TITLE: Is there a closed-form solution for $\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{nm(3n+m)}$?
QUESTION [30 upvotes]: 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.

REPLY [4 votes]: I can't comment,but I have a closed form about the sum
$$\sum_{n=1}^{\infty}\frac{H_{\alpha n}}{n^2}$$
It equals to
$$\frac{1}{2}\left(\alpha^{2}+\frac{3}{\alpha}\right) \zeta(3)-\frac{\pi}{2 \alpha} \sum_{k=1}^{\alpha-1} \cot \left(\frac{\pi k}{\alpha}\right) \psi^{(1)}\left(\frac{k}{\alpha}\right)$$
where
$\alpha\in\mathbb{N}^{+}$

For example,let $\alpha=1,2,3,4,5,6$,then we get
$$\begin{aligned}
&\sum_{n=1}^{\infty} \frac{H_{n}}{n^2}=2\zeta(3)\\
&\sum_{n = 1}^{\infty} \frac{H_{2n}}{n^2} = \frac{11}{4}\zeta(3)\\
&\sum_{n=1}^{\infty} \frac{H_{3n}}{n^2}=5\zeta(3)-
\frac{\pi}{6\sqrt{3} } \left [ \psi^{(1)} \left ( \frac{1}{3}  \right )
-\psi^{(1)} \left ( \frac{2}{3}  \right )
 \right ]\\
&\sum_{n=1}^{\infty} \frac{H_{4n}}{n^2}=\frac{67}{8} \zeta(3)-2\pi G\\
&\sum_{n=1}^{\infty} \frac{H_{5n}}{n^2}=\frac{64}{5} \zeta(3)-\frac{\pi}{10\sqrt{5}}
 \left\{ 
\sqrt{5+2\sqrt{5} }
\left [ \psi^{(1)}\left ( \frac{1}{5}  \right ) - \psi^{(1)}\left ( \frac{4}{5}  \right ) \right ]
 +
\sqrt{5-2\sqrt{5} } \left [  \psi^{(1)}\left ( \frac{2}{5}  \right ) - \psi^{(1)}\left ( \frac{3}{5}  \right ) \right ] \right\}\\
&\sum_{n=1}^{\infty} \frac{H_{6n}}{n^2}=\frac{73}{4}\zeta(3)-\frac{\pi}{12} \left \{ \sqrt{3}\left [
 \psi^{(1)}\left ( \frac{1}{6}  \right )- \psi^{(1)}\left ( \frac{5}{6}  \right ) \right ] 
+\frac{1}{\sqrt{3} } \left [  \psi^{(1)}\left ( \frac{1}{3}  \right )- \psi^{(1)}\left ( \frac{2}{3}  \right ) \right ] 
  \right \}
\end{aligned}$$

