Using recurrence formula to solve Legendre polynomial integral

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Discussion Overview

The discussion centers around proving an integral expression involving Legendre polynomials, specifically the relation between the integral of the polynomial and its recurrence relation. The scope includes mathematical reasoning and exploration of integral calculus techniques.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes to prove the expression $$ \int _{0}^{1}p_{l}(x)dx=\frac{p_{l-1}(0)}{l+1} $$ for $$ l \geq 1 $$ using a recurrence relation.
  • The same participant attempts to apply integration by parts but expresses confusion about how to handle the terms resulting from the integration.
  • Another participant suggests that the substitution in the integral was not performed correctly and encourages a reevaluation.
  • A third participant references an external source that provides a formula for the integral of Legendre polynomials and notes that $$ P_l(1) = 1 $$ for all $$ l $$, indicating that only one term is necessary in the numerator.
  • This participant also points out a potential typo in the original post regarding the term involving $$ x P_l(x) $$, suggesting it should be $$ x P_l'(x) $$ instead.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the initial substitution and the approach to solving the integral. There is no consensus on the resolution of the integral or the validity of the steps taken by the first participant.

Contextual Notes

There are unresolved aspects regarding the handling of integration by parts and the specific terms involved in the integral expressions. The discussion reflects uncertainty about the correct application of the recurrence relation and the properties of Legendre polynomials.

TheGreatDeadOne
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TL;DR
Using recurrence formula to solve Legendre polynomial integral
I am trying to prove the following expression below:

$$ \int _{0}^{1}p_{l}(x)dx=\frac{p_{l-1}(0)}{l+1} \quad \text{for }l \geq 1 $$

The first thing I did was use the following relation:

$$lp_l(x)+p'_{l-1}-xp_l(x)=0$$

Substituting in integral I get:

$$\frac{1}{l}\left[ \int_0^1 xp'_l(x)dx \quad- \quad \int_0^1 p'_{l-1}(x)dx\right]$$

For the first integral (from the left), I use integration by parts, making u = x and v'= p′l(x) ( I also tried to apply $p_1(x) = x$), and for the second I just integrate. But I end up getting stuck in that part, because I kind of don't know the best way to solve it. I don't know how to make the uv part disappear. I'm a little confused by the integral of $p′_{l−1}$ (My doubt is in relation to the term with upper limit, also I don't know how to disappear with it)
I'm stuck here(I think it's wrong):

$$ Il= p_l(x) - p_{l-1}(1) + p_{l-1}(0) - \int_0^1 p_l(x)dx \quad \rightarrow \frac{I(1+l)}{l}=p_l(x) - p_{l-1}(1) + p_{l-1}(0)$$
 
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doesn't look like you did the substitution correctly, try again.
 
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See http://physicspages.com/pdf/Mathematics/Legendre%20polynomials%20-%20recurrence.pdf
It might be worth checking your formulas with another source.
 
Using (25) in the above paper, I get ## \int\limits_{0}^{1}P_l(x) \, dx=(P_l(1)+P_{l-1}(0)-P_{l-1}(1))/(l+1) ##.
Edit: I also see somewhere in the literature that ##P_l(1)= 1## for all ## l ##, so only the middle term is needed in the numerator.
(25) shows the ## x P_l(x) ## term you have should be ## x P_l'(x) ##. I see that appears to be just a typo on your part. Do it carefully, you should get the correct result, with the numerator as described above.
 
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