# How do I apply the product (capital pi) symbol here?

I know $\prod$ is the product symbol but I'm not sure how to apply it in the result shown below. The $a$, $b$, and $d$ are constants.

$$\frac{\prod\left(\frac{(b - a)(b + a)}{d^2 + b^2}\,;x\,\middle|\,\frac{(b - a)(b + a)}{d^2 + b^2}\right)}{(d^2 + b^2)^\frac{3}{2}}$$

Any help appreciated. By the way, this result above came from the integration of this: $1/(d^2 + b^2\cos^2x+a^2\sin^2x)^\frac{3}{2}$

EDIT: How do I evaluate that result to a numerical result? (e.g. With integration limits $0$ to $2\pi$)

• How, then, did you come across that result, if you say that you're doing this by hand? Apr 16, 2017 at 1:23
• You didn't ask me how I came across the result, you asked me how I was going to be evaluating it. I arrived at the integral by working through the mechanics of deriving the magnetic field due to current flowing around a loop of wire in shape of ellipse. When I got to that integral I could not work it so I plugged it in to an online tool that integrates for you. Hence the result with the capital pi that I did not know how to evaluate to a numerical solution. This is not some school problem, I'm 50 years old. Apr 16, 2017 at 1:28
• I did not assume this was a school problem at all. :) I was just curious about the situation that led you to an elliptic integral. In any event: I totally missed that you mentioned the integration limits; in that case, it is possible that you have a complete elliptic integral, and you can use AGM if you really want to compute things by hand, and are alright with taking square roots. I'll need to think about it... Apr 16, 2017 at 1:35
• Thank you J.M. I would appreciate any direction you can afford me. No hurry. Apr 16, 2017 at 1:38
• Also, I have Mathcad and Matlab if there is a way to evaluate with those tools. I would much prefer to understand how to evaluate by hand too. I am not in a hurry here, I am trying to really get a good physical grasp of this problem I am working on. Apr 16, 2017 at 1:41

That is not the product symbol. It is the incomplete elliptic integral of the third kind (A non-elementary function).

It is defined as:

$$\Pi(n ; \varphi \,|\,m) = \int_{0}^{\sin \varphi} \frac{1}{1-nt^2} \frac{dt}{\sqrt{\left(1-m t^2\right)\left(1-t^2\right) }}$$

You will probably not be able to simplify your result further.

• Ok, so how do I evaluate that to a numerical result? e.g. Assume integrating between 0 and 2pi. Apr 15, 2017 at 23:16
• @relayman, depends. What computing environment are you using? Apr 16, 2017 at 0:41
• Pen and paper, doing it by hand. Apr 16, 2017 at 0:47
• @relayman, then, unless you have access to tables of the elliptic integral of the third kind (e.g. the ones in Abramowitz and Stegun), or willing to use a numerical quadrature method, then you can't really do much with pen and paper. Apr 16, 2017 at 1:22
• @J.M.isn'tamathematician Ok, I found a copy of "Handbook of Mathematical Functions" by Abramowitz and Stegun. It is a doosey at 1060 pages. cs.bham.ac.uk/~aps/research/projects/as/resources/…. Page 599 has Elliptic Integrals of the third Kind. Not sure what to do from there. Apr 16, 2017 at 2:15

This is not a product: it's an elliptic integral (of the third kind, according to Legendre's scheme), defined by $$\Pi(n;\varphi \mid m) = \int_0^{\sin{\varphi}} \frac{dt}{(1-nt^2)\sqrt{(1-t^2)(1-mt^2)}}.$$ More information about elliptic integrals may be found on Wikipedia, Mathworld, the DLMF pages, or the book Modern Analysis by Whittaker and Watson (published in the 1920s, now freely available online), Chapter 22. Be aware that the notation varies by source: some use $k^2$ instead of $m$, or something else instead of $\sin{\varphi}$.

• Here is a short discussion of the various (confusing) notations for elliptic integrals. In addition to what was said there: some computing systems use $x=\sin\varphi$ instead of the amplitude $\varphi$ as one of the arguments. And particularly in the case of the elliptic integral of the third kind, the characteristic $n$ is sometimes changed to $\alpha^2=n$. Apr 16, 2017 at 0:40