There are evaluations of the above integral using complex analytic techniques, but is there a real variables way to show it? I've tried the "Feynman differentiation under the integral trick" but it seems to give more complicated integrals.
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- 3$\begingroup$ Have you tried using this: $$\begin{align} \int_{-\infty}^\infty \frac{\sinh(ax)}{\sinh(\pi x)}\cos(bx)\,dx&=4\int_0^\infty \frac{\sinh(ax)}{e^{\pi x}-e^{-\pi x}}\cos(bx)\,dx\\\\ &=4\int_0^\infty \frac{e^{-\pi x}\sinh(ax)}{(1-e^{-2\pi x})}\cos(bx)\,dx\\\\ &=4\sum_{n=0}^\infty \int_0^\infty e^{-(2n+1)\pi x}\sinh(ax)\cos(bx)\,dx\\\\ \end{align}$$I'm not sure if it leads to a tractable solution, but it is worth a try $\endgroup$Mark Viola– Mark Viola2025-11-20 17:10:33 +00:00Commented Nov 20 at 17:10
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For $|\Re(\alpha)|<\pi,$ \begin{align} f(\alpha)&=\int_{-\infty}^{\infty}\frac{\sinh(\alpha x)}{\sinh(\pi x)}\mathrm{d}x,\\ &= 2\int_0^{\infty} \frac{e^{-\pi x}(e^{\alpha x}-e^{-\alpha x})}{1-e^{-2\pi x}}\mathrm{d}x,\\ &= \frac{1}{\pi}\int_0^1 \frac{t^{1/2-1}(t^{-\alpha/2\pi}-t^{\alpha/2\pi})}{1-t}\mathrm{d}t,\tag{1}\\ &= \frac{1}{\pi}\int_0^1 \frac{(1-t^{1/2+\alpha/2\pi-1})-(1-t^{1/2-\alpha/2\pi-1})}{1-t}\mathrm{d}t,\\ &= \frac{1}{\pi}\left(\psi\left(\frac{1}{2}+\frac{\alpha}{2\pi}\right)-\psi\left(\frac{1}{2}-\frac{\alpha}{2\pi}\right)\right),\tag{2}\\ &= \cot\left(\frac{\pi}{2}-\frac{\alpha}{2}\right),\tag{3}\\ &= \tan\left(\frac{\alpha}{2}\right),\end{align} where we have used\begin{align*}&(1): t=e^{-2\pi x},\\ &(2): \psi(z)+\gamma=\int_0^1 \frac{1-t^{z-1}}{1-t}\mathrm{d}t,\qquad (\text{integral representation for the digamma function})\\ &(3): \psi(1-z)-\psi(z)=\pi\cot(\pi z).\qquad (\text{digamma reflection formula})\end{align*} Now since $\sinh(ax)\cos(bx)=\Re \sinh(a+bi)x$, given that $|a|<\pi$, the original integral is given by \begin{align} I &= \int_{-\infty}^{\infty} \frac{\sinh(ax)\cos(bx)}{\sinh(\pi x)}\mathrm{d}x,\\ &= \Re f(a+bi),\\ &= \Re \tan\left(\frac{a+bi}{2}\right),\\ &= \Re \frac{\tan(a/2)+i\tanh(b/2)}{1-i\tan(a/2)\tanh(b/2)}\cdot \frac{1+i\tan(a/2)\tanh(b/2)}{1+i\tan(a/2)\tanh(b/2)},\\ &= \frac{\tan(a/2)\text{sech}^2(b/2)}{1+\tan^2(a/2)\tanh^2(b/2)}\cdot \frac{2\cos^2(a/2)\cosh^2(b/2)}{2\cos^2(a/2)\cosh^2(b/2)},\\ &= \frac{2\sin(a/2)\cos(a/2)}{\cos^2(a/2)\cdot2\cosh^2(b/2)+\sin^2(a/2)\cdot 2\sinh^2(b/2)},\\ &= \frac{\sin a}{\cos^2(a/2)(\cosh b+1)+\sin^2(a/2)(\cosh b-1)},\\ &= \frac{\sin a}{(\cos^2(a/2)-\sin^2(a/2))+(\cos^2(a/2)+\sin^2(a/2))\cosh b},\\ &= \frac{\sin a}{\cos a+\cosh b}. \end{align}
