let $s$ be a complex parameter. We have the integral :
$$\int_{0}^{\pi/2}\tan^{-1}\left[\frac{\tan(sx)}{\tanh(s\log[\sec(x)])} \right ]\frac{\sec^{2}(x)}{e^{2\pi \tan(x)}-1}dx$$
This is a 'simplified' version of the rather unsettling integral: $$\int_{0}^{\infty}\tan^{-1}\left[ \frac{ \tan\left( s\tan^{-1}(x)\right )}{\tanh\left(s\log\left(\sqrt{1+x^{2}} \right )\right)}\right ]\frac{dx}{e^{2\pi x}-1}$$
Any insight on how to do the integral is highly appreciated
EDIT1:
I am especially interested in the integral on the imaginary axis ($s=it, \Im(t)=0)$
EDIT2
It turns out that i have made an innocent mistake. The integral doesn't converge, and has to be modified : $$\int_{0}^{\pi/2}\left(\tan^{-1}\left[\frac{\tan(sx)}{\tanh(s\log[\sec(x)])} \right ]-\tan^{-1}\left[\frac{x}{\log\sec(x)} \right ]\right)\frac{\sec^{2}(x)}{e^{2\pi \tan(x)}-1}dx$$
$$\int_{0}^{\infty}\left(\tan^{-1}\left[ \frac{ \tan\left( s\tan^{-1}(x)\right )}{\tanh\left(s\log\left(\sqrt{1+x^{2}} \right )\right)}\right ]-\tan^{-1}\left[\frac{2\tan^{-1}(x)}{\log(1+x^{2}))} \right ]\right)\frac{dx}{e^{2\pi x}-1}$$