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

Zurück zu Bestimmte Integrale

0.1

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{1+x^2}\frac{x}{\sinh \pi x}dx=2\log 2-1$

ohne Beweis (Abels Integral)

1.1

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^{n-1}}{\sinh x}dx=\frac{2^n(2^n-1)|B_n|}{n}\frac{{\pi }^n}{2^{n-1}}n\in {\mathbb{Z}}^{\ge 2}$

ohne Beweis

1.2

${\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{\sinh x}dx=2\mathrm{\Gamma }(\alpha )\lambda (\alpha )\text{Re}(\alpha )>1$

Beweis

Für $x>0$ ist $\frac{1}{\sinh x}=\frac{2}{e^x-e^{-x}}=2e^{-x}\frac{1}{1-e^{-2x}}=2e^{-x}\sum _{k=0}^{\mathrm{\infty }}{e}^{-2kx}=2\sum _{k=0}^{\mathrm{\infty }}{e}^{-(2k+1)x}$.

Also ist $\frac{x^{\alpha -1}}{\sinh x}=2\sum _{k=0}^{\mathrm{\infty }}{x}^{\alpha -1}{e}^{-(2k+1)x}$ und somit ist

${\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{\sinh x}dx=2\sum _{k=0}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}{e}^{-(2k+1)x}dx=2\sum _{k=0}^{\mathrm{\infty }}\frac{\mathrm{\Gamma }(\alpha )}{(2k+1{)}^{\alpha }}=2\mathrm{\Gamma }(\alpha )\lambda (\alpha )$.

1.3

${\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{{\sinh }^{2}x}dx=\frac{\mathrm{\Gamma }(\alpha )\zeta (\alpha -1)}{2^{\alpha -2}}\text{Re}(\alpha )>2$

Beweis

Für $x>0$ ist $\frac{1}{{\sinh }^{2}x}=\frac{4}{(e^x-e^{-x}{)}^2}=\frac{4e^{-2x}}{(1-e^{-2x}{)}^2}=4\sum _{k=1}^{\mathrm{\infty }}k{e}^{-2kx}$.

Also ist ${\int }_0^{\mathrm{\infty }}\frac{x^{\alpha -1}}{{\sinh }^{2}x}dx=4\sum _{k=1}^{\mathrm{\infty }}k{\int }_0^{\mathrm{\infty }}{x}^{\alpha -1}{e}^{-2kx}dx=4\sum _{k=1}^{\mathrm{\infty }}k\frac{\mathrm{\Gamma }(\alpha )}{(2k{)}^{\alpha }}=\frac{\mathrm{\Gamma }(\alpha )\zeta (\alpha -1)}{2^{\alpha -2}}$.

1.4

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sinh \alpha x}{\sinh x}dx=\pi \tan \left(\frac{\alpha \pi }{2}\right)-1<\mathrm{R}\mathrm{e}(\alpha )<1$

Beweis

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sinh \alpha x}{\sinh x}dx$   $=2{\int }_0^{\mathrm{\infty }}\frac{e^{\alpha x}-e^{-\alpha x}}{e^x-e^{-x}}dx=\psi \left(\frac{1+\alpha }{2}\right)-\psi \left(\frac{1-\alpha }{2}\right)$   $=\pi \tan \left(\frac{\alpha \pi }{2}\right)$

1.5

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sinh \alpha x}{\sinh \pi x}\frac{1}{1+x^2}dx=-\alpha \cos \alpha +2\sin \alpha \log (2\cos \frac{\alpha }{2})-\pi <\mathrm{R}\mathrm{e}(\alpha )<\pi$

Beweis

$y(\alpha ):={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sinh \alpha x}{\sinh \pi x}\frac{1}{1+x^2}dx$

$\Rightarrow y^{″}(\alpha )+y(\alpha )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sinh \alpha x}{\sinh \pi x}dx=\tan \frac{\alpha }{2}$

mit $y(0)=0$ und $y^{\prime }(0)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sinh \pi x}\frac{x}{1+x^2}dx=-1+2\log 2$.


Ansatz (Variation der Konstante):

$y(x)=c(x)\sin x+d(x)\cos x$

$y^{\prime }(x)=c(x)\cos x-d(x)\sin x+\underset{=0\text{(Forderung)}}{\underbrace{c^{\prime }(x)\sin x+d^{\prime }(x)\cos x}}$

$y^{″}(x)=-c(x)\sin x-d(x)\cos x+c^{\prime }(x)\cos x-d^{\prime }(x)\sin x$

Also ist $y^{″}(x)+y(x)=c^{\prime }(x)\cos x-d^{\prime }(x)\sin x=\tan \frac{x}{2}$ und $c^{\prime }(x)\sin x+d^{\prime }(x)\cos x=0$.

$\left(\begin{array}{cc}\cos x& -\sin x\\ \sin x& \cos x\end{array}\right)\left(\begin{array}{c}{c}^{\prime }(x)\\ d^{\prime }(x)\end{array}\right)=\left(\begin{array}{c}\tan \frac{x}{2}\\ 0\end{array}\right)$

$\Rightarrow \left(\begin{array}{c}{c}^{\prime }(x)\\ d^{\prime }(x)\end{array}\right)=\left(\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right)\left(\begin{array}{c}\tan \frac{x}{2}\\ 0\end{array}\right)=\left(\begin{array}{c}\cos x\tan \frac{x}{2}\\ -\sin x\tan \frac{x}{2}\end{array}\right)$

$\Rightarrow \left(\begin{array}{c}c(x)\\ d(x)\end{array}\right)=\left(\begin{array}{c}-\cos x+2\log (2\cos \frac{x}{2})\\ \sin x-x\end{array}\right)$, wegen $\left(\begin{array}{c}c(0)\\ d(0)\end{array}\right)=\left(\begin{array}{c}{y}^{\prime }(0)\\ y(0)\end{array}\right)=\left(\begin{array}{c}-1+2\log 2\\ 0\end{array}\right)$.

Somit ist $y(x)=[-\cos x+2\log (2\cos \frac{x}{2})]\sin x+(\sin x-x)\cos x=-x\cos x+2\sin x\log (2\cos \frac{x}{2})$.

2.1

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x}{\sinh \alpha \pi x}\frac{1}{x^2+{\beta }^2}dx=-\frac{1}{\alpha \beta }+\psi (\frac{\alpha \beta }{2}+\frac{1}{2})-\psi \left(\frac{\alpha \beta }{2}\right)\text{Re}(\alpha ),\text{Re}(\beta )>0$

ohne Beweis