A closed form of the double sum $\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2} $ I want to evaluate the double sum $\displaystyle \sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2} $ where I know that the value is $-\frac{17\pi^4}{1440}$ , I have failed to evaluate to the inner finite sum cleverly so as to transform it into a reasonable infinite series.
 A: Hint. One may observe that, for $m^2\neq n^2$, we have
$$
\frac1{(m^2-n^2)^2}=\int_0^1\int_0^1x^{m+n-1}y^{m-n-1}\log x \log y\:dxdy \tag1
$$ giving, by using absolute convergence,
$$
\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}=\int_0^1\!\!\int_0^1\frac{-\log x \log y}{(1-x^2)(1+xy)}\:dxdy. \tag2
$$ Then, by partial fraction decomposition, one may write
$$
\begin{align}
&\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}
\\\\&=\int_0^1\!\!\int_0^1\frac{-\log x \log y}{(1-x^2)(1+xy)}\:dxdy
\\\\&=\int_0^1\!\!\int_0^1\left(\frac{-\log x \log y}{2 (1+x) (1-y)}+\frac{-\log x \log y}{2 (1-x)(1+y)}+\frac{y^2\log x \log y}{(1-y^2) (1+x y)}\right)\!dxdy \tag3
\\\\&=-2\cdot \frac{\pi^4}{144}+\frac{\pi^4}{480}
\\\\&=-\frac{17\pi^4}{1440}
\end{align}
$$ as announced.
A: Let $k\in\mathbb{N}$.  The function $\tau:\mathbb{N}\to\mathbb{N}$ is the divisor-counting function.  For an odd value of $k$, there are $\left\lceil\frac{1}{2}\tau(k)\right\rceil$ pairs $(m,n)$ with $m\in\mathbb{N}$ and $n\in\mathbb{N}_0$ such that $m>n$ and $k=m^2-n^2$, whereas there are $\left\lceil\frac{1}{2}\tau\left(\frac{k}{4}\right)\right\rceil$ such pairs if $k$ is divisible by $4$.  If $k$ is even but not divisible by $4$, then there are no such pairs $(m,n)$.  
The required sum is then
$$S:=\sum_{m=1}^\infty\,\sum_{n=0}^{m-1}\,\frac{(-1)^{m-n}}{\left(m^2-n^2\right)^2}=\sum_{k\equiv0\!\pmod{4}}\,\frac{\left\lceil\frac{1}{2}\tau\left(\frac{k}{4}\right)\right\rceil}{k^2}-\sum_{k\text{ odd}}\,\frac{\left\lceil\frac{1}{2}\tau(k)\right\rceil}{k^2}\,.$$
Now, by handling perfect-square values of $k$, we see that
$$S=\frac{1}{2}\,\sum_{k\equiv0\!\pmod{4}}\,\frac{\tau(k/4)}{k^2}-\frac{1}{2}\sum_{k\text{ odd}}\,\frac{\tau(k)}{k^2}+\frac{1}{2}\,\sum_{j=1}^\infty\,\frac{(-1)^j}{j^4}\,.$$
Hence, with $j:=\frac{k}{4}$ in the first sum, we have
$$S=\frac{1}{32}\,\sum_{j=1}^\infty\,\frac{\tau(j)}{j^2}-\frac{1}{2}\,\sum_{k\text{ odd}}\,\frac{\tau(k)}{k^2}-\frac{1}{2}\,\eta(4)\,,$$
where $\eta$ is the Dirichlet eta function.  Note that $$\eta(4)=\left(1-\frac{1}{2^3}\right)\,\zeta(4)=\frac{7}{8}\left(\frac{\pi^4}{90}\right)=\frac{7\pi^4}{720}\,,$$ where $\zeta$ is the Riemann zeta function.
Notice that $$\sum_{j=1}^\infty\,\frac{\tau(j)}{j^2}=\prod_{p\text{ prime}}\,\sum_{r=0}^\infty\,\frac{r+1}{p^{2r}}=\prod_{p\text{ prime}}\,\frac{p^4}{\left(p^2-1\right)^2}$$
and that
$$\sum_{k\text{ odd}}\,\frac{\tau(k)}{k^2}=\prod_{p\text{ odd prime}}\,\sum_{r=0}^\infty\,\frac{r+1}{p^{2r}}=\prod_{p\text{ odd prime}}\,\frac{p^4}{\left(p^2-1\right)^2}\,,$$
whence
$$\sum_{k\text{ odd}}\,\frac{\tau(k)}{k^2}=\frac{9}{16}\,\sum_{j=1}^\infty\,\frac{\tau(j)}{j^2}\,.$$
Now, observe that
$$\sum_{j=1}^\infty\,\frac{\tau(j)}{j^2}=\left(\prod_{p\text{ prime}}\,\frac{1}{1-p^{-2}}\right)^2=\big(\zeta(2)\big)^2=\left(\frac{\pi^2}{6}\right)=\frac{\pi^4}{36}\,.$$
Combining all the pieces, we have
$$S=\frac{1}{32}\left(\frac{\pi^4}{36}\right)-\frac{1}{2}\left(\frac{9}{16}\right)\left(\frac{\pi^4}{36}\right)-\frac{1}{2}\left(\frac{7\pi^4}{720}\right)=-\frac{17\pi^4}{1440}\,.$$
P.S. We can show that $$\sum_{j=1}^\infty\,\frac{\tau(j)}{j^s}=\big(\zeta(s)\big)^2$$
for all $s\in\mathbb{C}$ with $\text{Re}(s)>1$.
