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Some rumours point out that the integral you see might be evaluated in a straightforward way.
But rumours are sometimes just rumours. Could you confirm/refute it?

$$ \int_0^{\infty}\left[\frac{\log\left(x\right)\arctan\left(x\right)}{x}\right]^{2} \,{\rm d}x $$

EDIT W|A tells the integral evaluates $0$ but this is not true. Then how do I exactly compute it?

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    $\begingroup$ The function is non-negative, positive and continuous (except perhaps at the origin). I don't think that the integral should be zero, even if that is what WA is indicating. Then again, if you change the integral from 0 to 5, for example, you get a positive number. Perhaps this is an error on WA? $\endgroup$ Commented Mar 24, 2013 at 21:08
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    $\begingroup$ Mathematica agrees that the definite integral should be zero (yet clearly it should not be, as Isaac's comment explains). Cool! I've seen this before - usually it is due to Wolfram's integration methods (complex functions are employed). Report it to Wolfram, sometimes you can get a free T-shirt out of it! $\endgroup$
    – icurays1
    Commented Mar 24, 2013 at 21:15
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    $\begingroup$ @user68326 I would follow the suggestion in this comment if I were you. $\endgroup$
    – Git Gud
    Commented Mar 24, 2013 at 21:15
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    $\begingroup$ I don't understand why everyone is upvoting this question. It was posed as a riddle, without any indication that the "rumours" refer to Wolfram|Alpha; if WChargin hadn't told us about the Wolfram|Alpha bug, the question would have made no sense whatsoever. I downvoted. $\endgroup$
    – joriki
    Commented Mar 24, 2013 at 21:16
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    $\begingroup$ @user68326 Regardless of how good at maths people here are, there's no need for personal attacks. I suggest you delete your offensive comments towards other users. On a side note: I sure hope people here are good at maths, because if they aren't, then I must be a very, very, very dumb person as I know close to nothing when compared to the average user here. $\endgroup$
    – Git Gud
    Commented Mar 24, 2013 at 21:36

4 Answers 4

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Related problems: (I), (II), (III). Denoting our integral by $J$ and recalling the mellin transform $$ F(s)=\int_{0}^{\infty}x^{s-1} f(x)\,dx \implies F''(s)=\int_{0}^{\infty}x^{s-1} \ln(x)^2\,f(x)\,dx.$$

Taking $f(x)=\arctan(x)^2$, then the mellin transform of $f(x)$ is

$$ \frac{1}{2}\,{\frac {\pi \, \left( \gamma+2\,\ln\left( 2 \right) +\psi \left( \frac{1}{2}+\frac{s}{2} \right)\right) }{s\sin \left( \frac{\pi \,s}{2} \right)}}-\frac{1}{2}\,{\frac {{\pi }^{2}}{s\cos\left( \frac{\pi \,s}{2} \right) }},$$

where $\psi(x)=\frac{d}{dx}\ln \Gamma(x)$ is the digamma function. Thus $J$ can be calculated directly as

$$ J= \lim_{s\to -1} F''(s) = \frac{1}{12}\,\pi \, \left( 3\,{\pi }^{2}\ln \left( 2 \right) -{\pi }^{2}+24 \,\ln \left( 2 \right) -3\,\zeta \left( 3 \right) \right)\sim 6.200200824 .$$

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    $\begingroup$ You seem like the best around here (I saw some of your answers). Why aren't you upvoted? $\endgroup$
    – user68326
    Commented Mar 24, 2013 at 22:18
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    $\begingroup$ Because there is much envy on MSE as I can see. What kind of people come here, really? $\endgroup$
    – user68326
    Commented Mar 24, 2013 at 22:19
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    $\begingroup$ @user68326: Thank you very much for your comment. I really appreciate it. I am happy that many people are benefited of my answers. $\endgroup$ Commented Mar 24, 2013 at 22:25
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    $\begingroup$ How did you get the Mellin transform of $\large\arctan\left(x\right)$ ?. $\endgroup$ Commented Jul 10, 2014 at 23:47
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    $\begingroup$ @MhenniBenghorbal Thanks a lot. $\endgroup$ Commented Jul 12, 2014 at 6:06
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The function is positive and continuous (except perhaps at the origin) with isolated zeros. The integral is therefore some nonzero, positive quantity.

