TITLE: 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} $
QUESTION [14 upvotes]: 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.

REPLY [16 votes]: 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.