The Gudermannian function is related to the exponential function, see for example the MathWorld's article Gudermannian. From this idea I was playing with the integral $$\int_0^\infty e^{-nx}\operatorname{gd}(x)\mathrm dx$$ where $n\geq 1$ are integers, when by summation over all these $n$, I wondered about the integral $$\int_0^\infty\frac{\operatorname{gd}(x)}{e^x-1}\mathrm dx.$$
Using Wolfram Alpha online calculator I know the indefinite integral $\int e^{-nx}\operatorname{gd}(x)\mathrm dx$, and also approximations (see the code, or similars, 10 digits of int gd(x)/(e^x-1)dx, from x=0 to infinity
) for which searching with Wolfram Alpha online calculator for a closed-form of the mentioned approximations, I got the following conjecture.
Claim(?). Seems that $$\int_0^\infty \frac{\operatorname{gd}(x)}{e^x-1}\mathrm dx=2K-\frac{\pi \log 2}{4},\tag{C}$$ being $K$ the Catalan's constant.
Question. Is it possible to get a proof of previous conjecture $(\text{C})$? Many thanks.