Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,exp,sin,cot,coth)

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1.1

${\int }_0^{\mathrm{\infty }}\sin nx(\cot x+\coth x)e^{-nx}dx=\frac{\pi }{2}\{\begin{array}{ccc}\tanh \frac{n\pi }{2}& ,& n\text{ungerade}\\ \\ \coth \frac{n\pi }{2}& ,& n\text{gerade}\end{array}n\in {\mathbb{Z}}^{>0}$

Beweis (eine Formel nach Ramanujan)

Zum einen ist ${\int }_0^{\mathrm{\infty }}{e}^{-nx}\sin (nx)\cot xdx={\int }_0^{\mathrm{\infty }}{e}^{-nx}\frac{\sin nx}{\sin x}\cos xdx$

$=\underset{N\to \mathrm{\infty }}{lim}{\int }_0^{\mathrm{\infty }}{e}^{-nx}\frac{\sin nx}{\sin x}(\cos x-\cos (2N+1)x)dx\text{(Riemannsches Lemma)}$

$=2\cdot \underset{N\to \mathrm{\infty }}{lim}{\int }_0^{\mathrm{\infty }}{e}^{-nx}\frac{\sin nx}{\sin x}\sin (N+1)x\cdot \sin Nx\cdot dx=2\cdot \underset{N\to \mathrm{\infty }}{lim}{\int }_0^{\mathrm{\infty }}{e}^{-nx}\sin nx\sum _{k=1}^N\sin (2kx)dx$

$=2\cdot \underset{N\to \mathrm{\infty }}{lim}\sum _{k=1}^N{\int }_0^{\mathrm{\infty }}{e}^{-nx}\sin nx\cdot \sin (2kx)dx=\sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-nx}(\cos (n-2k)x-\cos (n+2k)x)dx$

$=\sum _{k=1}^{\mathrm{\infty }}\frac{n}{n^2+(n-2k{)}^2}-\sum _{k=1}^{\mathrm{\infty }}\frac{n}{n^2+(n+2k{)}^2}$.



Zum anderen ist ${\int }_0^{\mathrm{\infty }}{e}^{-nx}\sin (nx)\coth xdx=\frac{1}{2n}+2\sum _{k=1}^{\mathrm{\infty }}\frac{n}{n^2+(n+2k{)}^2}$

${\int }_0^{\mathrm{\infty }}{e}^{-nx}\sin (nx)\coth xdx={\int }_0^{\mathrm{\infty }}{e}^{-nx}\sin nx\cdot (1+\frac{2}{e^{2x}-1})dx$

$=\frac{n}{n^2+n^2}+2{\int }_0^{\mathrm{\infty }}\frac{e^{-nx}\sin nx}{e^{2x}-1}dx=\frac{1}{2n}+2\sum _{k=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(n+2k)x}\sin nxdx=\frac{1}{2n}+2\sum _{k=1}^{\mathrm{\infty }}\frac{n}{n^2+(n+2k{)}^2}$



Faßt man beide Teile zusammen, so ist${\int }_0^{\mathrm{\infty }}{e}^{-nx}\sin (nx)(\cot x+\coth x)dx$

$=\frac{1}{2n}+\sum _{k=1}^{\mathrm{\infty }}\frac{n}{n^2+(n-2k{)}^2}+\sum _{k=1}^{\mathrm{\infty }}\frac{n}{n^2+(n+2k{)}^2}=\frac{\pi }{2}\frac{\sinh \pi n}{\cosh \pi n-\cos \pi n}=\frac{\pi }{2}\{\begin{array}{ccc}\tanh \frac{n\pi }{2}& ,& n\text{ungerade}\\ \\ \coth \frac{n\pi }{2}& ,& n\text{gerade}\end{array}$.