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Math Help - Some integrals

  1. #1
    Forum Admin topsquark's Avatar
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    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
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by topsquark View Post
    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
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  3. #3
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    Quote Originally Posted by topsquark View Post
    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
    Last edited by NonCommAlg; August 20th 2008 at 10:53 AM. Reason: a couple of typos fixed!
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