real analysis - What about of this integral involving the Gudermannian function $int_0^infty frac{operatorname{gd}(x)}{e^x-1}mathrm dx$?

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.

Answer

We first notice that

$$\begin{align*} \int_{0}^{\infty}\frac{\operatorname{gd}(x)}{e^x-1}\,dx &= \int_{0}^{\infty} \frac{1}{e^x-1} \left( \int_{0}^{x} \frac{dy}{\cosh y} \right) \, dx \\ &= \int_{0}^{\infty} \frac{1}{\cosh y} \left( \int_{y}^{\infty} \frac{dx}{e^x - 1} \right) \,d y \\ &= -2 \int_{0}^{\infty} \frac{\log(1 - e^{-y})}{e^y + e^{-y}} \, dy \\ &=-2\int_{0}^{\frac{\pi}{4}}\log(1-\tan\theta)\,d\theta \tag{$e^{-x}=\tan\theta$}. \end{align*}$$

The last integral is our starting point. We introduce two tricks to evaluate this.

Step 1. Notice that $\tan(\frac{\pi}{4}-\theta)=\frac{1-\tan\theta}{1+\tan\theta}$. So by the substitution $\theta \mapsto \frac{\pi}{4}-\theta$, it follows that

$$\int_{0}^{\frac{\pi}{4}}\log(1+\tan\theta)\,d\theta = \int_{0}^{\frac{\pi}{4}}\log\left(\frac{2}{1+\tan\theta}\right)\,d\theta$$

and hence both integrals have the common value $\frac{\pi}{8}\log 2$. Applying the same idea to our integral, it then follows that

$$\begin{align*} -2\int_{0}^{\frac{\pi}{4}}\log(1-\tan\theta)\,d\theta &= -2\int_{0}^{\frac{\pi}{4}}\log\left(\frac{2\tan\theta}{1+\tan\theta}\right)\,d\theta \\ &= -2\int_{0}^{\frac{\pi}{4}}\log\tan\theta \, d\theta - \frac{\pi}{4}\log 2. \end{align*}$$

Step 2. In order to compute the last integral, we notice that for $\theta\in\mathbb{R}$ with $\cos\theta\neq0$, we have

$$\begin{align*} -\log\left|\tan\theta\right| &= \log\left|\frac{1+e^{2i\theta}}{1-e^{2i\theta}}\right| = \operatorname{Re} \log\left(\frac{1+e^{2i\theta}}{1-e^{2i\theta}}\right) \\ &= \operatorname{Re}\left( \sum_{n=1}^{\infty} \frac{1+(-1)^n}{n} e^{2in\theta} \right) \\ &= \sum_{k=0}^{\infty} \frac{2}{2k+1}\cos(4k+2)\theta. \end{align*}$$

So by term-wise integration, we obtain

$$\begin{align*} -2\int_{0}^{\frac{\pi}{4}}\log\tan\theta \, d\theta &= \sum_{k=0}^{\infty} \frac{4}{2k+1} \int_{0}^{\frac{\pi}{4}} \cos(4k+2)\theta \, d\theta \\ &= 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)^2} = 2K, \end{align*}$$

where $K$ is the Catalan's constant.