Normalization constant for Legendre Polynomials

[rafaelpol]

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]

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

Thanks

For the integral:

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

try integration by parts with:

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

 

[rafaelpol]

Thank you very much.