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The following integral was proposed by Cornel Ioan Valean and appeared as Problem $12054$ in the American Mathematical Monthly earlier this year.

Prove $$\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=\frac{\pi^3}{16}$$

I had small tries for it, such as writting:

$$I=\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx\overset{ x\to \tan \frac{x}{2}}=-\frac12 {\int_0^\frac{\pi}{2}\frac{x\ln(1-\sin x)}{\sin x} dx}$$

And with Feynman's trick we obtain: $$J(t)=\int_0^\frac{\pi}{2} \frac{x\ln(1-t\sin x)}{\sin x}dx\Rightarrow J'(t)=\int_0^\frac{\pi}{2} \frac{x}{1-t\sin x}dx$$ But I don't see a way to obtain a closed from for the above integral.


Also from here we have the following relation: $$\int_0^1 \frac{\arctan x \ln(1+x^2)}{x} dx =\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx$$ Thus we can rewrite the integral as: $$I=\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx -2\int_0^1 \frac{\arctan x \ln(1-x)}{x}dx$$


Another option might be to rewrite: $$\ln\left(\frac{1+x^2}{(1-x)^2}\right)= \ln\left(\frac{1+x}{1-x}\right)+\ln\left(\frac{1+x^2}{1-x^2}\right)$$ $$\Rightarrow I= \int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x}{1-x}\right)dx+\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{1-x^2}\right)dx$$ And now to use the power expansion of the log functions to obtain: $$\small I=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1 \frac{\arctan x}{x} \, \left(x^{2n+1}+x^{4n+2}\right)dx=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1\int_0^1 \frac{\left(x^{2n+1}+x^{4n+2}\right)}{1+y^2x^2}dydx$$


This seems like an awesome integral and I would like to learn more so I am searching for more approaches. Would any of you who also already solved it and submitted the answer to the AMM or know how to solve this integral kindly share the solution here?

Edit: In the meantime I found a nice solution by Roberto Tauraso here and another impressive approach due to Yaghoub Sharifi here.

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    $\begingroup$ I was able to break it down to an evaluation of harmonic sums $$I=\frac{3\pi^3}{32}-\sum_{n=0}^{\infty}\frac{\frac12\left[H_{n/2}-H_{(n-1)/2}\right]+\frac14\left[H_{n+1/4}-H_{n-1/4}\right]}{(2n+1)^2}$$ the latter sum should equal $\pi^3/32$ which seems to work out numerically but honestly speaking I am lost from hereon. Using the well-known result $\beta(3)=\pi^3/32$ one could conjecture that the combination of harmonic sums has to come out equal to $(-1)^n/(2n+1)$ in order to complete the representation of $\beta(3)$. $\endgroup$
    – mrtaurho
    Dec 24, 2018 at 1:48
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    $\begingroup$ I would say this solution here is quite impressive and convincing. $\endgroup$
    – mrtaurho
    Dec 25, 2018 at 15:38

5 Answers 5

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Another approach,

Perform integration by parts,

\begin{align*} I&=\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)\,dx\\ &=\Big[\ln (x) \ln\left(\frac{1+x^2}{(1-x)^2}\right)\arctan x\Big]_0^1 -\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-\int_0^1 \frac{2(1+x)\ln (x)\arctan (x)}{(1-x)(1+x^2)}dx\\ &=-\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-2\int_0^1 \frac{(1+x)\ln (x)\arctan (x)}{(1-x)(1+x^2)}dx\\ \end{align*}

For $x\in [0;1]$ define the function $R$ by,

\begin{align*} R(x)=\int_0^x \frac{(1+t)\ln t}{(1-t)(1+t^2)}dt=\int_0^1 \frac{x(1+tx)\ln (tx)}{(1-tx)(1+t^2x^2)}dt\\ \end{align*}

