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?

This was kind of tricky and the route I chose may not be justified, so I do expect to be chewed out.
Your sum is taken over $(m, n)$ in the first quadrant of $$C = \sum_{m \neq n} \frac{1}{n(m^2 - n^2)}$$
Start with partial fraction decomposition to get $$2C = \sum_{m \neq n } \Big(\frac{1}{n^2(m-n)} -\frac{1}{n^2(m+n)}\Big)$$ I will split the sum and "evaluate" the sums over each of the partial fractions.
Keep in mind the $(m, n)$ plane in the first quadrant, where m is the x-axis and n is the y-axis. Since $m \neq n$, we have that $$ m-n \neq 0$$ Consider each line $$ m - n = k$$ You will want to sum over these lines and adjust for n accordingly. When $k$ is positive, $n$ can be as small as one. When $k$ is negative, n must be at least $k+1$. Again, this can be seen if you keep the plane in mind. Since $m-n = k$:
$$\sum \frac{1}{n^2(m-n)} = \sum_{ k \geq 1, n \geq 1} \frac{1}{n^2k} +\sum_{ k \leq -1, n \geq k+1}\frac{1}{n^2k}$$ The sums on the RHS can be evaluated as
$$\sum_{ k \geq 1, n \geq 1} \frac{1}{n^2k} +\sum_{ k \leq -1, n \geq k+1}\frac{1}{n^2k} = \zeta (2)\sum_{k \geq 1}\frac{1}{k} - \sum_{k \geq 1}\frac{(\zeta(2)-H^{2}_{k})}{k} = \sum_{k \geq 1}\frac{H^{2}_{k}}{k}$$ where $H^{2}_{k}$ is the harmonic sum $$H^{2}_{k} = \sum_{r \leq k} \frac{1}{r^2}$$ This is a divergent sum, but can be worked around.
In this sum we had to be careful about stepping on the points $(m, n)$ along the diagonal. In the second sum though, when we step on a diagonal, we get $$\sum_{m \geq 1} \frac{1}{m^2(m+m)}=\frac{\zeta(3)}{2}$$ so we can sum over the entire quadrant (except the x and y axes) and subtract this quantity from it to get $$\sum_{k \geq 1, 1 \leq n \leq k-1} \frac{1}{n^2(m+n)} = \sum_{k \geq 1, 1 \leq n \leq k-1} \frac{1}{n^2k}$$ where, keep in mind, the plane, and the fact that we are summing over lines $m+n = k$. This gives $$\sum_{k \geq 1, 1 \leq n \leq k-1} \frac{1}{n^2k} = \sum_{k \geq 1}\frac{\big(1 +\frac{1}{2^2} + ... + \frac{1}{(k-1)^2}\big)}{k} = \sum_{ k \geq 1} \frac{H^{2}_{k-1}}{k} = \sum_{k \geq 1}\frac{H^2_{k} - \frac{1}{k^2}}{k} = -\zeta(3)+\sum_{k \geq 1}\frac{H^2_{k}}{k}$$ To find $2C$, we return to the original expression and plug things back in: $$2C = \sum_{m \neq n } \Big(\frac{1}{n^2(m-n)} -\frac{1}{n^2(m+n)}\Big) = \sum_{k \geq 1}\frac{H^2_{k}}{k} - \big(-\zeta(3) - \frac{\zeta(3)}{2} +\sum_{k \geq 1} \frac{H^2_{k}}{k} \big) = \frac{3\zeta(3)}{2}$$ Therefore, $$C = \frac{3\zeta(3)}{4}$$
