# Evaluate or express as a series $\int_0^1\int_0^1\int_0^1\frac{\log((x+z)(y+z))}{1+xyz}dxdydz$

I believe that it is possible to get in closed-form the integral $$\int_0^1\int_0^1\int_0^1\frac{\log((x+z)(y+z))}{1+xyz}dxdydz,\tag{1}$$ but I think that it is tedious. My belief is from the calculations that I tried using a CAS. This CAS know how to get the corresponding indefinite integrals but it is tedious to me to evaluate the limits of integration. My last attempt was write our integral as $$\sum_{n=0}^\infty\int_0^1\int_0^1(-1)^k(yz)^k\int_0^1x^k\log((x+z)(y+z))dxdydz.\tag{2}$$ The CAS tells me what are the corresponding antiderivatives, but it is difficult/tedious to me calculate the expressions for the evaluation of the mentioned limits.

Question. Can you express as a series or well calculate a closed-form (if it is possible) our integral $(1)$? Many thanks.

If you think that is feasible to provide me a sketch of the required calculations instead of all details feel free to do it, or well if this is a known integral from the literature adding the reference.

## 1 Answer

By symmetry the given integral equals $$2\iiint_{(0,1)^3}\frac{\log(x+y)}{1+xyz}\,dx\,dy\,dz = 2 \iint_{(0,1)^2}\frac{\log(x+y)\log(1+x y)}{xy}\,dx\,dy$$ and we may exploit symmetry again: the RHS equals

$$4\int_{0}^{1}\int_{0}^{x}\frac{\log(x+y)\log(1+xy)}{xy}\,dy\,dx=4\int_{0}^{1}\int_{0}^{1}\frac{\log(x+xz)\log(1+x^2 z)}{x z}\,dz\,dx$$ so the problem boils down to computing (via $\log(x+xz)=\log(x)+\log(1+z)$ and $x\mapsto\sqrt{x}$) $$\mathcal{J}_1=\iint_{(0,1)^2}\frac{\log(1+z)\log(1+x z)}{xz}\,dz\,dx,\qquad \mathcal{J}_2=\iint_{(0,1)^2}\frac{\log(x)\log(1+x z)}{xz}\,dz\,dx$$ and now it is a good moment for expanding $\log(1+xz)$ as a Maclaurin series. This yields

$$\begin{eqnarray*}\mathcal{J}_1&=&2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\iint_{(0,1)^2}\log(1+z) x^{n-1} z^{n-1}\,dx\,dz\\&=&2\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}\int_{0}^{1}\log(1+z)z^{n-1}\,dz\end{eqnarray*}$$

$$\begin{eqnarray*}\mathcal{J}_2&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\iint_{(0,1)^2}\log(x) x^{n-1} z^{n-1}\,dx\,dz\\&=&-\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^4}=-\frac{7\pi^4}{720}\end{eqnarray*}$$ and by recognizing $\mathcal{J}_1$ as a standard Euler sum (see Flajolet & Salvy, alternating Euler sums with weight $4$ and $5$), or by simply noticing the relation between $\mathcal{J}_1$ and $\left[\text{Li}_2(-x)^2\right]_{0}^{1}$, we have $\mathcal{J}_1=\frac{\pi^4}{144}$. Putting everything together $$\boxed{\iiint_{(0,1)^3}\frac{\log\left[(x+y)(y+z)\right]}{1+xyz}\,dx\,dy\,dz = \color{red}{-\frac{\pi^4}{360}}.}$$

• Many thanks tomorrow in the morning I am going to study it. This is a masterpiece of clarity since I was trying to calculate evauating polylogarithms and hypergeometrics. God point $\checkmark$. – user243301 Apr 3 '18 at 22:13