<s>[INST] Compute the double sum $\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)$ I am trying to compute the following double sum
$$\boxed{\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)}$$
I proceeded as following
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{\left(m+n \right)\left(m-n \right)}$$
$$=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{2m}\left[\frac{1}{\left(m+n \right)}+\frac{1}{\left(m-n \right)}\right]$$
$$=\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{m}\frac{1}{\left(m+n \right)}}_{2\zeta(3)}+   \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}  \sum_{m=1}^{\infty}\frac{1}{m}\frac{1}{\left(m-n \right)}$$
The first sum is given here  and I have also evaluated here, therefore we get that
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}  \sum_{m=1}^{\infty}\frac{1}{m}-\frac{1}{\left(m-n \right)}$$
$$=\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2}  \int_{0}^{1}\frac{1-x^{-n}}{1-x}dx$$
$$=\zeta(3)-\frac{1}{2}  \int_{0}^{1}\frac{1}{1-x}\left[ \sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{x^{-n}}{n^2}\right]dx$$
$$=\zeta(3)-\frac{1}{2}  \int_{0}^{1}\frac{\zeta(2)-Li_{2}\left( \frac{1}{x} \right)}{1-x}dx$$
From this post we can use the relation
$$Li_{2}\left( \frac{1}{x} \right)-Li_{2}\left( x \right)=\zeta(2)-\frac{1}{2}\log^2(-x)$$
To get
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\zeta(3)-\frac{\zeta(2)}{2}  \int_{0}^{1}\frac{1}{1-x}dx+\frac{1}{2}  \int_{0}^{1}\frac{Li_{2}\left( x \right)+\zeta(2)-\frac{1}{2}\log^2(-x)}{1-x}dx$$
$$=\zeta(3)+\frac{\zeta(2)}{2}  \int_{0}^{1}\frac{1}{1-x}dx+\frac{1}{2}  \int_{0}^{1}\frac{Li_{2}\left( x \right)}{1-x}dx-\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$=\zeta(3)+\frac{\zeta(2)}{2}  \log(1-x)\Big|_{0}^{1}+\frac{1}{2}\left\{ -Li_{2}(x)\log(1-x)\Big|_{0}^{1}-\int_{0}^{1}\frac{\log^2(1-x)}{x}dx\right\}  -\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$=\zeta(3)-\frac{1}{2}2\zeta(3)  -\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=-\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$
My questions are the following:
(1)Is the last equality correct?
(2)If so, what is the meaning of  $\log^2(-x)=\log^2(e^{i\pi}x)$?
(3) How do we solve the last integral?
 [/INST] This doesn't really answer your questions, however consider another approach perhaps as a way of side-lining somewhat difficult integrals
$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{n'=1}^{\infty }{\sum\limits_{m'=1}^{\infty }{\frac{1}{{{n}^{2}}\left( m-n \right)}}}-\frac{1}{2}\sum\limits_{n=1}^{\infty }{\sum\limits_{m=1}^{\infty }{\frac{1}{{{n}^{2}}\left( m+n \right)}}}+\frac{1}{4}\sum\limits_{m=1}^{\infty }{\frac{1}{{{m}^{3}}}}$$
Primes on sums denote excluding $n=m$.  Note
$$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}},n\ge 0$$
So we have by the theorem that the sum over all residues is zero
$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ -\underset{z=0}{\mathop{res}}\,\frac{\psi \left( -z \right)+\gamma }{{{z}^{2}}\left( m-z \right)}+\underset{z=-m,0}{\mathop{res}}\,\frac{\psi \left( -z \right)+\gamma }{{{z}^{2}}\left( m+z \right)}+\frac{1}{2{{m}^{3}}} \right\}}$$
The first residue calculation specifically excludes the residue at $z=m$ to avoid the singularities at $n=m$.  Using the expansion above to determine the residues we find
$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}\\=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ \frac{{{m}^{2}}{{\pi }^{2}}-6}{6{{m}^{3}}}+\frac{-1+\gamma +\psi \left( m \right)}{{{m}^{2}}}+\frac{6+6m-{{m}^{2}}{{\pi }^{2}}}{6{{m}^{3}}}+\frac{1}{2{{m}^{3}}} \right\}}$$
Cleaning this up
$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ \frac{\gamma +\psi \left( m \right)}{{{m}^{2}}}+\frac{1}{2{{m}^{3}}} \right\}}=\frac{1}{2}\left\{ \zeta \left( 3 \right)+\frac{1}{2}\zeta \left( 3 \right) \right\}=\frac{3}{4}\zeta \left( 3 \right)$$
Where we’ve used the reasonably well known representation
$$\zeta \left( 3 \right)=\sum\limits_{m=1}^{\infty }{\frac{\gamma +\psi \left( m \right)}{{{m}^{2}}}}$$
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