How would one evaluate the integral $$\int_0^\infty \frac{\arctan(x) }{x(1+x^2)}\,dx$$?
I was told it had a nice closed form and could have been solved with differentiation under the integral sign; however, I tried to set $$I(\alpha) = \int_0^\infty \frac{\arctan(\alpha x)}{x(1+x^2)}~dx$$ and got nowhere (the resulting integral was very messy). Is there a much more clever substitution that can be used to tackle the integral?
Best Answer
Rewrite
$$\frac{\arctan{x}}{x} = \int_0^1 \frac{du}{1+x^2 u^2} $$
Then plugging this in and reversing the order of integration, we get the integral; value as
$$\begin{align}\int_0^1 \frac{du}{u^2} \, \int_0^{\infty} dx \, \left (\frac1{\frac1{u^2}+x^2} \frac1{1+x^2} \right ) &= \int_0^1 \frac{du}{1-u^2} \, \int_0^{\infty} dx \left ( \frac1{1+x^2}-\frac1{\frac1{u^2}+x^2} \right )\\ &= \int_0^1 \frac{du}{1-u^2} \, \frac{\pi}{2} (1-u) \\ &= \frac{\pi}{2} \log{2}\end{align}$$