Proof!!!!!
Consider function
$$f(z)=\frac{\pi\cot(\pi z)(\gamma+\psi^{(0)}(-\alpha z))}{z^2} ,a\in\mathbb{N}^+$$
By the residue theorem!!!!!
$$\begin{aligned}
&\mathrm{Res}(f,-n)=\frac{H_{\alpha n}}{n^2} -\frac{1}{\alpha n^3}\\
&\mathrm{Res}(f,0)=-\alpha^2\zeta(3)\\
&\mathrm{Res}(f,n)=-\frac{2}{\alpha n^3} +\frac{H_{\alpha n}}{n^2} \\
&\mathrm{Res}(f,n/\alpha)=\cfrac{\pi\alpha\cot\left(\frac{\pi n}{\alpha} \right)}{n^2} 
\end{aligned}$$
Then
$$\sum_{n=1}^{\infty}\frac{H_{\alpha n}}{n^2} =\frac{1}{2} 
\left ( \alpha^2+\frac{3}{\alpha}  \right ) \zeta(3)
-\frac{\pi\alpha}{2} \sum_{n=1,n\not{{\equiv}}0(\mathrm{mod}\alpha)}^{\infty} 
\frac{\cot\left ( \frac{\pi n}{\alpha}  \right ) }{n^2}$$
And
$$\begin{aligned}
&\sum_{n=1,n\not{{\equiv}}0(\mathrm{mod}\alpha)}^{\infty} 
\frac{\cot\left ( \frac{\pi n}{\alpha}  \right ) }{n^2}\\&=
\frac{\cot\left ( \frac{\pi}{\alpha}  \right ) }{1^2} 
+\frac{\cot\left ( \frac{2\pi}{\alpha}  \right ) }{2^2}
+\frac{\cot\left ( \frac{3\pi}{\alpha}  \right ) }{3^2}+...
+\frac{\cot\left ( \frac{(\alpha-1)\pi}{\alpha}  \right ) }{(\alpha-1)^2}\\
&+\frac{\cot\left ( \frac{\pi}{\alpha}  \right ) }{(\alpha+1)^2}
+\frac{\cot\left ( \frac{2\pi}{\alpha}  \right ) }{(\alpha+2)^2}
+\frac{\cot\left ( \frac{3\pi}{\alpha}  \right ) }{(\alpha+3)^2}+...
+\frac{\cot\left ( \frac{(\alpha-1)\pi}{\alpha}  \right ) }{(2\alpha-1)^2}+...\\
&=\cot\left ( \frac{\pi}{\alpha}  \right ) 
\left [  \frac{1}{1^2}+\frac{1}{(\alpha+1)^2}+\frac{1}{(\alpha+2)^2}+\frac{1}{(\alpha+3)^2}+...  \right ]\\
&+\cot\left ( \frac{2\pi}{\alpha}  \right ) \left[\frac{1}{2^2}+ \frac{1}{(2\alpha+1)^2}+\frac{1}{(2\alpha+2)^2}+\frac{1}{(2\alpha+3)^2}+...  \right ]+...\\
&=\frac{1}{\alpha^2} \sum_{k=1}^{\alpha-1} \cot\left ( \frac{\pi k}{\alpha}  \right ) \psi^{(1)}\left (\frac{k}{\alpha}   \right ) 
\end{aligned}$$
So,
$$\boxed{\sum_{n=1}^{\infty} \frac{H_{\alpha n}}{n^{2}}=\frac{1}{2}\left(\alpha^{2}+\frac{3}{\alpha}\right) \zeta(3)-\frac{\pi}{2 \alpha} \sum_{k=1}^{\alpha-1} \cot \left(\frac{\pi k}{\alpha}\right) \psi^{(1)}\left(\frac{k}{\alpha}\right)}$$

We also have,$$\boxed{\sum_{n=1}^{\infty} \frac{H_{\alpha n}}{n^{4}}=\frac{1}{2}\left(\alpha^{4}+\frac{5}{\alpha}\right) \zeta(5)-\frac{\pi^2\alpha^2}{6}\zeta(3) -\frac{\pi}{12 \alpha} \sum_{k=1}^{\alpha-1} \cot \left(\frac{\pi k}{\alpha}\right) \psi^{(3)}\left(\frac{k}{\alpha}\right)}$$
They can make some interesting integrals,like
$$\begin{aligned}
&\int_{0}^{1} \frac{\ln^2(1-x)}{x} \text{d}x=2\zeta(3)\\
&\int_{0}^{1} \frac{\ln(1-x)\ln(1+x)}{x} \text{d}x=-\frac{5}{8}\zeta(3)\\
&\int_{0}^{1} \frac{\ln(1-x)\ln(1+x^2)}{x} \text{d}x=\frac{23}{32}\zeta(3)-\frac{\pi}{2}G \\
&\int_{0}^{1} \frac{\ln(1-x)\ln(1+x+x^2)}{x}\text{d}x=-\frac{1}{3} \zeta(3)-
\frac{\pi}{18\sqrt{3} } \left [ \psi^{(1)}\left ( \frac{1}{3}  \right )-\psi^{(1)}\left ( \frac{2}{3}  \right )\right ]\\ \\
& \int_{0}^{1} \frac{\text{Li}_2^2(x)}{x}\text{d}x=\frac{\pi^2}{3}\zeta(3)-3\zeta(5)\\
&\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^2)}{x}\text{d}x=\frac{\pi^2}{4}\zeta(3)-\frac{37}{16}\zeta(5)\\
&\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^3)}{x}\text{d}x=\frac{2\pi^2}{9}\zeta(3)-\frac{124}{27}\zeta(5)
+\frac{\pi}{324\sqrt{3} } \left [ \psi^{(3)}\left ( \frac{1}{3}  \right )-\psi^{(3)}\left ( \frac{2}{3}  \right )\right ]\\
&\int_{0}^{1} \frac{\text{Li}_2(x)\text{Li}_2(x^4)}{x}\text{d}x=\frac{5\pi^2}{24}\zeta(3)-\frac{1029}{128}\zeta(5)+2\pi\beta(4)
\end{aligned}$$