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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.

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  • $\begingroup$ I am sorry. I don't know how to format the expressions correctly. Thank you for your help. $\endgroup$ – und3rd06012 Dec 27 '15 at 17:30
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    $\begingroup$ This question is a bit similar. $\endgroup$ – mickep Dec 27 '15 at 21:07
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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,

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

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    $\begingroup$ That is actually very good! Thanks Olivier. $\endgroup$ – und3rd06012 Dec 27 '15 at 18:24
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    $\begingroup$ Solid answer. +1 - Mark $\endgroup$ – Mark Viola Dec 27 '15 at 18:38
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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

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

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  • $\begingroup$ Wow! I would have never thought of that.Thanks Ron. $\endgroup$ – und3rd06012 Dec 27 '15 at 18:16
  • $\begingroup$ @rongordon Happy Holidays! In the development, I believe you wanted to write $$\left.\frac{\partial I}{\partial a}\right|_{b=0}=0\implies f(a)=-\frac{\pi}{2}\log(a)$$and likewise for the partial of $I$ with respect to $b$ to find $g$. +1 for the answer - Mark $\endgroup$ – Mark Viola Dec 27 '15 at 18:37
  • $\begingroup$ @Dr.MV: yes, thanks. And HH to you as well. $\endgroup$ – Ron Gordon Dec 27 '15 at 18:38
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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.

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

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

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  • $\begingroup$ I am sorry I don't quite understand what you did. Is it possible for you to further explain please? $\endgroup$ – und3rd06012 Dec 29 '15 at 7:51
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    $\begingroup$ $\dfrac{1}{(1+a^2t^2)(1+b^2t^2)}=\dfrac{b^2}{(b^2-a^2)(1+b^2t^2)}-\dfrac{a^2}{(b^2-a^2)(1+a^2t^2)}$ $\endgroup$ – FDP Dec 29 '15 at 10:10
  • $\begingroup$ Oh ok. I completely missed the numerator in the integral. Thank you for the explanation. $\endgroup$ – und3rd06012 Dec 29 '15 at 13:54

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