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

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##### 1.1
${\displaystyle \int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2}+x^{2})}{\cosh \pi x}}\,dx=4\log \left({\sqrt {2}}\,\,{\frac {\Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)}}\right)\qquad {\text{Re}}(\alpha )>0}$
Beweis

Setzt man ${\displaystyle f(z)={\frac {2\alpha }{(\alpha ^{2}+z^{2})\cosh \pi z}}}$, so ist

${\displaystyle 2\pi i\,\;{\text{res}}(f,i\alpha )={\frac {2\pi }{\cos \alpha \pi }}=\pi \cot \left[\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\pi \right]-\pi \cot \left[\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\pi \right]}$.

Und ${\displaystyle 2\pi i\sum _{k=0}^{\infty }{\text{res}}\left(f,i\,{\frac {2k+1}{2}}\right)=\sum _{k=0}^{\infty }(-1)^{k}\left[{\frac {4}{2k+1+2\alpha }}-{\frac {4}{2k+1-2\alpha }}\right]}$

${\displaystyle =\left[\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]-\left[\psi \left({\frac {3}{4}}-{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}-{\frac {\alpha }{2}}\right)\right]}$,

wobei ${\displaystyle \psi \left({\frac {3}{4}}-{\frac {\alpha }{2}}\right)=\psi \left(1-\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right)=\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)+\pi \cot \left[\left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\pi \right]}$

und ${\displaystyle \psi \left({\frac {1}{4}}-{\frac {\alpha }{2}}\right)=\psi \left(1-\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\right)=\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)+\pi \cot \left[\left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)\pi \right]}$ ist.

${\displaystyle \int _{-\infty }^{\infty }f(x)\,dx=\lim _{N\to \infty }\oint _{\gamma _{N}}f\,dz=2\pi i\,\sum _{{\text{Im}}>0}{\text{res}}f=2\left[\psi \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\psi \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]}$.

Integriere nun beide Seiten nach ${\displaystyle \alpha \,}$:

${\displaystyle \underbrace {\int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2}+x^{2})}{\cosh \pi x}}\,dx} _{=:U(\alpha )}=\underbrace {4\left[\log \Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)-\log \Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)\right]} _{=:V(\alpha )}+C}$

Wegen ${\displaystyle U(\alpha )-\log(\alpha ^{2})=U(\alpha )-\int _{-\infty }^{\infty }{\frac {\log(\alpha ^{2})}{\cosh \pi x}}\,dx=\int _{-\infty }^{\infty }{\frac {\log \left(1+{\frac {x^{2}}{\alpha ^{2}}}\right)}{\cosh \pi x}}\,dx\to 0}$ für ${\displaystyle \alpha \to \infty \,}$

und ${\displaystyle V(\alpha )-\log(\alpha ^{2})\to -4\log {\sqrt {2}}}$ muss ${\displaystyle C=4\log {\sqrt {2}}}$ sein.

Daher lässt sich die rechte Seite auch schreiben als ${\displaystyle 4\log \left({\sqrt {2}}\,\;{\frac {\Gamma \left({\frac {3}{4}}+{\frac {\alpha }{2}}\right)}{\Gamma \left({\frac {1}{4}}+{\frac {\alpha }{2}}\right)}}\right)}$.

##### 1.2
${\displaystyle \int _{0}^{\infty }{\frac {\log x}{\cosh x+\cos \pi \alpha }}\,dx={\frac {\pi }{\sin \pi \alpha }}\log \left((2\pi )^{\alpha }\,\,{\frac {\Gamma \left({\frac {1+\alpha }{2}}\right)}{\Gamma \left({\frac {1-\alpha }{2}}\right)}}\right)\qquad 0<\alpha <1}$