Observe that,

\begin{align*} R(1)=\int_0^1 \frac{t\ln t}{1+t}dt+\int_0^1 \frac{\ln t}{1-t}dt \end{align*} Perform integration by parts,

\begin{align*} I&=-\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-2\Big[R(x)\arctan x\Big]_0^1+2\int_0^1\frac{R(x)}{1+x^2}dx\\ &=-\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-\frac{\pi}{2}R(1)+2\int_0^1 \int_0^1 \frac{x(1+tx)\ln (tx)}{(1-tx)(1+t^2x^2)(1+x^2)}dtdx\\ &=-\int_0^1 \frac{\ln x}{1+x^2}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx-\frac{\pi}{2}R(1)+\int_0^1 \ln x\left[\frac{1}{1+x^2}\ln\left(\frac{1+t^2x^2}{(1-tx)^2}\right)\right]_{t=0}^{t=1} dx+\\ &\int_0^1 \ln t\left[\frac{1}{1+t^2}\ln\left(\frac{1+x^2}{(1-tx)^2}\right)+\frac{2\arctan (tx)}{1-t^2}-\frac{2t\arctan x}{1+t^2}-\frac{2t\arctan x}{1-t^2}\right]_{x=0}^{x=1} dt\\ &=-\frac{\pi }{2}R(1)+\ln 2\int_0^1 \frac{\ln t}{1+t^2}dt-2\int_0^1 \frac{\ln (1-t)\ln t}{1+t^2}dt+2\int_0^1 \frac{\ln t\arctan t}{1-t^2}dt-\\ &\frac{\pi}{2} \int_0^1 \frac{t\ln t}{1+t^2}dt-\frac{\pi}{2} \int_0^1\frac{t\ln t}{1-t^2} dt\\ \end{align*}

For $x\in [0;1]$ define the function $S$ by,

\begin{align*} S(x)=\int_0^x \frac{\ln t}{1-t^2}dt=\int_0^1 \frac{x\ln(tx)}{1-t^2x^2} dt \end{align*}

Perform integration by parts,

\begin{align*} \int_0^1 \frac{\ln x\arctan x}{1-x^2}dx&=\Big[S(x)\arctan x\Big]_0^1-\int_0^1 \frac{S(x)}{1+x^2}dx\\ &=\frac{\pi}{4}S(1)-\int_0^1 \int_0^1 \frac{x\ln(tx)} {(1-t^2x^2)(1+x^2)} dtdx\\ &=\frac{\pi}{4}S(1)-\frac{1}{2}\int_0^1 \left[ \frac{\ln x}{1+x^2}\ln\left(\frac{1+tx}{1-tx} \right)\right]_{t=0}^{t=1} dx-\\ &\frac{1}{2}\int_0^1 \left[ \frac{\ln t}{1+t^2}\ln\left(\frac{1+x^2}{1-t^2x^2} \right)\right]_{x=0}^{x=1}dt\\ &=\frac{\pi}{4}S(1)-\frac{\ln 2}{2}\int_0^1 \frac{\ln t}{1+t^2}dt+\int_0^1 \frac{\ln(1-x)\ln x}{1+x^2}dx \end{align*}

Therefore,

\begin{align*}I&=\pi\int_0^1\frac{2t\ln t}{t^4-1} dt\end{align*}

Perform the change of variable $y=t^2$,

\begin{align*}I&=\frac{1}{2}\pi \int_0^1 \frac{\ln y}{y^2-1}dy\\ &=\frac{1}{2}\pi\times \frac{3}{4}\zeta(2)\\ &=\frac{\pi^3}{16} \end{align*}

