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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: Letting $x=\tan t$ which gives: $$I=\int_0^1 \frac{\arctan x}{x}\ln\left(\frac{1+x^2}{(1-x)^2}\right)dx=-\int_0^\frac{\pi}{4}\frac{t}{\sin t}\ln(1-\sin(2t))dt=$$ $$\overset{2t=x}=-\frac12 \underset{=J}{\int_0^\frac{\pi}{2}\frac{x}{\sin x} \ln(1-\sin x)dx}=-\frac12 \int_0^\frac{\pi}{2} x\ln(1-\sin x) \left(\ln\left(\tan \frac{x}{2}\right)\right)'dx=$$ $$\overset{IBP}=\frac12\int_0^\frac{\pi}{2} \ln\left(1-\sin x\right)\ln\left(\tan \frac{x}{2}\right)dx+\frac12 \int_0^\frac{\pi}{2} \frac{x\cos x}{\sin x-1}\ln\left(\tan \frac{x}{2}\right)dx$$ Or to employ Feynman's trick for the first integral $(J)$ in the second row. $$J(t)=\int_0^\frac{\pi}{2} \frac{x\ln(1-t\sin x)}{\sin x}dx\Rightarrow J'(t)=\int_0^\frac{\pi}{2} \frac{x}{1-t\sin x}dx$$ But even so I don't see a how to obtain a closed from for the last one. Also with a different parameter: $$J(t)=\int_0^\frac{\pi}{2} \frac{\text{arccot} (t \cot x)\ln(1-\sin x)}{\sin x}dx$$ $$\Rightarrow J'(t)=-\int_0^\frac{\pi}{2} \frac{\ln(1-\sin x)\cos x}{1+t^2 \cot^2x}\frac{dx}{\sin^2x}\overset{\sin x=y}=\int_0^1 \frac{\ln(1-y)}{1+t^2\left(1-\frac{1}{y^2}\right)}\frac{dy}{y^2}$$


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: $$I=\frac23 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx -2\int_0^1 \frac{\arctan x \ln(1-x)}{x}dx$$ $$=\frac23 \int_0^1 \int_0^1 \frac{\ln(1+x)-3\ln(1-x)}{1+x^2y^2}dydx=\frac23 \int_0^1 \int_t^1 \frac{\ln(1+x)-3\ln(1-x)}{1+t^2}dxdt $$


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 inside to obtain: $$I=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1 \frac{\arctan x}{x} \, \left(x^{2n+1}+x^{4n+2}\right)dx=\sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1\int_0^1 \frac{\left(x^{2n+1}+x^{4n+2}\right)}{1+y^2x^2}dydx$$


In the meantime I found one nice solution by Roberto Tauraso here.

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 solve it and submitted the answer to the AMM or know how to solve this integral kindly share the solution here?

Edit: Another impressive solution due to Yaghoub Sharifi is found 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 '18 at 1:48
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    $\begingroup$ I would say this solution here is quite impressive and convincing. $\endgroup$ – mrtaurho Dec 25 '18 at 15:38
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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 '18 at 17:59
  • $\begingroup$ Well done. (+1) $\endgroup$ – Mark Viola Dec 26 '18 at 4:26
  • $\begingroup$ Very nice solution and $\to +1$ $\endgroup$ – Claude Leibovici Dec 26 '18 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 '18 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 '18 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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