That Wolfram | Alpha indicates otherwise is quite possibly an error. If you change the upper limit of the integral from $\infty$ to, say $5$, Wolfram | Alpha gives you a positive quantity, further indicating that there may be an error in the way the site is evaluating this integral.

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    $\begingroup$ Oh, well: yet another bug in WA...what else's new? Good answer. +1 $\endgroup$
    – DonAntonio
    Commented Mar 24, 2013 at 21:37
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x:\ {\large ?}}$

$$ \mbox{With the identity}\quad\int_{0}^{1}{\dd y \over x^{2}y^{2} + 1} ={\arctan\pars{x} \over x}\quad\mbox{we'll have} $$

\begin{align} &\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x =\int_{0}^{\infty}\ln^{2}\pars{x} \int_{0}^{1}{\dd y \over x^{2}y^{2} + 1}\int_{0}^{1}{\dd z \over x^{2}z^{2} + 1}\,\dd x \\[3mm]&=\int_{0}^{1}\int_{0}^{1} \int_{0}^{\infty} {\ln^{2}\pars{x} \over \pars{x^{2}y^{2} + 1}\pars{x^{2}z^{2} + 1}}\,\dd x \,\dd y\,\dd z \\[3mm]&=\int_{0}^{1}\int_{0}^{1}\braces{ {y^{-2}z^{-2} \over z^{-2} - y^{-2}} \bracks{\int_{0}^{\infty} {\ln^{2}\pars{x} \over x^{2} + y^{-2}}\,\dd x -\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + z^{-2}}\,\dd x}}\,\dd y\,\dd z\tag{1} \end{align}

However, with $\ds{a > 0}$: \begin{align} &\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + a^{-2}}\,\dd x =a\int_{0}^{\infty}{\ln^{2}\pars{x/a} \over x^{2} + 1}\,\dd x =a\int_{0}^{\infty}{\bracks{\ln\pars{x} - \ln\pars{a}}^{2} \over x^{2} + 1}\,\dd x \\[3mm]&=a\bracks{\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x -2\ln\pars{a}\ \overbrace{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x}^{\ds{=\ 0}}\ +\ \ln^{2}\pars{a}\int_{0}^{\infty}{\dd x \over x^{2} + 1}} \\[3mm]&={\pi^{3} \over 8}\,a + {\pi \over 2}\,a\ln^{2}\pars{a}\tag{2} \end{align} since $\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x = {\pi^{3} \over 8}}$ is a well known result.

Replacing $\pars{2}$ in $\pars{1}$: \begin{align} &\int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x \\[3mm]&=\int_{0}^{1}\int_{0}^{1}\braces{{1 \over y^{2} - z^{2}} \bracks{{\pi^{3} \over 8}\,\pars{y - z} + {\pi \over 2}\,y\ln^{2}\pars{y} -{\pi \over 2}\,z\ln^{2}\pars{z}}}\,\dd y\,\dd z \\[3mm]&={\pi^{3} \over 8}\int_{0}^{1}\int_{0}^{1}{\dd y\,\dd z \over y + z} +{\pi \over 2}\int_{0}^{1}\int_{0}^{1} {y\ln^{2}\pars{y} - z\ln^{2}\pars{z} \over y^{2} - z^{2}}\,\dd y\,\dd z \end{align} Both integrals can be trivially evaluated: \begin{align} \int_{0}^{1}\int_{0}^{1}{\dd y\,\dd z \over y + z} &=2\ln\pars{2} \\[3mm]\int_{0}^{1}\int_{0}^{1} {z\ln^{2}\pars{z} - y\ln^{2}\pars{y} \over y^{2} - z^{2}}\,\dd y\,\dd z&= 4\ln\pars{2} - {1 \over 6}\,\pi^{2} - \half\,\zeta\pars{3} \end{align}

