Find this integral


My try: since $$\dfrac{1}{1-xy}=\sum_{n=0}^{\infty}(xy)^n$$ so $$I=\sum_{n=0}^{\infty}\int_{0}^{1}y^n\int_{0}^{1}x^n\ln{(1+xy)}dx$$ because $$\int_{0}^{1}x^n\ln{(1+xy)}dx=\dfrac{x^{n+1}\ln{(1+xy)}}{n+1}|_{0}^{1}-\dfrac{1}{n+1}\int_{0}^{1}\dfrac{x^{n+1}y}{1+xy}dx=\dfrac{\ln{(1+y)}}{n+1}-I_{1}$$ where $$I_{1}=y\int_{0}^{1}\dfrac{x^{n+1}}{1+xy}dx$$ even if $I_{1}$ can use the beta function, But then follow I can't it.Thank you


You have

$$ \int_{0}^{1}\!\int_{0}^{1}\dfrac{\ln{(1+xy)}}{1-xy}\mathrm dx \: \mathrm dy = \frac{\pi^2}{4}\ln 2 -\zeta(3). \tag1 $$

To obtain $(1)$ one may write $$ \begin{align} I=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln{(1+xy)}}{1-xy}dxdy &= \sum_{n=1}^{\infty}\int_{0}^{1}\int_{0}^{1} \frac{(-1)^{n-1}}{n}\dfrac{(xy)^n}{1-xy}dxdy \\ & = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_{0}^{1}\int_{0}^{1} \dfrac{(xy)^n}{1-xy}dxdy \\ & = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\Phi(1,2,n+1)\\ & = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(\sum_{k=1}^{\infty}\frac{1}{k^2} -\sum_{k=1}^{n}\frac{1}{k^2} \right)\\ & =\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(\frac{\pi^2}{6} \quad -\sum_{k=1}^{n}\frac{1}{k^2}\,\right) \\ & = \frac{\pi^2}{4}\ln 2 -\zeta(3) \end{align} $$ where we have used a result on double integrals from J. Guillera and J. Sondow (30): $$ \int_{0}^{1}\int_{0}^{1} \dfrac{(xy)^{u-1}}{1-xyz}(-\ln(xy))^s dxdy =\Gamma(s+2)\Phi(z,s+2,u) $$ $\displaystyle \Phi$ denoting the Lerch transcendent function: $$ \Phi(z,s,u)= \sum_{k=0}^{\infty}\frac{z^k}{(k+u)^{s}}.$$

Update: a proof of the last step may be found here.

  • $\begingroup$ edit the summation over $n$. The sums start at $n=1$ $\endgroup$ – Leucippus Aug 18 '14 at 16:47
  • $\begingroup$ @Leucippus Edited. Thanks! $\endgroup$ – Olivier Oloa Aug 18 '14 at 16:54
  • $\begingroup$ @OlivierOloa Mister Oloa, how did you compute $\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(\frac{\pi^2}{6} \quad -\sum_{k=1}^{n}\frac{1}{k^2}\,\right)$? It seems to me this is an important step that was skipped. $\endgroup$ – user 1357113 Sep 7 '14 at 9:18
  • $\begingroup$ @Chris'ssis You are right, I had used a special value of a standard integral I knew. Please, see the edit. Thanks! $\endgroup$ – Olivier Oloa Sep 7 '14 at 16:32

Though I'd actually prefer the solutions other users have already posted to the solution below, I thought it worth pointing out that there's really nothing stopping you from solving this integral by brute force.

The main non-trivial fact needed beforehand is the anti-derivative,


which can be verified via differentiation.

Here's a sketch of the rest of the calculation:

$$\begin{align} I &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\ln{(1+xy)}}{1-xy}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\int_{0}^{x}\mathrm{d}u\,\frac{\ln{(1+u)}}{1-u}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{x}\left[-\operatorname{Li}_2{\left(\frac{x+1}{2}\right)}-\ln{\left(\frac{1-x}{2}\right)}\ln{(1+x)}-\frac12\ln^2{2}+\frac12\zeta{(2)}\right]\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1-x}{2}\right)}\ln{(1+x)}}{x}-\int_{0}^{1}\mathrm{d}x\,\frac{\zeta{(2)}-\ln^2{2}-2\operatorname{Li}_2{\left(\frac{x+1}{2}\right)}}{2x}\\ &=\frac58 \zeta{(3)}+\frac12\zeta{(2)}\ln{2}-\frac{13}{8}\zeta{(3)}+\zeta{(2)}\ln{2}\\ &=\frac32\zeta{(2)}\ln{2}-\zeta{(3)}. \end{align}$$

  • $\begingroup$ I use Feynman's way by setting $$ \mathcal{I}(\alpha)=\int_0^1\int_0^1\frac{\ln(1+\alpha xy)}{1-xy}\ dx\ dy $$ then \begin{align} \frac{d\mathcal{I}}{d\alpha} &=\int_0^1\int_0^1\frac{1}{(1+\alpha xy)(1-xy)}\ dx\ dy\\ &=\frac1{1+\alpha}\int_0^1\int_0^1\left[\frac{1}{1-xy}+\frac\alpha{1+\alpha xy}\right]\ dx\ dy\\ &=\frac{\pi^2}{6}\cdot\frac1{1+\alpha}+\frac{\text{Li}_2(-\alpha)}{\alpha} + \frac{\text{Li}_2(-\alpha)}{1+\alpha}, \end{align} but I'm having trouble to evaluate the last term. Any idea? $\endgroup$ – Tunk-Fey Aug 18 '14 at 19:20
  • $\begingroup$ @Tunk-Fey I'm sorry, I'm not sure what exactly you're referring to. Are you asking how to evaluate $\int_{0}^{1}\frac{\text{Li}_2(-\alpha)}{1+\alpha}\mathrm{d}\alpha$? Or do you mean something else? $\endgroup$ – David H Aug 18 '14 at 19:45
  • $\begingroup$ Notice that in this and other similar contexts, $\ln2$ acts as a “normalized” value for $\zeta(1)$, since we are expecting results of the form $\displaystyle\sum a_k\cdot\zeta(k)\cdot\zeta(n-k)$, with $a_k\in\mathbb A$. However, in other contexts, it is $\gamma$ that fulfills this requirement, since $\dfrac{\zeta(1^+)+\zeta(1^-)}2=\gamma$. $\endgroup$ – Lucian Aug 19 '14 at 0:13
  • $\begingroup$ David: Yes, I'm looking a way to evaluate that integral. If possible, the indefinite integral. @Lucian I don't understand your comment, could you be more specific? Anyway, I found the source to evaluate the integral but it's too complicated to me. Perhaps, you both have a simple one. $\endgroup$ – Tunk-Fey Aug 19 '14 at 4:59
  • $\begingroup$ @Tunk-Fey I think the best way would be to use the two-dilog functional identity $\frac{\text{Li}_2{(-z)}}{1+z}=\frac{\zeta{(2)}-\ln{(-z)}\ln{(1+z)-\operatorname{Li}_2{(1+z)}}}{1+z}$ to split up the integrand into three bite-size morsels. None of the resulting integrals seem too difficult after that. $\endgroup$ – David H Aug 19 '14 at 5:36

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