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I am working on the problem of pde and I somehow get this infinite sum

$$\sum_{n=-\infty}^ \infty \frac{y/\pi}{(x-x_0+2nL)^2+y^2}-\sum_{n=-\infty}^ \infty \frac{y/\pi}{(x+x_0+2nL)^2+y^2}$$

Does this expression make sense? If it does, does it converge? And if so, is it possible to write down a simpler expression?

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  • $\begingroup$ In general, $$\sum_{n=-\infty}^\infty\frac1{n+a}~=~\pi~\cot(a\pi).$$ $\endgroup$ – Lucian Oct 24 '16 at 5:38
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From this answer one has the result

$$\sum_{n=-\infty}^{\infty} \frac1{(z+n)^2+a^2} = \frac{\pi}{2 a} \frac{\sinh{2 \pi a}}{\sin^2{\pi z}+\sinh^2{\pi a}}$$

In your case, we rewrite the expression as

$$\frac{y}{4 \pi L^2} \sum_{n=-\infty}^{\infty} \left [\frac1{\left (n+\frac{x-x_0}{2 L}\right )^2+\frac{y^2}{4 L^2}} - \frac1{\left (n+\frac{x+x_0}{2 L}\right )^2+\frac{y^2}{4 L^2}} \right ]$$

which, from the above result may be simplified to the form

$$\frac{1}{4 L} \left [\frac{\sinh{\left (\frac{\pi y}{L} \right )}}{\sin^2{\left (\frac{\pi (x-x_0)}{2 L} \right )}+\sinh^2{\left (\frac{\pi y}{2 L} \right )}} - \frac{\sinh{\left (\frac{\pi y}{L} \right )}}{\sin^2{\left (\frac{\pi (x+x_0)}{2 L} \right )}+\sinh^2{\left (\frac{\pi y}{2 L} \right )}}\right ] $$

This can be simplified in many different ways, depending on what you want from the final expression.

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