Double summation $\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$

Question

As a follow to this answer I came across the double sum $$\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}.$$ But unfortunately I do not have skills in techniques to handle double summation .

Help appreciated.

I've made some research in MSE and found several questions which could be helpful:

1) $\sum_{m=1}^{\infty}\sum_{n=0}^{m-1}\frac{(-1)^{m-n}}{(m^2-n^2)^2}=-\frac{17\pi^4}{1440}$

2) $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{\pi^6}{12960}$

3) $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}= \frac{1}{3}\zeta(6)$

Answer 1

The double summation is equal to $$\frac{11\zeta(4)}{8}=\frac{11\pi^4}{720}.$$

Note that $$\frac{m^2+n^2}{m n\left(m^2-n^2\right)^2}= \frac{1}{2 mn(m+n)^2}+ \frac{1}{2 mn(m-n)^2}.$$ Now consider the Tornheim double sums: $$T(a,b,c)= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^an^b(m+n)^c}.$$ Then $$ \sum_{m,n=1\, m\neq n}^{\infty} \frac{1}{mn(m+n)^2}=T(1,1,2)-\sum_{m=1}^{\infty}\frac{1}{4m^4}=T(1,1,2)-\frac{\zeta(4)}{4},$$ $$\begin{align} \sum_{m=1}^{\infty} \sum_{n=1}^{m-1} \frac{1}{mn(m-n)^2} &= \sum_{n=1}^{\infty} \sum_{m=n+1}^{\infty} \frac{1}{mn(m-n)^2}\\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{nk^2(n+k)}=T(1,2,1)\end{align}$$ and again $$ \sum_{m=1}^{\infty} \sum_{n=m+1}^{\infty} \frac{1}{mn(m-n)^2}=T(1,2,1).$$ Hence $$\begin{align}\sum_{m,n=1\, m\neq n}^{\infty}{\frac{m^2+n^2}{mn(m^2-n^2)^2}} &=\frac{T(1,1,2)-\zeta(4)/4}{2} +T(1,2,1)\\ &=\frac{11\zeta(4)}{8} \end{align}$$ where we used $$T(1,1,2)=\zeta(4)/2\quad,\quad T(1,2,1)=5\zeta(4)/2$$ (see page 31 in The evaluation of Tornheim double sums).

Answer 2

From Robert Z's answer, we have $$\frac{m^2+n^2}{mn(m^2-n^2)^2}=\frac{1}{2mn(m+n)^2}+\frac{1}{2mn(m-n)^2}.$$ Hence, the required sum $$S=\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m\neq n}}}\frac{1}{mn(m^2-n^2)^2}$$ can be split into two parts: $$S_1=\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m\neq n}}}\frac{1}{2mn(m+n)^2}=\sum_{m,n\in\Bbb N_1}\frac{1}{2mn(m+n)^2}-\sum_{n=1}^\infty\frac{1}{8n^4}$$ and $$S_2=\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m\neq n}}}\frac{1}{2mn(m-n)^2}=\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m>n}}}\frac{1}{2mn(m-n)^2}+\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m< n}}}\frac{1}{2mn(n-m)^2}.$$

Let's look at $S_2$ more closely. The first summand satisfies $$\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m>n}}}\frac{1}{2mn(m-n)^2}=\sum_{m,k\in\Bbb{N}_1}\frac{1}{2m(m+k)k^2},$$ where $k=m-n$. The second summand satisfies $$\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m>n}}}\frac{1}{2mn(m-n)^2}=\sum_{n,l\in\Bbb{N}_1}\frac{1}{2(n+l)nl^2},$$ where $l=n-m$. Therefore, we can rename the dummy variables and get \begin{align}S_2&=\sum_{m,k\in\Bbb{N}_1}\frac{1}{2m(m+k)k^2}+\sum_{n,l\in\Bbb{N}_1}\frac{1}{2(n+l)nl^2}\\&=\sum_{m,n\in\Bbb{N}_1}\frac{1}{2mn^2(m+n)}+\sum_{m,n\in\Bbb{N}_1}\frac{1}{2m^2n(m+n)}.\end{align} So, $$S_2=\sum_{m,n\in\Bbb{N}_1}\left(\frac{1}{2mn^2(m+n)}+\frac{1}{2m^2n(m+n)}\right)=\sum_{m,n\in \Bbb{N}_1}\frac{1}{2m^2n^2}.$$ This proves that $$S_2=\frac{1}{2}\sum_{m\in\Bbb{N}_1}\frac{1}{m^2}\sum_{n\in\Bbb{N}_1}\frac{1}{n^2}=\frac{1}{2}\big(\zeta(2)\big)\big(\zeta(2)\big)=\frac{1}{2}\left(\frac{\pi^2}{6}\right)^2=\frac{\pi^4}{72}.$$

Now, for $S_1$, we see that the last term is $$\sum_{n=1}^\infty\frac{1}{8n^4}=\frac{1}{8}\zeta(4)=\frac{1}{8}\left(\frac{\pi^4}{90}\right)=\frac{\pi^4}{720}.$$ We are left with the first term, which is $$\sum_{m,n\in\Bbb{N}_1}\frac{1}{2mn(m+n)^2}=\zeta(3,1)=\frac{1}{4}\zeta(4)=\frac{1}{4}\left(\frac{\pi^4}{90}\right)=\frac{\pi^4}{360},$$ by Robert Z's answer. Thus, $$S_1=\frac{\pi^4}{360}-\frac{\pi^4}{720}=\frac{\pi^4}{720}.$$

Combining the two results above, we have $$S=S_1+S_2=\frac{\pi^4}{720}+\frac{\pi^4}{72}=\frac{11\pi^4}{720}=\frac{11}{8}\zeta(4).$$ Here is another great resource to learn about the multiple zeta function. Corollary 1.64 on page 26 and discussions that lead to this corollary are important to this problem.

