- #1

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## 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):

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

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

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):

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