4
$\begingroup$

I need to solve this sum: $$\sum_{n=-\infty}^\infty\frac{y^2}{[(x-n\pi)^2+y^2]^{3/2}}.$$ Do you have any ideas for how I could do this?

I know that this sum: $$\sum_{n=-\infty}^\infty\frac{y}{(x-n\pi)^2+y^2},$$ can be simplified by using the following identities (see this Wikipedia article): $$\cot(x+iy) = \sum_{n=-\infty}^\infty\frac{1}{x-n\pi + iy},$$ $$\sum_{n=-\infty}^\infty\frac{y}{(x-n\pi)^2+y^2}=\frac{i}{2}\sum_{n=-\infty}^\infty\frac{1}{x-n\pi+iy}-\frac{1}{x-n\pi-iy}.$$ Hence, $$\begin{aligned} \sum_{n=-\infty}^\infty\frac{y}{(x-n\pi)^2+y^2}&=\frac{i}{2}[\cot(x+iy)-\cot(x-iy)] \\ &=\frac{\sinh(2y)}{\cosh(2y)-\cos(2x)}. \end{aligned}$$

Do you know if a similar trick can be used to solve the original sum?

$\endgroup$
12
  • 2
    $\begingroup$ Is this from an electrostatics problem? This is more-or-less the $y$-component of the force on a charge located at $(x,y)$ due to an infinite array of unit charges on the $x$-axis spaced a distance $\pi$ apart. $\endgroup$ Sep 22, 2021 at 21:06
  • 1
    $\begingroup$ Yes, it is. I am actually trying to calculate the magnetic field along the axis of symmetry for an infinite number of current-carrying loops spaced $\pi$ apart. You can see the formula I am using here. Replace $z$ with $x$ and $R$ with $y$ and set all other quantities to 1. $\endgroup$
    – Peanutlex
    Sep 22, 2021 at 21:09
  • 2
    $\begingroup$ You can also use the Poisson summation formula to write this as a sum over a modified Bessel function of the second kind. (I'm just spitballing here.) $\endgroup$ Sep 22, 2021 at 21:14
  • 1
    $\begingroup$ That's good I want as many ideas as I can get. I'm not sure how the Poisson summation formula works. I'll look it up now. $\endgroup$
    – Peanutlex
    Sep 22, 2021 at 21:19
  • 1
    $\begingroup$ I find that Mathematica is unable to solve $\sum_{n = -\infty}^{+\infty}{\left(n^2 + 1\right)}^{-3/2}$, so I don't hold high hopes for this to have a closed-form solution. $\endgroup$ Sep 22, 2021 at 21:32

0

Your Answer

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

Browse other questions tagged or ask your own question.