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    $\begingroup$ That's impressive, thank you! I've seen you use this approach alot and it's quite useful, let me a few time to understand it's working better. $\endgroup$
    – Zacky
    Dec 25, 2018 at 17:59
  • $\begingroup$ Well done. (+1) $\endgroup$
    – Mark Viola
    Dec 26, 2018 at 4:26
  • $\begingroup$ Very nice solution and $\to +1$ $\endgroup$ Dec 26, 2018 at 6:07
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    $\begingroup$ I compute $\int_0^1 F(t,x)\ln t\,dx$ and $\int_0^1 F(t,x)\ln x\,dt$ and one can compute an antiderivative $U(t,x)$ of $F(t,x)$ wrt $x$, and on the other hand an antiderivative $V(t,x)$ of $F(t,x)$ wrt $t$. $\endgroup$
    – FDP
    Dec 26, 2018 at 17:07
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    $\begingroup$ @Zacky: remember in the double integrals you can choose to integrate wrt $x$ or wrt $t$. If there is a factor $\ln x$ you don't want to integrate wrt $x$ first. If there is a factor $\ln t$ you don't want to integrate wrt $t$ first. And, $\ln(tx)=\ln x +\ln t$ $\endgroup$
    – FDP
    Dec 26, 2018 at 17:25
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Put \begin{equation*} I=\int_{0}^1\dfrac{\arctan x}{x}\ln\left(\dfrac{1+x^2}{(1-x)^2}\right)\, \mathrm{d}x. \end{equation*} Via the substitution $ x=\dfrac{z}{z+1}$ we get \begin{equation*} I = \int_{0}^{\infty}\dfrac{\arctan \frac{z}{z+1}\ln(2z^2+2z+1)}{z^2+z}\, \mathrm{d}z. \end{equation*} Put \begin{equation*} \log z=\ln|z|+i\arg z, \quad -\pi<\arg z <\pi. \end{equation*} Then \begin{equation*} \arctan \frac{z}{z+1}\ln(2z^2+2z+1) = \text{Im}\left(\log^2(1+z+iz)\right). \end{equation*} Consequently \begin{equation*} I = \text{Im}\left(\int_{0}^{\infty}\dfrac{\log^2(1+z+iz)}{z^2+z}\right)\mathrm{d}z. \end{equation*} However, $ \log(z) $ is an analytic function in $ \text{Re} z>0 $. According to Cauchys integral theorem we get the same value if we integrate along the curve with the parametrization $ z=(1-i)s, s>0 $. \begin{gather*} I = \text{Im}\left(\int_{0}^{\infty}\dfrac{\ln^2(2s+1)}{s(s+1-is)}\, \mathrm{d}s\right) = \int_{0}^{\infty}\dfrac{\ln^2(2s+1)}{2s^2+2s+1}\, \mathrm{d}s = \\[2ex] \int_{0}^{\infty}\dfrac{2\ln^2(2s+1)}{(2s+1)^2+1}\, \mathrm{d}s = [t=2s+1] = \\[2ex] \int_{1}^{\infty}\dfrac{\ln^2(t)}{t^2+1}\, \mathrm{d}t =[u= 1/t] = \int_{0}^{1}\dfrac{\ln^2(u)}{u^2+1}\, \mathrm{d}u. \end{gather*} Thus \begin{equation*} 2I = \int_{0}^{\infty}\dfrac{\ln^2(u)}{u^2+1}\, \mathrm{d}u \end{equation*} In order to evaluate this integral we integrate $ \dfrac{\log^3(z)}{z^2+1} $ along a keyhole contour and use residue calculus. In this case $ \log z =\ln |z|+i\arg z, \quad 0<\arg z < 2\pi $. We get \begin{equation*} I = \dfrac{\pi^3}{16}. \end{equation*}

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Starting with breaking the integral

$\displaystyle I=\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)\ dx=\int_0^1\frac{\arctan x}{x}\ln(1+x^2)dx-2\int_0^1\frac{\arctan x}{x}\ln(1-x)dx$

then using the identity$\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}$ for the first integral and series-expanding $\displaystyle\arctan x$ of the second integral, we get \begin{align*} I&=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{2n+1}\int_0^1x^{2n}\ dx-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\int_0^1x^{2n}\ln(1-x)\ dx\\ &=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{(2n+1)^2}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(-\frac{H_{2n+1}}{2n+1}\right)\\ &=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}}{(2n+1)^2}-2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}\left(-\frac{H_{2n}}{2n+1}-\frac{1}{(2n+1)^2}\right)\\ &=2\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^3}=2\beta(3)=\frac{\pi^3}{16} \end{align*}

where $\beta(3)=\frac{\pi^3}{32}$ is the Dirichlet beta function.

Note that we used the classical result $\int_0^1 x^{n-1}\ln(1-x)dx=-\frac{H_n}{n}$ which can be proved as follows:

$$\int_0^1 x^{n-1}\ln(1-x)dx=-\sum_{k=1}^\infty\frac1k\int_0^1 x^{n+k-1}dx=-\sum_{k=1}^\infty\frac{1}{k(n+k)}\\=-\frac1n\sum_{k=1}^\infty\left(\frac1k-\frac1{n+k}\right)=-\frac1n\sum_{k=1}^n\frac1k=-\frac{H_n}{n}$$

