How find this integral $I=\int_{0}^{1}\int_{0}^{1}\frac{\ln{(1+xy)}}{1-xy}dxdy$

Solution 1:

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.

Solution 2:

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,

$$\int\mathrm{d}u\,\frac{\ln{(1+u)}}{1-u}=-\operatorname{Li}_2{\left(\frac{x+1}{2}\right)}-\ln{\left(\frac{1-x}{2}\right)}\ln{(1+x)}+constant,$$

which can be verified via differentiation.

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

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