Evaluating $\int_{-\infty}^{+\infty}\frac{\sin{(\cosh{x})}\cosh{x}}{1+\cosh^{2}{x}}dx$

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 suspect that the equivalent form: $$2\int_1^\infty\frac{t\sin t}{1+t^2}\frac{1}{\sqrt{t-1}\sqrt{t+1}}\,\mathrm{d}t$$Is amenable to a dogbone method May 29 at 21:44
• The integral itself is Feynman trickable. From there a rectangular contour is easier. May 29 at 23:50
• @FShrike can you please give me a hint about the correct contour? i can't figure it out myself, each one I try has some problem. Thanks May 30 at 14:30
• Yeah, I haven't so far been able to find a direct contour method for this integral. I'll keep trying when I get time. So far, I have only been able to use contour methods to deduce seemingly useless equalities such as: $$\frac{1}{2}J=\int_0^1\frac{x\cos x}{(1+x^2)\sqrt{1-x^2}}\,\mathrm{d}x-\operatorname{P.V.}\,\int_0^\infty\frac{x\exp(-x)}{(x^2-1)\sqrt{1+x^2}}\,\mathrm{d}x$$Where $J$ is the desired integral May 30 at 14:44
• Now if you had $\cos(\cosh(x))$ in the numerator I think I could get it. But the sine is somehow annoying to deal with May 30 at 14:48

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.