# Elliptic Integral

by G01
Tags: elliptic, integral
 HW Helper P: 2,685 1. The problem statement, all variables and given/known data Sub problem from a much larger HW problem: From previous steps we arrive at a complete elliptic integral of the second kind: $$E(k)=\int_0^{\pi/2} dx \sqrt{1+k^2\sin^2x}$$ In the next part of the problem, I need to expand this integral and approximate it by truncating at the first order term. (k is large) 2. Relevant equations $$E(k) = \frac{\pi}{2} \sum_{n=0}^{\infty} \left[\frac{(2n)!}{2^{2 n} (n!)^2}\right]^2 \frac{k^{2n}}{1-2 n}$$ 3. The attempt at a solution I believe I should use the expansion quoted above. Here is my question. Based of the previous steps I know that k^2 has to be large. Also, the sign of k^2 is opposite of what is is in the standard form of E(k). So, 1. Does this expansion truncated at first order approximate the integral well if k^2 is large? I think not. Is there another expansion, one for large k^2, that I can potentially use? 2. Can I just change the sign in the odd terms of the expansion to account for the sign change of k? I think this should work.
 Sci Advisor HW Helper PF Gold P: 2,602 The expansion above is not valid for large $$k$$. Simply rewrite $$E(k)=\int_0^{\pi/2} dx \sqrt{1+k^2\sin^2x} = \pm \int_0^{\pi/2} dx k \sin x\sqrt{1+\frac{1}{k^2\sin^2x}}$$ and expand this in $$1/k$$ yourself. Note that taking the square root introduces a sign ambiguity that should be chosen to give the correct sign to $$E(k)$$.
HW Helper
P: 2,685
 Quote by fzero The expansion above is not valid for large $$k$$. Simply rewrite $$E(k)=\int_0^{\pi/2} dx \sqrt{1+k^2\sin^2x} = \pm \int_0^{\pi/2} dx k \sin x\sqrt{1+\frac{1}{k^2\sin^2x}}$$ and expand this in $$1/k$$ yourself. Note that taking the square root introduces a sign ambiguity that should be chosen to give the correct sign to $$E(k)$$.
Hmm. This makes sense, however, it is only a good approximation when sin(x) is larger than 1/m.

However, the contributions to the integral in [0,1/m] are small, so can I fix the approximation by changing the lower bound on the integral to 1/m?