Formelsammlung Mathematik: Identitäten: Integralidentitäten nach Ramanujan

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

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

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

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

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

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