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Here is an elementary way to show that $\zeta(3,1)=\frac{1}{4}\zeta(4)$. Recall that $$\zeta(s_1,s_2,\ldots,s_t)=\sum_{\substack{{n_1,n_2,\ldots,n_t\in\Bbb{N}_1} \\{n_1>n_2>\ldots>n_t}}} \frac{1}{n_1^{s_1}n_2^{s_2}\cdots n_t^{s_t}}.$$ First, observe that $$\frac{1}{m^2n^2}=\frac{1}{(m+n)^2n^2}+\frac{1}{(m+n)^2m^2}+\frac{2}{(m+n)^3m}+\frac{2}{(m+n)^3n}.$$ Taking the sum of the expression above over $m,n\in\Bbb{N}_1$, we get $$\big(\zeta(2)\big)^2=\zeta(2,2)+\zeta(2,2)+2\zeta(3,1)+2\zeta(3,1)=2\big(\zeta(2,2)+2\zeta(3,1)\big),\tag{1}$$ so $$\zeta(2,2)+2\zeta(3,1)=\frac{1}{2}\big(\zeta(2)\big)^2=\frac{5}{4}\zeta(4).$$ Now, note that $$\frac{5}{2}\zeta(4)=\big(\zeta(2)\big)^2=\sum_{m,n\in\Bbb{N}_1}\frac{1}{m^2n^2}=\sum_{n\in\Bbb{N}_1}\frac{1}{n^4}+\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m>n}}}\frac{1}{m^2n^2}+\sum_{\substack{{m,n\in\Bbb{N}_1}\\{m<n}}}\frac{1}{m^2n^2}.$$ Hence, $$\frac{5}{2}\zeta(4)=\zeta(4)+\zeta(2,2)+\zeta(2,2),$$ making $$\zeta(2,2)=\frac{3}{4}\zeta(4).\tag{2}$$ Combining (1) and (2), we obtain $$\zeta(3,1)=\frac{1}{2}\left(\frac{5}{4}\zeta(4)-\zeta(2,2)\right)=\frac{1}{4}\zeta(4).$$

Answer 3

Partial Fractions gives $$ \frac{m^2+n^2}{\left(m^2-n^2\right)^2}=\frac{1/2}{(m-n)^2}+\frac{1/2}{(m+n)^2}\tag1 $$ First, we will compute $$ \begin{align} \sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{mn(m-n)^2} &=2\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{mn(m-n)^2}\tag{2a}\\ &=2\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{(m+n)nm^2}\tag{2b}\\ &=2\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac1{m^3}\tag{2c}\\ &=2\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac1{nm^2}\tag{2d}\\ &=\sum_{n=1}^\infty\sum_{m=1}^\infty\left(\frac1n-\frac1{m+n}\right)\frac{m+n}{nm^3}\tag{2e}\\ &=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{m^2n^2}\tag{2f}\\[3pt] &=\frac{\pi^4}{36}\tag{2g} \end{align} $$ Explanation:
$\text{(2a)}$: symmetry between $m\lt n$ and $n\lt m$
$\text{(2b)}$: substitute $m\mapsto m+n$
$\text{(2c)}$: partial fractions
$\text{(2d)}$: swap $m$ and $n$ in $\text{(2b)}$ then partial fractions
$\text{(2e)}$: average $\text{(2c)}$ and $\text{(2d)}$
$\text{(2f)}$: simplify
$\text{(2g)}$: evaluate $\zeta(2)^2$

Next, we will compute $$ \begin{align} \sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{mn(m+n)^2} &=\sum_{m=1}^\infty\sum_{n=1}^\infty\left(\color{#C00}{\frac1{nm^2(n+m)}}-\color{#090}{\frac1{m^2(n+m)^2}}\right)\tag{3a}\\ &=\color{#C00}{\frac{\pi^4}{72}}-\color{#090}{\frac{\pi^4}{120}}\tag{3b}\\[3pt] &=\frac{\pi^4}{180}\tag{3c} \end{align} $$ Explanation:
$\text{(3a)}$: partial fractions
$\text{(3b)}$: the red sum is $\frac12$ of $\text{(2b)}$, the green sum is $\frac12\left(\zeta(2)^2-\zeta(4)\right)$
$\text{(3c)}$: simplify

Therefore, $$ \begin{align} \sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{mn(m+n)^2} &=\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{mn(m+n)^2}-\frac14\zeta(4)\tag{4a}\\ &=\frac{\pi^4}{180}-\frac{\pi^4}{360}\tag{4b}\\[9pt] &=\frac{\pi^4}{360}\tag{4c} \end{align} $$ Explanation:
$\text{(4a)}$: subtract the terms where $m=n$
$\text{(4b)}$: apply $\text{(3c)}$
$\text{(4c)}$: simplify

Thus. $$ \begin{align} \sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{m^2+n^2}{mn\left(m^2-n^2\right)^2} &=\frac12\frac{\pi^4}{36}+\frac12\frac{\pi^4}{360}\tag{5a}\\ &=\frac{11\pi^4}{720}\tag{5b} \end{align} $$ Explanation:
$\text{(5a)}$: apply $(1)$, $(2)$, and $(4)$
$\text{(5b)}$: simplify