# Elliptic Integrals

In my homework I had to solve the following integral

$\displaystyle\int_0^\pi \mathrm{d}\Psi \frac{\cos\Psi}{\sqrt{1+2s(1-\cos\Psi)}}$ with some constant $s\ll1$

The solution said this is an "elliptic integral" which cannot be solved analytically, thus we expanded the square root in order to solve it.

Question 1: How do I recognize this as an elliptic integral? Wikipedia gives the following definition:

"An elliptic integral is an integral of the form

$$\int R\left(x,\sqrt{P(x)}\, \right)$$

where R is a rational function in two variables and P(x) is a polynomial of degree three or four with no repeated roots."

However, $1+2s(1-\cos\Psi)$, the argument of the square root of the given integral, is definitely not a polynomial of degree three or four (or is it?). So why is it an elliptic integral?

Question 2:: What is the dfinition of the notation $R\left(\cdot,\cdot \right)$? Am I save to assume $R\left(x,\sqrt{P(x)} \right)\propto \frac{x}{\sqrt{P(x)}}$?

Let $x=\cos \frac{t}{2}$, then $dx=-\frac{1}{2} \sin \frac{t}{2} \, dt \implies dt=-\frac{2\, dx}{\sqrt{1-x^{2}}}$.

Now $I=\int_{0}^{\pi} \frac{\cos t \, dt}{\sqrt{1+2s(1-\cos t)}} =2\int_{0}^{1} \frac{2x^{2}-1}{\sqrt{1+4s-4sx^{2}}} \frac{dx}{\sqrt{1-x^{2}}}$,

so the integrand has the denominator as a square root of a quartic.

• Bah, you swooped in just as I was about to post this (well, with $x=\sin(t/2)$ instead but same difference.) Nice job. It's worth pointing out that some additional manipulations are required to place it in the form shown below; appealing to the definition of complete elliptic integrals directly gives the (equivalent) form $(2+s^{-1})K(-4s)-s^{-1}E(-4s)$. – Semiclassical Dec 22 '15 at 20:37

Here two pages of Elliptic Integrals of Harris Hancock, I hope be useful for you.  Using Mathematica, show that it equals $\frac{(1+2s)E \left( \sqrt{\frac{4s}{1+4s}} \right)- (1+4s)K \left( \sqrt{\frac{4s}{1+4s}} \right)} {s\sqrt{1+4s}}$.

• The question is how you recognize the given integral as elliptic, and what the definition of $R(\cdot,\cdot)$ is, not what the result of the integral is. – Alice Ryhl Dec 22 '15 at 12:04