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  • $\begingroup$ (+1) The best answer in my opinion. $\endgroup$ Sep 9, 2021 at 3:43
  • $\begingroup$ @Jack D'Aurizio thank you Jack. $\endgroup$ Sep 9, 2021 at 8:12
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I continue your second try with the method FDP provide

$$\begin{aligned} I & = \frac2{3} \left( \int_{0}^{1} {\frac{\arctan x \ln(1+x)}{x} \mathrm{d}x} - 3\int_{0}^{1} {\frac{\arctan x \ln(1-x)}{x} \mathrm{d}x} \right)\\ & = \frac2{3} \left( -\int_{0}^{1} {\frac{\arctan x}{x} \left( \ln\frac{1-x}{1+x} \right) \mathrm{d}x} - 2\int_{0}^{1} {\frac{\arctan x \ln(1-x)}{x} \mathrm{d}x} \right) \end{aligned}$$

let $y=\tfrac{1-x}{1+x}$ in first integral and notice that $\arctan\tfrac{1-y}{1+y} + \arctan y = \tfrac{\pi}{4}$

$$\begin{aligned} \int_{0}^{1} {\frac{\arctan x}{x} \left( \ln\frac{1-x}{1+x} \right) \mathrm{d}x} & = 2\int_{0}^{1} {\frac{\arctan \tfrac{1-y}{1+y} \ln y}{1-y^2} \mathrm{d}y}\\ & = \frac{\pi}{2} \int_{0}^{1} {\frac{\ln y}{1-y^2} \mathrm{d}y} - 2\int_{0}^{1} {\frac{\arctan y \ln y}{1-y^2} \mathrm{d}y} \end{aligned}$$

second can be integrate by parts

$$\ \int_{0}^{1} {\frac{\arctan x \ln(1-x)}{x} \mathrm{d}x} = -\int_{0}^{1} {\frac{\ln x \ln(1-x)}{1+x^2} \mathrm{d}x} + 2\int_{0}^{1} {\frac{\arctan x \ln x}{1-x^2} \mathrm{d}x} - \int_{0}^{1} {\frac{\arctan x \ln x}{1+x} \mathrm{d}x}$$

using the method as FDP done, set

$$\begin{aligned} P(x) & = \int_{0}^{x} {\frac{\ln u}{1-u^2} \mathrm{d}u} = \int_{0}^{1} {\frac{x\ln tx}{1-t^2x^2} \mathrm{d}t}\\ Q(x) & = \int_{0}^{x} {\frac{\ln u}{1+u} \mathrm{d}u} = \int_{0}^{1} {\frac{x\ln tx}{1+tx} \mathrm{d}t} \end{aligned}$$

deduce

$$\int_{0}^{1} {\frac{\arctan x \ln x}{1-x^2} \mathrm{d}x} = \frac{\pi}{4}\int_{0}^{1} {\frac{\ln u}{1-u^2} \mathrm{d}u} - \frac{\ln2}{2}\int_{0}^{1} {\frac{\ln t}{1+t^2} \mathrm{d}t} + \int_{0}^{1} {\frac{\ln t \ln(1-t)}{1+t^2} \mathrm{d}t}$$

and (this part is a same question like here)

$$\begin{aligned} \int_{0}^{1} {\frac{\arctan x \ln x}{1+x} \mathrm{d}x} = &\> \arctan x \cdot Q(x) \big|_{x=0}^{1} - \int_{0}^{1} {\frac{Q(x)}{1+x^2} \mathrm{d}x}\\ = &\> \frac{\pi}{4}\int_{0}^{1} {\frac{\ln u}{1+u} \mathrm{d}u} - \int_{0}^{1} {\int_{0}^{1} {\frac{x\ln tx}{(1+x^2)(1+tx)} \mathrm{d}t} \mathrm{d}x}\\ = &\> \frac{\pi}{4}\int_{0}^{1} {\frac{\ln u}{1+u} \mathrm{d}u} - \int_{0}^{1} {\frac{\ln t}{1+t^2} \left( \frac1{2} \ln\frac{1+x^2}{(1+tx)^2} + t\arctan x \right)\biggr|_{x=0}^{1} \mathrm{d}t}\\ & - \int_{0}^{1} {\frac{\ln x}{1+x^2} \ln(1+tx) \biggr|_{t=0}^{1} \mathrm{d}x}\\ = &\> \frac{\pi}{4}\int_{0}^{1} {\frac{\ln u}{1+u} \mathrm{d}u} - \frac{\ln2}{2}\int_{0}^{1} {\frac{\ln t}{1+t^2} \mathrm{d}t} + \int_{0}^{1} {\frac{\ln t \ln(1+t)}{1+t^2} \mathrm{d}t}\\ & - \frac{\pi}{4}\int_{0}^{1} {\frac{t\ln t}{1+t^2} \mathrm{d}t} - \int_{0}^{1} {\frac{\ln x \ln(1+x)}{1+x^2} \mathrm{d}x}\\ = &\> \frac{3\pi}{16}\int_{0}^{1} {\frac{\ln u}{1+u} \mathrm{d}u} - \frac{\ln2}{2}\int_{0}^{1} {\frac{\ln t}{1+t^2} \mathrm{d}t} \end{aligned}$$

