$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \frac{ \sin[ka(m-n)]}{(m-n)} , m \neq n $
where $k$ and $a$ are constants.
How to treat this double sum?
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1Do you see why you don't need separate constants $k$ and $a$ but can replace their product by simply $b$? – David G. Stork Nov 21 '20 at 01:20
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Yes I can see it, but still can´t achieve to the solution. – Relax Environment Nov 21 '20 at 01:27
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So as a courtesy to the community, simplify your equations so we're not presented with irrelevant material. – David G. Stork Nov 21 '20 at 04:54
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Hint: Use the fact that $$\frac{\sin[(m-n)ka]}{m-n} = \frac 1 2\int_{-ka}^{ka}e^{i(m-n)t}dt$$
Stefan Lafon
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This is what i got. If there is any error, let me know, please.