Formelsammlung Mathematik: Identitäten: Integralidentitäten nach Ramanujan

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1

${\sqrt {\alpha }}\int _{0}^{\infty }{\frac {e^{-x^{2}}}{\cosh \alpha x}}\,dx={\sqrt {\beta }}\int _{0}^{\infty }{\frac {e^{-x^{2}}}{\cosh \beta x}}\,dx\qquad \alpha \beta =\pi$

Beweis

Beim Integral $I:=\int _{-\infty }^{\infty }{\frac {e^{-x^{2}}}{\cosh \alpha x}}\,dx$ ersetze ${\frac {1}{\cosh \alpha x}}$ durch $\int _{-\infty }^{\infty }{\frac {\cos 2\alpha xt}{\cosh \pi t}}\,dt$.

$I=\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }{\frac {e^{-x^{2}}\cos 2\alpha xt}{\cosh \pi t}}\,dt\,dx$

Vertausche die Integrationsreihenfolge und ersetze $\int _{-\infty }^{\infty }e^{-x^{2}}\cos 2\alpha xt\,dx$ durch ${\sqrt {\pi }}\,e^{-\alpha ^{2}t^{2}}$.

$I={\sqrt {\pi }}\int _{-\infty }^{\infty }{\frac {e^{-\alpha ^{2}t^{2}}}{\cosh \pi t}}\,dt$

Nach Substitution $t={\frac {x}{\alpha }}$ ist $I={\sqrt {\pi }}\int _{-\infty }^{\infty }{\frac {e^{-x^{2}}}{\cosh {\frac {\pi x}{\alpha }}}}\,{\frac {dx}{\alpha }}$.

Ersetzt man $\pi \,$ durch $\alpha \beta \,$, so ist $I={\sqrt {\frac {\beta }{\alpha }}}\,\int _{-\infty }^{\infty }{\frac {e^{-x^{2}}}{\cosh \beta x}}\,dx$.

Daraus folgt unmittelbar die Behauptung.

 

2

Setzt man $u(z)=e^{i\,{\frac {p}{q}}\,\pi z^{2}}\,{\frac {e^{2\pi pz}}{2\sinh 2\pi pz}}$   ,   $g(z)={\frac {1}{\sinh ^{2}\pi z}}$   ,   $f(z)=u(z)\cdot g(z)$

und $S=\int _{-\infty }^{\infty }e^{i\,{\frac {p}{q}}\,\pi x^{2}}\,g(x)\,dx=2\pi i\sum _{k=0}^{4pq}{\text{res}}\left(f,i\cdot {\frac {k}{2p}}\right)-i\pi \,{\text{res}}(f,0)-i\pi \,{\text{res}}(f,i\cdot 2q)$, so gilt

$\int _{0}^{\infty }{\frac {8\,{\frac {p}{q}}\,x\cdot \cos {\frac {p}{q}}\,\pi x^{2}}{e^{2\pi x}-1}}\,dx=\int _{-\infty }^{\infty }{\frac {\sin {\frac {p}{q}}\,\pi x^{2}}{\sinh ^{2}\pi x}}\,dx={\text{Im}}(S)$

$\int _{0}^{\infty }{\frac {8\,{\frac {p}{q}}\,x\cdot \sin {\frac {p}{q}}\,\pi x^{2}}{e^{2\pi x}-1}}\,dx=\int _{-\infty }^{\infty }{\frac {1-\cos {\frac {p}{q}}\,\pi x^{2}}{\sinh ^{2}\pi x}}\,dx=-{\frac {2}{\pi }}-{\text{Re}}(S)$

ohne Beweis

 

3

$\int _{0}^{\infty }{\frac {z^{x}}{\Gamma (1+x)}}\,dx=e^{z}-\int _{0}^{\infty }{\frac {e^{-zx}}{x\,{\big (}\pi ^{2}+\log ^{2}x{\big )}}}\,dx\qquad {\text{Re}}(z)>0$

ohne Beweis