thus

$$\begin{aligned} I & = -\frac{\pi}{3} \int_{0}^{1} {\frac{\ln y}{1-y^2} \mathrm{d}y} + \frac4{3} \left( -\frac{\pi}{4}\int_{0}^{1} {\frac{\ln u}{1-u^2} \mathrm{d}u} + \frac{3\pi}{16}\int_{0}^{1} {\frac{\ln u}{1+u} \mathrm{d}u} \right)\\ & = -\frac{2\pi}{3} \int_{0}^{1} {\frac{\ln u}{1-u^2} \mathrm{d}u} + \frac{\pi}{4} \int_{0}^{1} {\frac{\ln u}{1+u} \mathrm{d}u} = \frac{\pi^3}{16} \end{aligned}$$

this may be a simplified version of FDP's first answer.

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We have $\log\left(\frac{1+x^2}{(1-x)^2}\right)=\log(1-x^4)-\log(1-x^2)-2\log(1-x)$, hence by integration by parts

$$ I_1 = \frac{3\pi^3}{32}-2\int_{0}^{1}\frac{1}{1+x^2}\sum_{n\geq 1}\frac{\chi(n) x^n}{n^2}\,dx\qquad \chi(n)=\left\{\begin{array}{rcl}1&\text{if}&n\equiv 1\pmod{2}\\ 2 & \text{if} & n\equiv 2\pmod{4}\\ 0 &\text{if}&n\equiv 0\pmod{4}\end{array}\right. $$ where $$\int_{0}^{1}\frac{1}{1+x^2}\sum_{n\geq 1}\frac{\chi(n) x^n}{n^2}\,dx=\iint_{(0,1)^2}\frac{-\log(y)}{y(1+x^2)}\sum_{n\geq 1}\chi(n)(xy)^n\,dx\,dy=\iint_{(0,1)^2}\frac{-\log(y)}{y(1+x^2)}\cdot\frac{(1+xy)^2 xy}{1-(xy)^4}\,dx\,dy$$ equals $$ \iint_{(0,1)^2}\frac{-\log(y)x(1+xy)}{(1+x^2)(1-x y)(1+x^2 y^2)}\,dx\,dy. $$ By performing a partial fraction decomposition this integral is reduced to four integrals in the $y$-variable, with two of them (namely $\int_{0}^{1}\frac{y\log y}{1-y^4}\,dy = -\frac{\pi^3}{32}$ and $\int_{0}^{1}\frac{\log(2)}{1+y^2}\,dy = \frac{\pi}{4}\log(2)$) being elementary and the remaining ones being $$ J_1 = \int_{0}^{1}\frac{\arctan(y)\log(y)}{1-y^2}\,dy,\qquad J_2=\int_{0}^{1}\frac{\log(y)\log(1-y)}{1+y^2}\,dy. $$ $J_2$ can be tackled by performing the substitution $y=\tan\theta$ and exploiting the Fourier series of $\log \sin$ and $\log\cos$. By performing the substitution $y\mapsto\frac{1+y}{1-y}$ $J_1$ is reduced to $$ \int_{0}^{1}\frac{\arctan(x)\operatorname{arctanh}(x)}{x}\,dx $$ then, by computing $\int_{0}^{1}\frac{x^{2k}}{2k+1}\operatorname{arctanh}(x)\,dx$, to the series (also appearing here) $$ \sum_{k\geq 0}\frac{(-1)^k H_k}{(2k+1)^2}=\int_{0}^{1}\frac{\log(1+z^2)\log(z)}{1+z^2}\,dx. $$ Via $z\to\tan\theta$ we have that both $J_1$ and $J_2$ are reduced to the integrals $\int_{0}^{\pi/4}\log^2(\sin\theta)\,d\theta$ and $\int_{0}^{\pi/4}\log(\sin\theta)\log(\cos\theta)\,d\theta$, which are well-known and related to Euler sums with weight $3$. Luckily the contributions related to $\pi\log^2(2),\pi^2\log(2),K\log(2)$ and $\text{Im}\,\text{Li}_3\left(\frac{1+i}{2}\right)$ cancel out and only leave a rational multiple of $\pi^3$.

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