# Evaluating a Polynomic-Trigonometric-Hyperbolic Integral

Within this AoPS thread it is asked to evaluate the following integral

$$\mathfrak I~=~\int_0^\infty \frac{x\sin x}{\cos x+\cosh^2 x}\mathrm dx\tag1$$

In order to be precise there is also a possible closed-form conjectured which is given by

$$\mathfrak I~=~G-\frac12\tag2$$

But as it is pointed out within the linked thread this seems to be only a reasonable approximation off after the $$5$$th decimal digit.

I have to admit that it is highly improbable that there exists a nice looking closed-form for $$(1)$$ since the integrand involves polynomials, trigonometric aswell as hyperbolic functions. I am not even sure how to get started, i.e. which substitution to choose or which technique at all to start with.

A related, but perhaps more handable integral, would be the following

$$\mathfrak J~=~\int_0^\infty \frac{\sin x}{\cos x+\cosh^2 x}\mathrm dx\tag{1'}$$

Out of experience I could imagine that $$(1')$$ may has a closed-form in terms of known constants $$($$or series$$)$$ since it only contains the two closely connected trigonometric and hyperbolic functions.

Is it in fact possible to deduce a closed-form for $$(1)$$ and $$(1')$$? For myself I cannot offer an approach since everything I tried was not helpful at all hence I was not even able to perform one or two steps in order to simplify the given integrals. I would be glad to see a full solution or even attempts in evaluating $$(1)$$ and $$(1')$$ since I have no idea how to deal with such integrands.

$$\int_0^\infty\frac{x\sin^2x}{\cosh x+\cos x}\mathrm dx=1$$
Which on the other hand motivates me to believe that there may be a closed-form for $$(1)$$.
• Note. Since the integrand is even, we ave $$2 \mathfrak I~=~\int_{-\infty}^\infty \frac{x\sin x}{\cos x+\cosh^2 x}\mathrm dx$$ and there is a chance this could be done using a contour in the complex plane. – GEdgar Apr 5 at 21:18
• The integral also appears within this list: en.wikiversity.org/wiki/User:Integrals123 (I typed in ctrl + F cosh and scrolled down till the $54$-th integral), which unfortunately has many wrong answers since it was done with Inverse Symbolic calculator and that result kinda played us. But yes a closed form would be interesting even though it's not $G-\frac12$. – Zacky Apr 26 at 9:22