Using recurrence formula to solve Legendre polynomial integral

  • #1

Summary:

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):

$$ Il= p_l(x) - p_{l-1}(1) + p_{l-1}(0) - \int_0^1 p_l(x)dx \quad \rightarrow \frac{I(1+l)}{l}=p_l(x) - p_{l-1}(1) + p_{l-1}(0)$$
 

Answers and Replies

  • #2
Dr Transport
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doesn't look like you did the substitution correctly, try again.
 
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  • #4
Charles Link
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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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