- #1

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## 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.

[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.