5
$\begingroup$

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.

Thanks in advance!

EDIT

Out of pure chance I just stumbled upon a related MSE question dealing with the integral

$$\int_0^\infty\frac{x\sin^2x}{\cosh x+\cos x}\mathrm dx=1$$

Which on the other hand motivates me to believe that there may be a closed-form for $(1)$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – GEdgar Apr 5 at 21:18
  • $\begingroup$ 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 $G-\frac12$. $\endgroup$ – Zacky Apr 26 at 9:22
  • 1
    $\begingroup$ @Zacky I am aware of this list but has not thought about searching within for the given integral. As you already mentioned this list is unreliable but cotains some "good" approximations but way to many completely wrong values. $\endgroup$ – mrtaurho Apr 26 at 10:08
  • $\begingroup$ @GEdgar I have problems with applying the residue theorem for complicated integrals correctly. Are you here able to do so? $\endgroup$ – mrtaurho Apr 26 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.