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

Question

Find this integral

$$I=\int_{0}^{1}\int_{0}^{1}\dfrac{\ln{(1+xy)}}{1-xy}dxdy$$

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

Answer 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.

Answer 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:

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

Answer 3

An elementary evaluation: \begin{align} I &=\int_{0}^{1}\int_{0}^{1}\frac{\ln{(1+xy)}}{1-xy}\overset{t=xy}{dy}\ dx \\ &=\int_{0}^{1} \int_{0}^{x}\frac{\ln{(1+t)}}{1-t}dt\ {d} {(\ln x)}\ \overset{ibp} =-\int_{0}^{1} \frac{\ln{(1+x)}\ln x}{1-x} dx\\ \end{align} Then, consider $K(a)= \int_0^1 \frac{ \ln x\ln(1-ax)}{x(1+x)}dx$ \begin{align} K’(a) &=-\int_0^1\frac{\ln x}{(1+x)(1-ax)}dx\\ &= -\frac{1}{1+a}\left(\int_0^1 \frac{\ln x}{1+x}dx + \int_0^1\frac{a\ln x}{1-ax} \overset{ax\to y}{dx }\right)\\ &= \frac{\zeta(2)}{2(1+a)} - \frac{\ln a\ln(1-a)}{1+a}- \frac1{1+a}\int_0^a\frac{\ln y}{1-y}dy\\ \end{align} to establish

\begin{align} 0&=K( 0) =K(1)-\int_0^1 K’(a)da \\ &= \int_0^1 \frac{\ln x\ln(1-x)}{x(1+x)} {dx} - \frac12\zeta(2)\int_0^1 \frac1{1+a}da \\ & \hspace{10mm} +\int_0^1 \frac{\ln x\ln(1-x)}{1+x} dx+\int_0^1 \frac1{1+a}\int_0^a\frac{\ln y}{1-y}dy \ \overset{ibp}{da}\\ &= \int_0^1 \frac{\ln x\ln(1-x)}{x}dx - \frac12\zeta(2)\ln2 +\ln2\int_0^1 \frac{\ln y}{1-y}dy -I\\ &= \zeta(3)-\frac32 \zeta(2)\ln2-I \end{align}

which leads to

$$I= \zeta(3) -\frac32 \zeta(2)\ln2 $$