How can one evaluate the basic Tornheim Sum? I recently came across the claim that the double sum from $1$ to $\infty$ of $\frac{1}{mn(m+n)} = 2 \zeta(3)$.
I can show it equals the sum from $1$ to $\infty$ of $\frac{H_n}{n^2}$, where $H_n$ is the $n^{th}$ Harmonic number.
Does anyone know the full proof?
 A: Here is another proof of the identity
$$s = \sum _{n=1}^{\infty } \sum _{m=1}^{\infty } \frac{1}{m n (m+n)} = 2 \zeta (3)$$
Letting
$$\frac{1}{n+m} = \int_0^1 t^{m+n-1} \, dt$$
the double sum factorizes nicely under the $t$-integral leading to
$$s =\int_0^1 t^{-1} \sum _{n=1}^{\infty }\frac{t^{n}}{n}  \sum _{m=1}^{\infty }\frac{t^{m}}{m} \, dt$$
The sums are just $-\log (1-t)$ so that we have
$$s = \int_0^1 t^{-1} {\log (1-t)}^{2} \, dt$$
Letting now 
$$t \to 1 - e^{-u}$$
gives
$$s = \int_0^\infty u^{2} \frac{e^{-u}}{1-e^{-u}} \, du$$
Expanding into a geometric sum gives
$$s =  \sum _{k=1}^{\infty } \int_0^\infty u^{2}e^{- k\; u} \, du$$
The integral can easily be evaluated letting $k\; u \to z$ to give 
$$\frac {1}{k^3} \Gamma(3)$$
Hence observing that $\Gamma(3)=2$ and
$$\sum _{k=1}^{\infty }  \frac{1}{k^3} = \zeta(3)$$
completes the proof.
A: Consider the following three sums 
\begin{eqnarray*}
T(1,1,1) & =&  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)} \\
\zeta(2,1) & =&  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m(m+n)^2} \\
\zeta(3) & =&  \sum_{m=1}^{\infty}  \frac{1}{m^3}. \\
\end{eqnarray*}
Using partial fractions, you already know
\begin{eqnarray*}
  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)}  =  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2} \left(\frac{1}{n} -\frac{1}{m+n} \right)=  \sum_{m=1}^{\infty}  \frac{H_m}{m^2} = \zeta(3) + \zeta(2,1) .
\end{eqnarray*}
Now the trick you need is this
\begin{eqnarray*}
  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{mn(m+n)}  =  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \color{red}{\frac{1}{(m+n)^2} \left(\frac{1}{m} +\frac{1}{n} \right) } \\ =  \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m(m+n)^2}  +\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n(m+n)^2}  = 2 \zeta(2,1) 
\end{eqnarray*}
So $\zeta(2,1)=\zeta(3)$ & your sum is $T(1,1,1) =2 \zeta(3)$.
