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I am stuck at this summation. $$\sum_{i\neq j}^{\infty}\frac{1}{(i^2-j^2)^2}$$
$i, j$ start from $1$.
I am kind of expecting an answer in terms of $\pi^4$ but that might not be true.
Also this is a sum I came across while researching something so I do not expect it to have an answer.

Miracle Invoker
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  • looks like it may well be $\pi^4 / 160 \approx 0.608806819$, numerically. – mjqxxxx Aug 31 '23 at 05:02
  • @mjqxxxx Nice! I was kind of leaning in that direction. How did you calculate it numerically by the way. – Miracle Invoker Aug 31 '23 at 05:03
  • Similar to https://math.stackexchange.com/questions/2986764/double-summation-sum-m-n-1-m-neq-n-infty-fracm2n2mnm2-n22?rq=1 – Blitzer Aug 31 '23 at 05:06
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    The beginning of this answer gives (but doesn't prove it?) an integral expression for the general term, in case that somehow helps. – Bruno B Aug 31 '23 at 12:09
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    Coming back to this: the proof for the series-integral equality is actually straightforward as long as you know how to justify integrability, and I was able to arrive to $$\sum_{i \neq j} \frac{1}{(i^2 - j^2)^2} = 2\int_0^1 \int_0^1 \frac{x}{y-x} \left(\frac{1}{1 - xy} - \frac{1}{1 - x^2}\right) \log(x)\log(y) ,\mathrm{d}x \mathrm{d} y$$ which Wolfram confirms (at least numerically) does go the value mentioned by @mjqxxxx. – Bruno B Aug 31 '23 at 16:24
  • Now there only remains to actually prove that the double integral is indeed $\pi^4/160$ by hand, I'll see what I can do and make it an answer if I succeed. – Bruno B Aug 31 '23 at 16:25
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    I've determined through calculating the integral from my above comment a different way, after simplifying that integrand into $\frac{x^2 \log x \log y}{(1 - xy) (1 - x^2)}$, that your double series is equal to twice the double series $\sum_{k,l = 1}^\infty \frac{1}{k^2(k+2l)^2}$, however I couldn't find a way to calculate this due to the $2l$ instead of $l$. In case someone knows how to handle such series. I'll stop here though. – Bruno B Sep 01 '23 at 07:27
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    Re: how to numerically find it ... I calculated it directly for $i,j < 50, 100, 200, 400, 800$. The increments between these decrease by a factor of about $2$ per step. (That is, e.g., $S(200) - S(100) \approx (1/2)(S(100) - S(50))$.) So the sequence was converging exponentially, with the remaining error always approximately equal to the last increment. So $2S(2N) - S(N)$ converges much faster. For $N=8192$, it's $\approx 0.6088068$, digits that can be looked up in the ISC to get $\pi^4/16$: https://wayback.cecm.sfu.ca/cgi-bin/isc/lookup?number=0.6088068&lookup_type=simple . – mjqxxxx Sep 01 '23 at 16:12

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$$ \begin{align} \sum_{i\neq j}^{\infty}\frac{1}{(i^2-j^2)^2} &= 2\sum_{j=1}^{\infty}\sum_{i=j+1}^{\infty}\frac{1}{(i+j)^2(i-j)^2} \\ &= 2\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{1}{(i+2j)^2i^2} \\ &= \sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\left(\frac{1}{(i+j)^2i^2} + \frac{(-1)^j}{(i+j)^2i^2}\right) \\ &= \sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\left(\frac{1}{(i+j)^2i^2} + \frac{(-1)^{i+j}(-1)^{i}}{(i+j)^2i^2}\right) \\ &= \sum_{i=1}^{\infty}\sum_{j=i+1}^{\infty}\left(\frac{1}{j^2i^2} + \frac{(-1)^{j}(-1)^{i}}{j^2i^2}\right) \\ &= \frac{1}{2}\left(\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\left(\frac{1}{j^2i^2} + \frac{(-1)^{j}(-1)^{i}}{j^2i^2}\right)-\sum_{i=1}^{\infty}\left(\frac{1}{i^4} + \frac{(-1)^{2i}}{i^4}\right)\right) \\ &= \frac{1}{2}\left(\sum_{i=1}^{\infty}\frac{1}{i^2}\sum_{j=1}^{\infty}\frac{1}{j^2} + \sum_{i=1}^{\infty}\frac{(-1)^{i}}{i^2}\sum_{j=1}^{\infty}\frac{(-1)^{j}}{j^2}-2\sum_{i=1}^{\infty}\frac{1}{i^4}\right) \\ &= \frac{1}{2}\left(\frac{\pi^4}{36} + \frac{\pi^4}{144}-2\frac{\pi^4}{90}\right) \\ &= \frac{\pi^4}{160}. \end{align} $$

user23571113
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