Schlaefli integral satisfies Legendre equation

  • #1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
56
8
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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