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

Zurück zu Bestimmte Integrale

1.1

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

Beweis

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

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

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

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

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

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



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

Integriere nun beide Seiten nach $\alpha$:

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

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

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

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

1.2

${\int }_0^{\mathrm{\infty }}\frac{\log x}{\cosh x+\cos \pi \alpha }dx=\frac{\pi }{\sin \pi \alpha }\log ((2\pi {)}^{\alpha }\frac{\mathrm{\Gamma }\left(\frac{1+\alpha }{2}\right)}{\mathrm{\Gamma }\left(\frac{1-\alpha }{2}\right)})0<\alpha <1$

Beweis

In der Formel

${\int }_0^1\frac{\log \log \left(\frac{1}{x}\right)}{1+2\cos \alpha \pi \cdot x+x^2}dx=\frac{\pi }{2\sin \alpha \pi }(\alpha \log 2\pi +\log \frac{\mathrm{\Gamma }(\frac{1}{2}+\frac{\alpha }{2})}{\mathrm{\Gamma }(\frac{1}{2}-\frac{\alpha }{2})})0<\alpha <1$

substituiere $x\mapsto e^{-x}$.