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I am seeking a closed-form solution for this double sum:

\begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{3}n+m)}= ?. \end{eqnarray*}

I will turn it into $3$ tough integrals in a moment. But first I will state some similar results:

\begin{eqnarray*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(n+m)} &=& 2 \zeta(3) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{2}n+m)} &=& \frac{11}{8} \zeta(3) \\ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(\color{blue}{4}n+m)} &=& \frac{67}{32} \zeta(3) -\frac{G \pi}{2}. \\ \end{eqnarray*}

where $G$ is the Catalan constant. The last result took some effort ...

Now I know most of you folks prefer integrals to sums, so lets turn this into an integral. Using

\begin{eqnarray*} \frac{1}{n} &=& \int_0^1 x^{n-1} dx\\ \frac{1}{m} &=& \int_0^1 y^{m-1} dy\\ \frac{1}{3n+m} &=& \int_0^1 z^{3n+m-1} dz \\ \end{eqnarray*} and summing the geometric series, we have the following triple integral \begin{eqnarray*} \int_0^1 \int_0^1 \int_0^1 \frac{z^3 dx dy dz}{(1-xz^3)(1-yz)}. \end{eqnarray*}

Now doing the $x$ and $y$ integrations we have \begin{eqnarray*} I=\int_0^1 \frac{\ln(1-z) \ln(1-z^3)}{z} dz. \end{eqnarray*}

Factorize the argument of the second logarithm ...

\begin{eqnarray*} I= \underbrace{\int_0^1 \frac{\ln(1-z) \ln(1-z)}{z} dz}_{=2\zeta(3)} + \int_0^1 \frac{\ln(1-z) \ln(1+z+z^2)}{z} dz. \end{eqnarray*}

So if you prefer my question is ... find a closed form for:

\begin{eqnarray*} I_1 = - \int_0^1 \frac{\ln(1-z) \ln(1+z+z^2)}{z} dz. \end{eqnarray*}

Integrating by parts gives:

\begin{eqnarray*} I_1 = - \int_0^1 \frac{\ln(z) \ln(1+z+z^2)}{1-z} dz + \int_0^1 \frac{(1+2z)\ln(z) \ln(1-z)}{1+z+z^2} dz. \end{eqnarray*}

and let us call these integrals $I_2$ and $I_3$ respectively.

All $3$ of these integrals are not easy for me to evaluate and any help with their resolution will be gratefully received.