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Normalization constant for Legendre Polynomials

  • Thread starter rafaelpol
  • Start date
  • #1
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Homework Statement



I am following a derivation of Legendre Polynomials normalization constant.

Homework Equations



[tex]

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

[/tex]

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

[tex]

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

[/tex]

The Attempt at a Solution



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

[tex]

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

[/tex]

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
 

Answers and Replies

  • #2
SammyS
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[tex]
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
[/tex]
...

Thanks
For the integral:

[tex]\displaystyle - \int_{-1}^{1}x^2(1-x^2)^{\ell-1}\,dx[/tex]

try integration by parts with:

[tex]\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\,.[/tex]
 
  • #3
17
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Thank you very much.
 

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