Elliptic integral power series expansion

**hairymclairy** · November 25th 2011, 07:25 AM

Hi,

I've attached a problem I've come across and was wondering if anyone could give me a hand?

Cheers.

**Opalg** · November 25th 2011, 08:30 AM

You need to evaluate $I_k = \int_0^{\pi/2}\sin^{2k}\phi\,d\phi.$ Integrate by parts:

$\begin{aligned}I_k &= \int_0^{\pi/2}\sin\phi\sin^{2k-1}\phi\,d\phi \\ &= \Bigl[-\cos\phi\sin^{2k-1}\phi\Bigr]_0^{\pi/2} + \int_0^{\pi/2}(2k-1)\cos^2\phi\sin^{2k-2}\phi\,d\phi \\ &= \int_0^{\pi/2}(2k-1)(1-\sin^2\phi)\sin^{2k-2}\phi\,d\phi = (2k-1)(I_{k-1}-I_k).\end{aligned}$

Therefore $I_k = \frac{2k-1}{2k}I_{k-1} = \frac{k-\frac12}kI_{k-1}.$

Now check that $I_1=\pi/4$ and use induction to show that $I_k = \frac{\pi}2\,\frac{\bigl(\frac12\bigr)_k}{k!}.$

Thread: Elliptic integral power series expansion

LinkBack

Thread Tools

Display

Elliptic integral power series expansion

Re: Elliptic integral power series expansion

Similar Math Help Forum Discussions

power series expansion

power series expansion II

Power series expansion

power series expansion

series expansion of an elliptic integral

Search Tags