Evaluate the double sum $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}$

Question

As a follow up of this nice question I am interested in

$$ S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2} $$

Furthermore, I would be also very grateful for a solution to

$$ S_2=\sum_{m=1}^{\infty}\sum_{n=m+1}^{\infty}\frac{ 1}{m n\left(m^2-n^2\right)^2} $$

Following my answer in the question mentioned above and the numerical experiments of @Vladimir Reshetnikov it's very likely that at least

$$ S_1+S_2 = \frac{a}{b}\pi^6 $$

I think both sums may be evaluated by using partial fraction decomposition and the integral representation of the Polygamma function but I don't know how exactly and I guess there could be a much more efficient route.

The inner summation in the first line is summed from $n=0$ to $n-1$. This needs to be adjusted to make sense. As it should also be adjusted in the problem from whence this is first presented. — Leucippus, May 22 '15 at 14:06
Sorry for the cross-edits, just making sure the title was covered too — abiessu, May 22 '15 at 14:13
@RenatoFaraone Makes no difference (for $S_1$), since the inner sum is an empty sum for $m = 1$. — Daniel Fischer, May 22 '15 at 14:28

score 8 · Accepted Answer · edited Apr 13 '17 at 12:21

Clearly, $S_1$=$S_2$ (this can be shown by reversing the order of summation, as was noted above). Using $$ \frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{ (m+n)^2-(m-n)^2}{4 m^2 n^2\left(m^2-n^2\right)^2} $$ we get $$ S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m-n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}, $$ and after reversing the order of summation in the first sum $$ S_1=\frac{1}{4}\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty}\frac{ 1}{m^2 n^2\left(m-n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}=\\ \frac{1}{4}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}. \qquad\qquad (1) $$

Let's introduce a third sum $$ S_3=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2} =\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}=\\ \frac{1}{2}\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}-\frac{1}{8}\zeta(6). $$ Using An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$? we get $$ S_3=\frac{1}{2}\cdot\frac{1}{3}\zeta(6)-\frac{1}{8}\zeta(6)=\frac{1}{24}\zeta(6).\qquad\qquad\qquad (2) $$ From (1) and (2) we get

$$ S_1=\frac{1}{4}\cdot\frac{1}{3}\zeta(6)-\frac{1}{4}\cdot\frac{1}{24}\zeta(6)=\frac{7}{96}\zeta(6)=\frac{7}{96}\frac{\pi^6}{945}=\frac{\pi^6}{12960} $$

excellent work (+1) , i was failing after your formula one back then....! — tired, Oct 18 '15 at 10:48

score 1 · Answer 2 · answered May 22 '15 at 15:24

1

Numerically, I get $$ S_1+S_2 = 0.14836252987273216621 $$ which agrees with $$ \frac{\pi^6}{6480} $$ Also numerically, $$ S_1 = 0.074181264936366083104 \\ S_2 = 0.074181264936366083104 $$ are seemingly equal.

answered May 22 '15 at 15:24

GEdgar

111,679

very well! multiplying this by $8\pi$ and adding $\pi^7/70$gives us the result suggested by @Vladimir Reshetnikov – tired May 22 '15 at 15:31
Do you have an idea where this symmerty come from? I'm not getting my head around it :/ – tired May 22 '15 at 15:54
2

Reverse the order of summation in $S_2$ and see what you get. – GEdgar May 22 '15 at 16:41
thanks, oh man today is not my brightest day! – tired May 22 '15 at 17:25

Evaluate the double sum $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}$

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