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.(adsbygoogle = window.adsbygoogle || []).push({});

[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.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Schlaefli integral satisfies Legendre equation

Loading...

Similar Threads for Schlaefli integral satisfies |
---|

B Calculate the expression of the antiderivative |

A Newmark-Beta vs Predictor Corrector |

A Can Newton's method work with an approximated integral |

I Three different integration schemes |

B Simple double integration of square wave question |

**Physics Forums | Science Articles, Homework Help, Discussion**