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

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##### 1.1
${\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,\cos(2ax)\,dx={\sqrt {\pi }}\cdot e^{-a^{2}}\qquad a\in \mathbb {C} }$
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 ${\displaystyle {\sqrt {\pi }}=\int _{-\infty }^{\infty }f(x)\,dx=\int _{-\infty }^{\infty }f(x+ia)\,dx}$

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

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

##### 3.1
${\displaystyle \int _{0}^{\infty }e^{-ax}\,\cos(bx)\,x^{s-1}\,dx={\frac {\Gamma (s)}{{\sqrt {a^{2}+b^{2}}}^{s}}}\,\cos \left(s\,\arctan {\frac {b}{a}}\right)\qquad a>0\,,\,b\in \mathbb {R} \,,\,{\text{Re}}(s)>0}$
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