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. 
 A: Consider
$$I(a,b) = \int_0^{\infty} dt \frac{\arctan{a t}}{t} \frac{\arctan{b t}}{t}  $$
Then
$$\begin{align}\frac{\partial^2 I}{\partial a \, \partial b} &= \int_0^{\infty}  \frac{dt}{(1+a^2 t^2)(1+b^2 t^2)}\\ &= \frac{\pi}{2 (a+b)} \end{align}$$
This result may be derived using Parseval's theorem from Fourier transforms.
$$\implies \frac{\partial I}{\partial a} = \frac{\pi}{2}\log{(a+b)} + f(a)$$
$$\frac{\partial I}{\partial b} = \frac{\pi}{2}\log{(a+b)} + g(b)$$
$$I_a(a,0)=0 \implies f(a)=-\frac{\pi}{2} \log{a} $$
$$I_b(0,b)=0 \implies g(b)=-\frac{\pi}{2} \log{b} $$
Then
$$I(a,b) = \frac{\pi}{2} \left [(a+b) \log{(a+b)} - (a+b) - a \log{a} + a \right ] + F(b)$$
but 
$$I(a,b) = \frac{\pi}{2} \left [(a+b) \log{(a+b)} - (a+b) - b \log{b} + b \right ] + G(a)$$
Accordingly, $F(b) = -\frac{\pi}{2}(b \log{b}-b)$ and $G(a) = -\frac{\pi}{2}(a \log{a}-a)$ and therefore
$$I(a,b) = \frac{\pi}{2} \left [(a+b) \log{(a+b)} - a \log{a}  - b \log{b}  \right ] $$
and the integral we want is
$$I(1,1) = \int_0^{\infty} dt \frac{\arctan^2{t}}{t^2} = \pi \log{2} $$
A: One may first integrate by parts,
$$
\begin{align}
\int_0^{+\infty}\!\! \left(\frac{\arctan t}{t}\right)^2\!\!dt=\color{#365A9E}{\left.-\frac1{t}\left(\arctan t\right)^2\right|_0^{+\infty}}\!+2\!\!\int_0^{+\infty} \!\frac{\arctan t}{t(1+t^2)}dt=\color{#365A9E}{0}+2\!\!\int_0^{+\infty}\! \frac{\arctan t}{t(1+t^2)}dt
\end{align}
$$ then making the change of variable $x=\arctan t$, $dx=\dfrac{dt}{1+t^2}$, in the latter integral gives
$$
\begin{align}
2\!\int_0^{+\infty}\! \frac{\arctan t}{t(1+t^2)}dt=2\!\int_0^{\pi/2} \!\!x\:\frac{\cos x}{\sin x}dx=\left.2x\log(\sin x)\right|_0^{\pi/2}-2\!\int_0^{\pi/2} \!\log(\sin x)dx=\pi\log 2
\end{align}
$$  where we have used the classic result:$\color{#365A9E}{\displaystyle \int_0^{\pi/2} \!\log(\sin x)\:dx=\!\int_0^{\pi/2} \!\log(\cos x)\:dx=-\frac{\pi}2\log 2.}$
Finally,

$$
\int_0^{+\infty}\! \left(\frac{\arctan t}{t}\right)^2dt=\pi\log 2.
$$

A: 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.
A: $J=\displaystyle\int_0^{\infty}  \dfrac{dt}{(1+a^2 t^2)(1+b^2 t^2)}=\int_0^{\infty}\dfrac{b^2}{(b^2-a^2)(1+b^2t^2)}dt-\int_0^{\infty}\dfrac{a^2}{(b^2-a^2)(1+a^2t^2)}dt$
If $a>0$, one obtains (change of variable $y=at$):
$\displaystyle \int_0^{\infty}\dfrac{a}{1+a^2t^2}dx=\dfrac{\pi}{2}$
Thus, $\displaystyle J=\dfrac{\pi}{2}\left(\dfrac{b}{b^2-a^2}-\dfrac{a}{b^2-a^2}\right)=\dfrac{\pi}{2(a+b)}$
