I am trying to evaluate the following double sum
\begin{eqnarray*}
\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{n+m}}{n(3n+m)}.
\end{eqnarray*}
Using the integral trick
\begin{eqnarray*}
\frac{1}{3n+m} =\int_0^1 y^{3n+m-1} dy,
\end{eqnarray*}
the sum can be transformed into integral
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1+y^3)}{1+y} dy.
\end{eqnarray*}
Now "half" of this is easy (IBP & rearrange)
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1+y)}{1+y} dy = \frac{1}{2} (\ln 2)^2.
\end{eqnarray*}
So we are left with
\begin{eqnarray*}
\int_0^1 \frac{ \ln(1-y+y^2)}{1+y} dy = \int_0^1 \frac{ (1-2y)\ln(1+y)}{1-y+y^2} dy.
\end{eqnarray*}
Now, apart from the obvious IBP done above, this integral has me stumped.
An exact evaluation would be nice, barring that, an expression in terms of dilogrithmic values or something similar would be helpful. Any comments or answers, gratefully received.
If this integral has been seen before a reference & advise on how I would search & find something similar in the future would be gratefully appreciated.
Best Answer
Call the integral $\mathfrak{I}$ and make the transformation $x\mapsto\tfrac {1-x}{1+x}$. What's left is$$\mathfrak{I}=\log^22+\int\limits_0^1\mathrm dx\,\frac {\log(1+3x^2)}{1+x}-3\int\limits_0^1\mathrm dx\,\frac {\log(1+x)}{1+x}$$ The last integral is trivially equal to$$\int\limits_0^1\mathrm dx\,\frac {\log(1+x)}{1+x}\color{red}{=\frac 12\log^22}$$The middle integral is difficult and I am still not aware of a full solution to the integral by hand. Replace the three with a parameter say, $a$, and differentiate with respect to $a$$$\mathfrak{I}(a)=\int\limits_0^1\mathrm dx\space\frac {\log(1+ax^2)}{1+x}$$Thus$$\begin{align*}\mathfrak{I}'(a) & =\int\limits_0^1\mathrm dx\space\left[\frac 1{(1+a)(1+x)}+\frac x{(1+a)(1+ax^2)}-\frac 1{(1+a)(1+ax^2)}\right]\\ & =\frac 1{1+a}\log 2+\frac {\log(1+a)}{2a(1+a)}-\frac 1{\sqrt a(1+a)}\arctan\sqrt a\end{align*}$$Integrate both sides with respect to $a$ and from zero to three to get (I used Wolfram Alpha on this step)$$\int\limits_0^1\mathrm dx\space\frac {\log(1+3x^2)}{1+x}\color{brown}{=\log^22+\frac 12\operatorname{Li}_2\left(-\frac 13\right)+\frac 14\log^23-\frac {\pi^2}{36}}$$Now add everything together and you should get the result I stated at the beginning of my answer.