An integral with log. and arctan

Prove that:

$$\int_{0}^{1}\ln (1+x^2) \arctan x \, {\rm d}x = \ln 2 - \frac{\pi}{2} - \frac{1}{4}\ln^2 2 + \frac{\pi^2}{16} + \frac{\pi \ln 2}{4}$$

Solution

We are needing a basic result from complex analysis.

Lemma: It holds that:

$$\arctan = \mathfrak{Im} \left [ \ln (1+ix) \right ]$$

Proof:We simply note that:

$$\ln (1+ix) = \ln \left | 1+ix \right | +i \arg \left ( 1+ix \right )= \frac{1}{2} \ln (1+x^2) +i \arctan x$$

Hence:

 \begin{align*}
\int_{0}^{1}\ln (1+x^2) \arctan x \, {\rm d}x &=\mathfrak{Im} \left ( \int_{0}^{1} \ln^2 (1+ix)\, {\rm d}x \right ) \\
 &\overset{y=1+ix}{=\! =\! =\! =\! =\!} \;  \mathfrak{Im} \left [ -i \int_{1}^{1+i} \ln^2 y \, {\rm d}y \right ]\\
 &= \mathfrak{Im} \left [ -i \left ( 2y - 2y\ln y +y \ln^2 y \right ) \right ]_1^{1+i}\\
 &= \mathfrak{Im} \left [ \left ( 2 -\ln 2 - \frac{\pi}{2}+\frac{\pi \ln 2}{4}+ \frac{\ln^2 2}{4} - \frac{\pi^2}{16}\right ) + \\
  i \left ( \ln 2 - \frac{\pi}{2} - \frac{1}{4}\ln^2 2 + \frac{\pi^2}{16} + \frac{\pi \ln 2}{4} \right ) \right ]\\
 &= \ln 2 - \frac{\pi}{2} - \frac{1}{4}\ln^2 2 + \frac{\pi^2}{16} + \frac{\pi \ln 2}{4}
\end{align*}