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I have a question that I need to prove that for $n≥1$,

$$\frac{1}{2n}\int_{-1}^{1}x\frac{d}{dx}(P_n(x)^2)dx=\frac{2}{2n+1}$$

I have to evaluate the integral instead of using the orthogonality property of Legendre's polynomial which I am having difficulty doing. I have tried completing the integral using integration by parts however I get to a step and need to use the orthogonality property to simply it further.

I have used the relation $xP'_n(x)-P'_{n-1}(x) = nP_n(x)$ and the fact $\int_{-1}^{1}f(x)P_n(x)dx=0$ for any polynomial $f(x)$ of degree less than $n$ which is permitted in this question.

user00134857693
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1 Answers1

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Indeed we want to prove that $\int_{-1}^1 x P_n(x)P_n'(x)dx=\frac {2n}{2n+1}.$ By putting $P_n'(x)dx=dv$ and $x P_n(x)=u$, and using integration by parts we reach:

$$\int_{-1}^1 x P_n(x)P_n'(x)dx=\left.xP_n^2(x)\right\rvert_{-1}^{1}- \int_{-1}^1 P_n^2(x)+x P_n(x)P_n'(x)dx;$$

hence:

$$2\int_{-1}^1 x P_n(x)P_n'(x)dx=2-\int_{-1}^1 P_n^2(x)dx.$$

By this link, we know $\int_{-1}^1 P_n^2(x)dx=\frac{2}{2n+1}$. So,

$$\int_{-1}^1 x P_n(x)P_n'(x)dx=\frac {2-\frac{2}{2n+1}}{2}=\frac{2n}{2n+1}.$$

Reza Rajaei
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