# Math Help - Some integrals

1. ## Some integrals

These cropped up in one of my calculations. I have forgotten if they are even possible to integrate and I can't find the reference, so I'd appreciate a hand. (I will note that at least one of these looks suspiciously like an elliptic integral.)
$\int_0^{2 \pi} (b - a~cos( \phi - \alpha ) )^{3/2} \left \{ \begin{matrix} 1 \\ sin( \alpha ) \\ cos( \alpha ) \end{matrix} \right \} ~ d \alpha$
We may take b > a > 0 WLOG.

Any help/thoughts would be appreciated.

-Dan

2. Originally Posted by topsquark
These cropped up in one of my calculations. I have forgotten if they are even possible to integrate and I can't find the reference, so I'd appreciate a hand. (I will note that at least one of these looks suspiciously like an elliptic integral.)
$\int_0^{2 \pi} (b - a~cos( \phi - \alpha ) )^{3/2} \left \{ \begin{matrix} 1 \\ sin( \alpha ) \\ cos( \alpha ) \end{matrix} \right \} ~ d \alpha$
We may take b > a > 0 WLOG.

Any help/thoughts would be appreciated.

-Dan
Not much help but you can simplify this slightly to:

$b^{3/2}\int_0^{2 \pi} ( 1- c~\cos( \phi - \alpha ) )^{3/2} \left \{ \begin{matrix} 1 \\ \sin( \alpha ) \\ \cos( \alpha ) \end{matrix} \right \} ~ d \alpha$

where $1>c=a/b$.

RonL

3. Originally Posted by topsquark
These cropped up in one of my calculations. I have forgotten if they are even possible to integrate and I can't find the reference, so I'd appreciate a hand. (I will note that at least one of these looks suspiciously like an elliptic integral.)
$\int_0^{2 \pi} (b - a~cos( \phi - \alpha ) )^{3/2} \left \{ \begin{matrix} 1 \\ sin( \alpha ) \\ cos( \alpha ) \end{matrix} \right \} ~ d \alpha$
We may take b > a > 0 WLOG.

Any help/thoughts would be appreciated.

-Dan
all three integrals are linear combinations of elliptic integrals of the first and the second kind.

i'll give you the solution for the first integral. the other two have the same idea:

let $b=ca, \ \phi - \alpha = x, \ J=\int_0^{\pi} \sqrt{c-\cos x} \ dx, \ K=\int_0^{\pi} \frac{dx}{\sqrt{c-\cos x}},$ and $L=\int_0^{\pi} \sqrt{(c-\cos x)^3} \ dx.$

it's easy to see that $L= \frac{4cJ +(1-c^2)K}{3},$ which we'll call it $(1).$ now we have the following:

$I=\int_0^{2\pi}\sqrt{(b-a\cos(\phi - \alpha))^3} \ d \alpha=a\sqrt{a} \int_{\phi - 2\pi}^{\phi} \sqrt{(c - \cos x)^3} dx$

$=a\sqrt{a} \int_0^{2\pi} \sqrt{(c - \cos x)^3} \ dx=2 a\sqrt{a}\int_0^{\pi} \sqrt{(c - \cos x)^3} \ dx.$ thus by $(1)$ we have:

$I=\frac{2a\sqrt{a}(4cJ + (1-c^2)K)}{3}. \ \ \ \ \ \ \ (2)$

now we find $J$ and $K$ in terms of elliptic integrals of the second and first kind respectively: put $x=2\theta.$ then:

$J=2\int_0^{\frac{\pi}{2}} \sqrt{c-1+2\sin^2 \theta} \ d \theta = 2\sqrt{c-1} \int_0^{\frac{\pi}{2}} \sqrt{1 + \frac{2}{c-1} \sin^2 \theta} \ d \theta$

$=2\sqrt{c-1}E \left(\frac{\pi}{2} \mid \frac{2}{1-c} \right).$ similarly we have: $K=\frac{2}{\sqrt{c-1}} F\left(\frac{\pi}{2} \mid \frac{2}{1-c} \right).$ thus by $(2)$ we will have:

$I=\frac{4a \sqrt{a(c-1)}}{3}\left[4c E \left(\frac{\pi}{2} \mid \frac{2}{1-c} \right) -(c+1) F \left(\frac{\pi}{2} \mid \frac{2}{1-c} \right) \right]. \ \ \ \ \square$

.................................................. .................................................. .

Remark: you can also find $I$ in the form of an infinite series:

since $\sigma_n=\int_0^{\pi} \cos^n x \ dx = \frac{(1+(-1)^n)\sqrt{\pi} \Gamma \left(\frac{n+1}{2} \right)}{2\Gamma \left(\frac{n+2}{2} \right)},$ we'll get:

$I=2a\sqrt{a}L=2a\sqrt{a}\int_0^{\pi}\sum_{n=0}^{\i nfty} \binom{\frac{3}{2}}{n}c^{\frac{3}{2}-n}(-\cos x)^n \ dx$

$=2ac \sqrt{ac}\sum_{n=0}^{\infty}(-1)^n\binom{\frac{3}{2}}{n}c^{-n}\int_0^{\pi} \cos^n x \ dx=2ac\sqrt{ac}\sum_{n=0}^{\infty} (-1)^n \binom{\frac{3}{2}}{n}\sigma_nc^{-n}. \ \ \ \square$