I am attempting to solve the following definite integral $$\int_{-\infty}^{+\infty}\frac{\sin{(\cosh{x})}\cosh{x}}{1+\cosh^{2}{x}}dx.$$
I was trying passing to the complex plane and use the residue theorem, but I did not manage to find a satisfying contour. The half-circle seem to not work because the poles accumulate at infinity, while a rectangle does not because the integrand diverges on the vertical sides.
Which one is the right approach to tackle this problem?
I don't believe the integral has a closed form. The substitution $t=\cosh x$ gives \begin{align}I&=\int_{-\infty}^\infty\frac{\cosh x\sin\cosh x}{1+\cosh^2x}\,dx\\&=2\int_1^\infty\frac{t\sin t}{(t^2+1)\sqrt{t^2-1}}\,dt\\&=2\int_1^\infty\frac{t\sin t+\sinh1}{(t^2+1)\sqrt{t^2-1}}\,dt-2\int_1^\infty\frac{\sinh1}{(t^2+1)\sqrt{t^2-1}}\,dt\\&=2\int_1^\infty\int_0^1\frac{\cos ut\cosh(u-1)}{\sqrt{t^2-1}}\,du\,dt-\sqrt2\log(1+\sqrt2)\sinh1\\&=-\pi\int_0^1Y_0(u)\cosh(u-1)\,du-\sqrt2\log(1+\sqrt2)\sinh1\end{align} We therefore seek a closed form for $\int_0^1Y_0(u)e^u\,du$. However, as the upper limit of $1$ is not a special value of the Bessel function of the second kind, this amounts to a closed form for the indefinite integral which I doubt is known.