The book gave the integration of a function with the legendre polynomial formula:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_{-1}^1 f(x)P_n(x)dx = \frac{1}{2^n n!}\int_{-1}^{1} f^n(x)(x^2-1)^n dx[/tex]

It just said the formula can be obtained by repeat using Rodrigues formula and integral by parts but did not go into detail. I want to work out the steps and I got stuck. This is what I have:

Using Rodrigues:

[tex]\int_{-1}^{1} f(x)P_n(x)dx = \frac{1}{2^n n!}\int_{-1}^{1} f(x)\frac{d^n}{dx^n}[(x^2-1)^n] dx[/tex]

After the first integral by parts:

[tex]\frac{1}{2^n n!}\int_{-1}^{1} f(x)\frac{d^n}{dx^n}[(x^2-1)^n] dx = \frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 \;-\; \frac{1}{2^n n!}\int_{-1}^{1} f^{(1)}(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n] dx[/tex]

In order for this to continue to the next integration by parts, the following has to be true:

[tex]\frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 = 0[/tex]

[tex]\Rightarrow\; [\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 = 0[/tex]

I don't know whether my assumption is correct. If so, I still don't know how it is equal to zero. can anyone give me some guidance.

**Physics Forums - The Fusion of Science and Community**

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

# Please help integration using rodrigues formula.

Loading...

Similar Threads - Please help integration | Date |
---|---|

A Please help with 2-variable partial differential equation | Nov 4, 2016 |

Please help with this differential equation | Sep 20, 2013 |

Weird double integral. Please help | Jun 7, 2012 |

Integrating product of bessel function,please help! | Oct 20, 2010 |

Please help in integration of Associate Legendre function | Jun 20, 2010 |

**Physics Forums - The Fusion of Science and Community**