Originally Posted by

**mr fantastic** There are many possible approaches. Here's a simple one:

Let $\displaystyle f(a) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx$.

Then $\displaystyle f'(a) = \frac{d}{da} \left[\int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx\right] = \int_{0}^{\infty} \frac{d}{da} \left[ \frac{\tan^{-1}(ax)}{x} \right] \, dx$

$\displaystyle = \int_{0}^{\infty} \frac{1}{a^2 x^2 + 1} \, dx = \frac{1}{a} \, \tan^{-1}(ax) \bigg{|}_{0}^{\infty} = \frac{\pi}{2a}$.

Therefore $\displaystyle f(a) = \int \frac{\pi}{2a} \, da = \frac{\pi}{2} \ln (a) + C$.

Therefore $\displaystyle f(a) - f(b) = \int_{0}^{\infty} \frac{\tan^{-1}(ax)}{x} \, dx - \int_{0}^{\infty} \frac{\tan^{-1}(bx)}{x} \, dx$

$\displaystyle = \left(\frac{\pi}{2} \ln (a) + C\right) - \left(\frac{\pi}{2} \ln (b) + C\right) = \frac{\pi}{2} \ln\left( \frac{a}{b}\right)$.