Prove $\int_0^\infty \frac{\ln \tan^2 (ax)}{1+x^2}\,dx = \pi\ln \tanh(a)$ $$
\mbox{How would I prove}\quad
\int_{0}^{\infty}
{\ln\left(\,\tan^{2}\left(\, ax\,\right)\,\right) \over 1 + x^{2}}\,{\rm d}x
=\pi
\ln\left(\,\tanh\left(\,\left\vert\, a\,\right\vert\,\right)\,\right)\,{\Large ?}.
\qquad a \in {\mathbb R}\verb*\* \left\{\,0\,\right\}
$$

Another approach:
$$ I(a) = \int_{0}^{+\infty}\frac{\log\tan^2(ax)}{1+x^2}\,dx$$
first use "lebniz integral differentiation"
and then do change of variable to calculate the integral.
Hope it helps.
