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?

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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.