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.

  • 1
    $\begingroup$ i have found $$-C+\frac{1}{2} i (\text{Li}_2(-i e)-\text{Li}_2(i e))$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 3 '16 at 16:16
  • 1
    $\begingroup$ C is the Catalan-constant and LI the polylog-function $\endgroup$ – Dr. Sonnhard Graubner Sep 3 '16 at 16:17
  • $\begingroup$ Yes I compute it with previous code, but I don't know how calculate this, if it is a well known identity, you can add only some hints: how one can start to explore and deduce the statement? Very thanks much @Dr.SonnhardGraubner $\endgroup$ – user243301 Sep 3 '16 at 17:55

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}


  1. $\mathrm{G}$ is Catalan's constant
  2. $\mathrm{Li}_{2}(z)$ is the dilogarithm.
  3. Expressions for $\mathrm{Li}_{2}(\pm i)$ can be found here.
  • $\begingroup$ Very thanks much for your answer. $\endgroup$ – user243301 Oct 17 '16 at 11:31

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