I've conjectured the following closed-form: $$ I = \int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx = -2\,G\,\ln2, $$ where $G$ is Catalan's constant. Numerically $$ I \approx -1.2697979381877088371491554851603117320986537271546606092465\dots$$ How to prove it?

  • $\begingroup$ How to prove it ? - Believe, and do not question. $\endgroup$ – Lucian Jul 31 '15 at 11:03
  • 1
    $\begingroup$ @Lucian I've long suspected that to be one of the central algorithms employed by Mathematica ;) $\endgroup$ – David H Jul 31 '15 at 13:22
  • $\begingroup$ Your numerical evaluation is incorrect. Here: wolframalpha.com/input/… Combining the results from Jack's answer and an answer from the linked problem, the closed form seems to be: $$\frac{\pi^3}{32}+\frac{\pi}{8}\ln^2 2 -4\Im\left[\text{Li}_3\left(\frac{1+i}{2}\right)\right]-\frac{5}{16}\zeta(3)$$ $\endgroup$ – Pranav Arora Aug 1 '15 at 13:22
  • $\begingroup$ @PranavArora Sorry. There was a typo in the problem. $\endgroup$ – user153012 Aug 1 '15 at 15:07

Apply the obvious substitution $x\mapsto\sin{x}$ to the first integral $$I=\int^\frac{\pi}{2}\limits_0\frac{x\ln(\sin{x})}{\sin{x}}\ {\rm d}x+\int\limits^1_0\frac{2\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x$$ The latter integral has been addressed here and is equivalent to \begin{align} \int\limits^1_0\frac{2\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x=4\Im{\rm Li}_3(1-i)-2\mathbf{G}\ln{2}+\frac{3\pi^3}{16}+\frac{\pi}{4}\ln^2{2} \end{align} As for the first integral, \begin{align} \int\limits^\frac{\pi}{2}_0\frac{x\ln(\sin{x})}{\sin{x}}\ {\rm d}x &=2\int\limits^1_0\frac{\arctan{x}\ln\left(\frac{2x}{1+x^2}\right)}{x}\ {\rm d}x\\ &=2{\rm Ti}_2(1)\ln{2}+2\int\limits^1_0\frac{\arctan{x}\ln{x}}{x}\ {\rm d}x-2\int\limits^1_0\frac{\arctan{x}\ln(1+x^2)}{x}\ {\rm d}x\\ &=2\mathbf{G}\ln{2}-2\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^3}+2\int\limits^1_0\frac{\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x+4\int\limits^1_0\frac{x\arctan{x}\ln{x}}{1+x^2}\ {\rm d}x \end{align} and the integral \begin{align} 4\int\limits^1_0\frac{x\arctan{x}\ln{x}}{1+x^2}\ {\rm d}x=-8\Im{\rm Li}_3(1-i)-\frac{5\pi^3}{16}-\frac{\pi}{2}\ln^2{2} \end{align} has also been established in the link above. (Both integrals were covered in the evaluation of $\mathscr{J}_2$.) Hence the closed form is indeed \begin{align} I=-2\mathbf{G}\ln{2} \end{align}


Quoting this famous question, we have that: $$ \sum_{n\geq 1}\frac{2^{2n} x^{2n}}{n^2\binom{2n}{n}}=2\,\arcsin^2(x)\tag{1}$$ hence: $$\begin{eqnarray*} \int\limits_{0}^{1}\frac{\arcsin(x)}{\sqrt{1-x^2}}\cdot\frac{\log x}{x}\,dx &=& -\frac{1}{4}\sum_{n\geq 1}\frac{4^n}{n(2n-1)^2\binom{2n}{n}}\\&=&-\frac{1}{2}\sum_{n\geq 1}\frac{4^n\,B(n,n)}{(2n)^2(2n-1)^2}\\&=&-2\int\limits_{0}^{1}\sum_{n\geq 1}\frac{\left(4x(1-x)\right)^{n-1}}{(2n)^2(2n-1)^2}\,dx.\tag{2}\end{eqnarray*}$$ On the other hand, $ \log^2(1-x) = \sum_{n\geq 1}\frac{2H_{n-1}}{n}x^n $ implies: $$ \log^2(1+x^2) = \sum_{n\geq 1}\frac{2(-1)^n H_{n-1}}{n}x^{2n}\tag{3} $$ hence: $$ \int\limits_{0}^{1}\frac{\log(1+x^2)\log(x)}{(1+x^2)x}\,dx = \frac{1}{4}\sum_{n\geq 1}\frac{(-1)^n H_{n-1}}{n^2}=-\frac{5}{32}\zeta(3)\tag{4}$$ and the problem boils down to evaluating $(2)$. We have:

$$\begin{eqnarray*} \sum_{n\geq 1}\frac{y^{n-1}}{(2n)^2(2n-1)^2}&=&\sum_{n\geq 1}\left(\frac{1}{(2n)^2}+\frac{1}{(2n-1)^2}+\frac{1}{n}-\frac{2}{2n-1}\right)y^{n-1}\tag{5}\\&=&\frac{\text{Li}_2(y)}{4y}-\frac{\log(1-y)}{y}-\frac{2\,\text{arctanh}(\sqrt{y})}{\sqrt{y}}+\frac{\text{Li}_2(\sqrt{y})-\text{Li}_2(-\sqrt{y})}{2\sqrt{y}}\end{eqnarray*}$$ and, according to Mathematica: $$ \int\limits_{0}^{1}\frac{\text{Li}_2(4x(1-x))}{x(1-x)}\,dx=2\pi^2\log 2-7\zeta(3), $$ $$ \int\limits_{0}^{1}\frac{\log(1-4x(1-x))}{x(1-x)}\,dx = -\pi^2,$$ $$ \int\limits_{0}^{\pi/2}\text{arctanh}(\sin \theta)\,d\theta = 2K,\tag{6} $$ and the last two integrals left to compute are: $$ I_+ = \int\limits_{0}^{\pi/2}\text{Li}_2(\sin\theta)\,d\theta, \qquad I_-=\int\limits_{0}^{\pi/2}\text{Li}_2(-\sin\theta)\,d\theta.$$ They can be computed through integration by parts, then by exploiting the Fourier series of $t\cot t$ and $\log(\sin t),\log(\cos t)$. For the latter, see this question.

Footnote: I still need to complete this answer by properly explaining the last part and proving the identities kindly provided by Mathematica. Since it is quite a lot to typeset, I leave this answer here even if not complete: maybe someone finds one or more shortcuts.

  • $\begingroup$ btw, substituting $x=\sin\theta$ in the second integral gives an integral exactly similar to the one asked in this problem: math.stackexchange.com/questions/1164183/… but it doesn't seem to have a nice closed form. $\endgroup$ – Pranav Arora Jul 31 '15 at 15:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.