Then, \begin{align}&\color{#66f}{\large \int_{0}^{\infty}\bracks{\ln\pars{x}\arctan\pars{x} \over x}^{2}\,\dd x} ={\pi^{3} \over 8}\,\bracks{2\ln\pars{2}} +{\pi \over 2}\bracks{4\ln\pars{2} - {1 \over 6}\,\pi^{2} - \half\,\zeta\pars{3}} \\[3mm]&=\color{#66f}{\large{1 \over 4}\bracks{\ln\pars{2} - {1 \over 3}}\pi^{3} + \bracks{2\ln\pars{2} - {1 \over 4}\,\zeta\pars{3}}\pi} \approx 6.200200822 \end{align}

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Using the fact that $\displaystyle \int_0^{1} \frac{1}{1+u^2x^2}\,du = \frac{\arctan x}{x}$,

\begin{align*}I = \int_0^{\infty} \frac{\arctan^2 x \log^2 x}{x^2}\,dx &= \int_0^{\infty}\int_0^{1}\int_0^{1} \frac{\log^2 x}{(1+u^2x^2)(1+v^2x^2)}\,du\,dv\,dx\\&= \int_0^{1}\int_0^{1} \frac{1}{u^2-v^2}\left(\int_0^{\infty}\frac{u^2\log^2 x}{1+u^2x^2} - \frac{v^2\log^2 x}{1+x^2v^2}\,dx\right)\,du\,dv\end{align*} Now, \begin{align*} f(u) = \int_0^{\infty} \frac{u^2\log^2 x}{1+u^2x^2}\,dx &= u\int_0^{\infty} \frac{\log^2 (x/u)}{1+x^2}\,dx \\&= u\int_0^{\infty} \frac{\log^2 x}{1+x^2}\,dx + \frac{\pi}{2}u\log^2 u \qquad (\text{since, } \int_0^{\infty} \frac{\log x}{1+x^2}\,dx = 0)\\&= \frac{\pi^3}{8}u + \frac{\pi}{2}u\log^2 u\end{align*} Thus, splitting the inner integral at $v$ and interchanging the order of integration while exploiting symmetry, \begin{align*}I &= \int_0^{1}\int_0^{1} \frac{f(u) - f(v)}{u^2-v^2}\,du\,dv\\&= \int_0^{1}\int_0^{v} \frac{f(u) - f(v)}{u^2-v^2}\,du\,dv + \int_0^{1}\int_v^{1} \frac{f(u) - f(v)}{u^2-v^2}\,du\,dv\\&= 2\int_0^{1}\int_0^{v} \frac{f(u) - f(v)}{u^2-v^2}\,du\,dv\\& \underbrace{=}_{u = tv} 2\int_0^1 \int_0^1 \frac{f(v) - f(tv)}{v^2(1-t^2)}\cdot v \,dt\,dv\\&= 2\int_0^1 \int_0^1 \frac{\frac{\pi^3}{8}(1-t)+\frac{\pi}{2}(\log^2v - t\log^2 (tv))}{1-t^2} \,dt\,dv\\&= \frac{\pi^3}{4}\log 2 + \pi\int_0^1 \int_0^1 \frac{\log^2v - t\log^2 (tv)}{1-t^2} \,dv\,dt\\&= \frac{\pi^3}{4}\log 2 + \pi\int_0^1 \frac{2(1-t) + 2t\log t - t\log^2 t}{1-t^2} \,dt\\&= \frac{\pi^3}{4}\log 2 + 2\pi\log 2 + \frac{\pi}{2}\int_0^1 \frac{\log t}{1-t}\,dt - \frac{\pi}{8}\int_0^1 \frac{\log^2 t}{1-t}\,dt \\&= \frac{\pi^3}{4}\log 2 + 2\pi\log 2 - \frac{\pi}{2}\zeta(2) - \frac{\pi}{4}\zeta(3)\end{align*}

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