# Evaluate the Integral $$\int_{0}^{1} \tan^{-1}(\mathrm{e}^{x}) \mathrm{d}x$$

How to evaluate

I = \int\limits_{0}^{1} \tan^{-1}(\mathrm{e}^{x}) \mathrm{d}x

was a question posed at Mathematics Stack Exchange.

Here is my solution. We express the inverse tangent as logarithms:

\tan^{-1}(z) = \frac{i}{2} [\ln(1-iz) – \ln(1+iz)]

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:

1. $$\mathrm{G}$$ is Catalan’s constant.
2. Expressions for $$\mathrm{Li}_{2}(\pm i)$$ can be found here.