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Schlaefli integral satisfies Legendre equation

  1. Sep 17, 2014 #1
    The integral representation of Legendre functions is [itex] P_\nu(z) = \oint_{\Gamma} \frac{(w^2-1)^\nu}{(w-z)^{\nu+1}} dw[/itex]. I'm trying to show that this satisfies Legendre's equation. When I take the derivatives and plug it into the equation, I just get a nasty expression with nasty integrals times functions of z. I don't see how I can combine any of the terms, since they have different powers of z, and all the integrals look different.

    [itex]P'_\nu(z) = \oint_{\Gamma} \frac{(w^2-1)^\nu(\nu+1)}{(w-z)^{\nu+2}} dw[/itex]
    and
    [itex]P''_\nu(z) = \oint_{\Gamma} \frac{(w^2-1)^\nu(\nu+1)(\nu+2)}{(w-z)^{\nu+3}} dw[/itex]
    So the expression is
    [itex](1-z^2)\oint_{\Gamma} \frac{(w^2-1)^\nu(\nu+1)(\nu+2)}{(w-z)^{\nu+3}} dw
    -2z \oint_{\Gamma} \frac{(w^2-1)^\nu(\nu+1)}{(w-z)^{\nu+2}} dw
    +\nu(\nu+1)\oint_{\Gamma} \frac{(w^2-1)^\nu}{(w-z)^{\nu+1}} dw
    [/itex]
    Which I need to somehow show is equal to zero. I tried to collect all the integrals together, hoping they would combine together to give me a nice term. At first I was hoping it would give me [itex] \frac{d}{dw} [/itex] of the integrand, so I could just use Cauchy's theorem to show that the integral is zero. But the terms didn't seam to collect that way.
     
  2. jcsd
  3. Sep 19, 2014 #2
    Your guess was correct, you get a derivative inside the integral. Just substitute in the differential equations and arrange terms:

    [tex]
    (1-z^2)\frac{d^2 P_\nu(z)}{d z^2} - 2z\frac{d P_\nu(z)}{dz} + \nu(\nu +1)P_\nu(z) = (\nu + 1)\oint_{\Gamma} \frac{(w^2 -1)^\nu}{(w-z)^{\nu + 3}}(-(\nu+2)(w^2 -1) + 2(\nu + 1)w(w-z)) d\,w,
    [/tex]
    which equals

    [tex](\nu +1)\oint_{\Gamma} \frac{d}{dw}\left( \frac{(w^2 - 1)^{\nu +1}}{(w-z)^{\nu +2}} \right) d\,w.[/tex]

    When [itex]\nu [/itex] is an integer, this equals zero. Otherwise you have too be a little bit careful with the contour.
     
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