Evaluating $\int_0^{+\infty}(\frac{\arctan t}{t})^2dt$

Question

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.

Answer 1

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

$$I(1,1) = \int_0^{\infty} dt \frac{\arctan^2{t}}{t^2} = \pi \log{2} $$

Answer 2

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,

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

Answer 3

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.

Answer 4

$J=\displaystyle\int_0^{\infty} \dfrac{dt}{(1+a^2 t^2)(1+b^2 t^2)}=\int_0^{\infty}\dfrac{b^2}{(b^2-a^2)(1+b^2t^2)}dt-\int_0^{\infty}\dfrac{a^2}{(b^2-a^2)(1+a^2t^2)}dt$

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, $\displaystyle J=\dfrac{\pi}{2}\left(\dfrac{b}{b^2-a^2}-\dfrac{a}{b^2-a^2}\right)=\dfrac{\pi}{2(a+b)}$