Evaluating $\int_0^{+\infty}(\frac{\arctan t}{t})^2dt$ Is it possible to calculate $$\int_0^{+\infty}\Big(\frac{\arctan t}{t}\Big)^2 dt$$ without using complex analysis? 
I found this on a calculus I book and I don't know how to solve it. 
I tried to set $t = \tan u$ but it didn't help. 

The answer can be derived from a tailor-made version of Parseval's theorem, as already suggested by Ron Gordon. The Fourier sine series of $x$ over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ is given by:
$$ x = \sum_{k\geq 1}\frac{(-1)^{k+1}}{k}\,\sin(2k x) \tag{1} $$
and if $k,j\in\mathbb{N}^+$ we have:
$$ \int_{0}^{\pi/2}\frac{\sin(2kt)}{\sin(t)}\cdot\frac{\sin(2jt)}{\sin(t)}\,dt = \pi\cdot\min(j,k).\tag{2} $$
That gives:
$$ \int_{0}^{\pi/2}\frac{x^2}{\sin^2 x}\,dx = \pi \sum_{j,k\geq 1}\frac{(-1)^{j+k}\min(j,k)}{jk}\tag{3} $$
and by reindexing over $j+k$, then using partial fraction decomposition, it is not difficult to check that the RHS of $(3)$ equals $\color{red}{\pi\log 2}$. On the other hand, the LHS of $(3)$ equals $\int_{0}^{+\infty}\frac{\arctan^2 t}{t^2}\,dt$ through the obvious substitution.
