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Gradshteyn&Ryzhik $3.514.4$ states that $$\int_0^{\infty } \frac{\sinh (a x) \sinh (b x)}{(\cosh (a x)+\cos (t))^2} \, dx=\frac{\pi b \csc (t) \csc \left(\frac{\pi b}{a}\right) \sin \left(\frac{b t}{a}\right)}{a^2}$$ Whenever $0<\left| b\right| <a,0<t<\pi$. My bet is on Feynman's trick or contour integration but haven't figure out the exact way. Any help will be appreciated!

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Integration by parts (just to simplify the evaluation) gives $$\int_0^\infty\frac{\sinh ax\sinh bx}{(\cosh ax+\cos t)^2}\,dx=\frac{b}{a}\int_0^\infty\frac{\cosh bx\,dx}{\cosh ax+\cos t}.$$ This can be evaluated using the residue theorem. Consider $I_R=\displaystyle\int_{C_R}\frac{e^{bz}\,dz}{\cosh az-\cos t}$ where $C_R$ is the rectangular contour with vertices at $\pm R\pm i\pi/a$; the integrand has simple poles at $z=\pm it/a$, thus $$I_R=2\pi i\left(\operatorname*{Res}_{z=it/a}+\operatorname*{Res}_{z=-it/a}\right)\frac{e^{bz}}{\cosh az-\cos t}=\frac{4\pi i}{a}\frac{\sin(bt/a)}{\sin t},$$ and on the other hand, $\lim\limits_{R\to\infty}I_R$ is equal to $$\int_{-\infty}^\infty\frac{e^{b(x-i\pi/a)}\,dx}{-\cosh ax-\cos t}+\int_\infty^{-\infty}\frac{e^{b(x+i\pi/a)}\,dx}{-\cosh ax-\cos t}=4i\sin\frac{b\pi}{a}\int_0^\infty\frac{\cosh bx\,dx}{\cosh ax+\cos t}.$$

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