How to evaluate $\int_{0}^{1}\int_{0}^{1}\frac{(x+y+xy)\log{(x+y+xy)}}{x+y}dxdy$?

Question

I am trying to evaluate this integral: $$\int_{0}^{1}\int_{0}^{1}\frac{(x+y+xy)\log{(x+y+xy)}}{x+y}dxdy.$$ It looks like this integral is symmetry with variables $x$ and $y$ but I can't find the way to exploit it. I tried to use this: $$\int_{0}^{1}\int_{0}^{1}\frac{(x+y+xy)\log{(x+y+xy)}}{x+y}dxdy\\=\int_{0}^{1}\int_{0}^{1}2x\frac{(x+xy+x^2y)\log{(x+xy+x^2y)}}{x(1+y)}dxdy,$$ but still can't isolate $x$ and $y$. I really need some advices here, thank you.

EDIT After trying more, i end up with this:$$\int_{0}^{1}\int_{0}^{1}\frac{(xy)\log{(x+y+xy)}}{x+y}dxdy.$$ I cant process more, any advices, thank you.

Answer 1

Using the simmetry with respect to $y=x$, your second integral can be written using polar coordinates as follows. \begin{eqnarray} \mathcal I &=& \int_0^1\int_0^1\frac{xy\log(x+y+xy)}{x+y}dxdy=\\ &=&2\int_0^{\pi/4}\int_0^{1/\cos\theta}\frac{\rho^3\cos\theta\sin\theta\log(\rho\cos\theta + \rho\sin\theta + \rho^2\cos\theta\sin\theta)}{\rho\cos\theta+\rho\sin\theta}d\rho d\theta=\\ &=&2\int_0^{\pi/4}\frac{\cos\theta\sin\theta}{\cos\theta + \sin\theta}\int_0^{1/\cos\theta}\rho^2\log(\rho\cos\theta + \rho \sin\theta + \rho^2 \cos\theta \sin\theta)d\rho d\theta. \end{eqnarray} Integrating by parts yields \begin{eqnarray} \mathcal I &=&2\int_0^{\pi/4}\frac{\cos\theta\sin\theta}{\cos\theta + \sin\theta}\left\{\left[\frac{\rho^3}3\log(\rho\cos\theta + \rho \sin\theta + \rho^2 \cos\theta \sin\theta)\right]_0^{1/\cos\theta}\right.+\\ & &\left.-\frac13\int_0^{1/\cos\theta}\frac{\rho^2(\cos\theta+\sin\theta+2\rho\cos\theta\sin\theta)}{\cos\theta+\sin\theta+\rho\cos\theta\sin\theta}d\rho\right\}d\theta=\\ &=&2\int_0^{\pi/4}\frac{\cos\theta\sin\theta}{\cos\theta + \sin\theta}\left\{\frac{\log(1+2\tan\theta)}{3\cos^3\theta}+\right.\\ & & -\frac13\int_0^{1/\cos\theta}\left(2\rho^2-\rho\frac{\cos\theta+\sin\theta}{\cos\theta\sin\theta}+\frac{(\cos\theta+ \sin\theta)^2}{\cos^2\theta\sin^2\theta}+\right.\\ & &-\left.\left.\frac{(\cos\theta+\sin\theta)^3}{\cos^2\theta\sin^2\theta}\cdot\frac1{\cos\theta+\sin\theta+\rho\cos\theta\sin\theta}\right)d\rho\right\}d\theta=\\ &=&2\int_0^{\pi/4}\left[\frac13\cdot\frac{\sin\theta\log(1+2\tan\theta)}{\cos^2\theta(\cos\theta+\sin\theta)}-\frac29\cdot \frac{\sin\theta}{\cos^2\theta(\cos\theta +\sin\theta)}+\frac16\cdot\frac1{\cos^2\theta}+\right.\\ & &\left.-\frac13\cdot \frac{\cos\theta + \sin\theta}{\cos^2\theta\sin\theta}+\frac13\cdot\frac{(\cos\theta+\sin\theta)^2}{\cos^2\theta\sin^2\theta}\cdot \log\left(\frac{\cos\theta+2\sin\theta}{\cos\theta+\sin\theta}\right)\right]d\theta. \end{eqnarray} Grouping into three integrals and replacing $\tan \theta$ with $t$ gives \begin{eqnarray} \mathcal I &=& \frac23\underbrace{\int_0^1\frac{t\log(1+2t)}{1+t}dt}_{\mathcal I_1}+\frac13\underbrace{\int_0^1\frac{1-t}{1+t}dt}_{\mathcal I_2}+\\& &-\frac23\underbrace{\int_0^1\left[ \frac{1+t}t-\frac{(1+t)^2}{t^2}\log\left(\frac{1+2t}{1+t}\right)\right]dt}_{\mathcal I_3}.\tag{*}\label{1} \end{eqnarray} $\mathcal I_1$ can be computed as follows \begin{eqnarray} \mathcal I_1 &=& \int_0^1 \frac{t \log(1+2t)}{1+t}dt=\\ &=&\int_0^1 \log(1+2t)dt -\int_0^1\frac{\log(1+2t)}{1+t}dt=\\ &=&\frac32\log 3 -1 -\int_0^1 \log(1+2t)d[\log(2+2t)]=\\ &=&\frac32\log 3 -1 -2\log3\log2+\int_0^1 \frac{\log[1+(1+2t)]}{1+2t}d(1+2t)=\\ &=&\frac32\log3 -1-2\log3\log2+[-\mbox{Li}_2(-1-2t)]_0^1=\\ &=&\frac32\log 3 -1 -2\log3\log2-\mbox{Li}_2(-3)-\frac{\pi^2}{12}. \end{eqnarray} Trivially $\mathcal I_2 = 2\log2-1$. When finding the primitive for $\mathcal I_3$ note that all the divergent terms $\log t$ cancel out. We are left with \begin{eqnarray} \mathcal I_3 &=& \int_0^1\left[1+\frac1 t-\frac{\log(1+2t)}{t^2}-\frac{2\log(1+2t)}{t} - \log(1+2t)\right.+\\ & &\left.+\frac{\log(1+t)}{t^2} + \frac{2\log(1+t)}{t} + \log(1+t)\right]dt=\\ &=&\left[t+\log t - 2\log t + \frac{(1+2t)\log(1+2t)}t+2\mbox{Li}_2(-2t)\right.+\\ & &-t\log(1+2t) - \frac12\log(1+2t) + t+\\ & &+\log t -\frac{(1+t)\log(1+t)}t-2\mbox{Li}_2(-t)+\\ & &\left. +(1+t)\log(1+t) - t\right]_0^1=\\ &=& 1+3\log3 +2\mbox{Li}_2(-2)-\log 3 -\frac12 \log 3+1+\\ & &-2\log2 -2\mbox{Li}_2(-1)+2\log 2 -1-2+1=\\ &=&\frac32\log3 +\frac{\pi^2}{6}+2\mbox{Li}_2(-2). \end{eqnarray} Plugging in these results into \eqref{1} completes the computation.