1Bearbeiten

{\displaystyle \int _{0}^{\infty }e^{-ax}\,\cos(bx)\,x^{s-1}\,dx={\frac {\Gamma (s)}{{\sqrt {a^{2}+b^{2}}}^{s}}}\,\cos \left(s\,\arctan {\frac {b}{a}}\right)\qquad a>0\,,\,b\in \mathbb {R} \,,\,{\text{Re}}(s)>0}

ohne Beweis