Calculations concerning the Gudermannian function I did calculations to get some simple expressions involving the Gudermannian function $$\text{gd}(x)=\int_0^x\frac{dt}{\cosh t}.$$
I don't sure if these statements are rights, but I know how do the calculations, thus my mistake should be a little detail.

Question 1. One has $$\lim_{x\to\infty}e^{-x}\int_0^xe^t\text{gd}(t)dt=\frac{\pi}{2},$$ and defining $$F(x)=\int_0^xe^t\text{gd}(t)dt,$$ then $$F''(x)-F'(x)=e^x(\text{gd}(x))'.$$
Are rights these claims? Only is required a yes, or where was my mistake.

And for the following, I know that there is a relationship satisfied by the Gudermannian function and the inverse tangent function (see in previous link to Wikipedia), then when I was exploring the calculation of the following integral with Wolfram Alpha $$\int_0^1\left(\frac{\pi}{2}+\text{gd}(x)\right)dx$$ I've asked to me

Question 2. Can you provide us hints to calculate $$\int_0^1 \arctan(e^x)dx?$$

Here is the code for previous online calculator

int arctan(e^x) dx, from x=0 to x=1
int arctan(e^x) dx

Many thanks.
 A: Here is a solution to question 2. To evaluate
\begin{equation}
I = \int\limits_{0}^{1} \tan^{-1}(\mathrm{e}^{x}) \mathrm{d}x
\end{equation}
we begin by expressing the inverse tangent as logarithms:
\begin{equation}
\tan^{-1}(z) = \frac{i}{2} [\ln(1-iz) - \ln(1+iz)]
\end{equation}
thus our integral becomes
\begin{align}
I &= \frac{i}{2} \int\limits_{0}^{1} \left(\ln(1-i\mathrm{e}^{x}) - \ln(1+i\mathrm{e}^{x}) \right) \mathrm{d}x \\
&= \frac{i}{2} (I_{1} + I_{2})
\end{align}
To evaluate $I_{1}$ we make the substitution $z = i\mathrm{e}^{x}$
\begin{align}
I_{1} &= \int\limits_{0}^{1} \ln(1-ie^{x}) \mathrm{d}x \\
&= \int\limits_{i}^{i\mathrm{e}} \frac{\ln(1-z)}{z} \mathrm{d}z \\
&= \int\limits_{0}^{i\mathrm{e}} \frac{\ln(1-z)}{z} \mathrm{d}z - \int\limits_{0}^{i} \frac{\ln(1-z)}{z} \mathrm{d}z\\
&= \mathrm{Li}_{2}(i) - \mathrm{Li}_{2}(i\mathrm{e}) \\
&= i\mathrm{G} - \frac{\pi^{2}}{48} - \mathrm{Li}_{2}(i\mathrm{e})
\end{align}
To evaluate $I_{2}$ we make the substitution $-z = i\mathrm{e}^{x}$
\begin{align}
I_{2} &= \int\limits_{0}^{1} \ln(1+ie^{x}) \mathrm{d}x \\
&= \int\limits_{-i}^{-i\mathrm{e}} \frac{\ln(1-z)}{z} \mathrm{d}z \\
&= \int\limits_{0}^{-i\mathrm{e}} \frac{\ln(1-z)}{z} \mathrm{d}z - \int\limits_{0}^{-i} \frac{\ln(1-z)}{z} \mathrm{d}z\\
&= \mathrm{Li}_{2}(-i\mathrm{e}) - \mathrm{Li}_{2}(-i) \\
&= \mathrm{Li}_{2}(-i\mathrm{e}) - \left(-i\mathrm{G} - \frac{\pi^{2}}{48} \right)
\end{align}
Putting the pieces together, we obtain our final result
\begin{align}
I &= \frac{i}{2} (I_{1} + I_{2}) \\
&= \frac{i}{2} [\mathrm{Li}_{2}(-i\mathrm{e}) - \mathrm{Li}_{2}(i\mathrm{e})] - \mathrm{G} \\
&= \int\limits_{0}^{1} \tan^{-1}(\mathrm{e}^{x}) \mathrm{d}x
\end{align}
Notes:


*

*$\mathrm{G}$ is Catalan's constant

*$\mathrm{Li}_{2}(z)$ is the dilogarithm.

*Expressions for $\mathrm{Li}_{2}(\pm i)$ can be found here.

