# Normalization constant for Legendre Polynomials

## Homework Statement

I am following a derivation of Legendre Polynomials normalization constant.

## Homework Equations

$$I_l = \int_{-1}^{1}(1-x^2)^l dx = \int_{-1}^{1}(1-x^2)(1-x^2)^{l-1}dx = I_{l-1} - \int_{-1}^{1}x^2(1-x^2)^{l-1}dx$$

The author then gives that we get the following relationship after integrating by parts the last integral:

$$I_l = I_{l-1} - I_l/(2l)$$

## The Attempt at a Solution

I can't get the last relationship by applying integration by parts. For the integration by parts, I set $$u = (1-x^2)^l-1$$ and $$dv = x^2dx$$. The $$u.v$$ term is equal to zero, since the integration goes from -1 to +1. However, for the - $$\int vdu$$ term I get

$$\int_{-1}^{1} x^3/3(1-x^2)^{l-2}(l-1)(2x)dx$$

I really cannot see how to get the final relationship by using this result of the integration by parts.

Any help will be appreciated.

Thanks

SammyS
Staff Emeritus
Homework Helper
Gold Member
$$I_l = \int_{-1}^{1}(1-x^2)^l dx = \int_{-1}^{1}(1-x^2)(1-x^2)^{l-1}dx = I_{l-1} - \int_{-1}^{1}x^2(1-x^2)^{l-1}dx$$
...

Thanks
For the integral:

$$\displaystyle - \int_{-1}^{1}x^2(1-x^2)^{\ell-1}\,dx$$

try integration by parts with:

$$\displaystyle u=x\ \to\ du=dx \ \text{ and }\ dv=-2x(1-x^2)^{\ell-1}\,dx\ \to\ v={{1}\over{\ell}}(1-x^2)^\ell\,.$$

Thank you very much.