Formelsammlung Mathematik: Identitäten: Integralidentitäten nach Ramanujan

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1

$\sqrt{\alpha }\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x^2}}{\cosh \alpha x}dx=\sqrt{\beta }\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x^2}}{\cosh \beta x}dx\alpha \beta =\pi$

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

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

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

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3

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{z^x}{\mathrm{\Gamma }(1+x)}dx=e^z-\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-zx}}{x({\pi }^2+{\log }^{2}x)}dx\text{Re}(z)>0$