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I want to evaluate $$\int_0^1x^2P_n(x)\,dx$$ but the only way out i see is by using $$xP_n(x)=\frac{nP_{n-1}+(n+1)P_{n+1}}{2n+1}$$ twice and since i got many $\int_0^1P_l(x)dx$, i use $$P_l(x)=\frac{P'_{n+1}-P'_{n-1}}{2l+1}.$$ In the endm there a lot of terms with $P_k(0)$ with big constants with factorials before. Furthermore the $P_k(0)$ are known as $$\frac{(-1)^{l/2}l!}{2^l(\frac{l}{2}!)^2}$$ for (l) even, and zero to (l) odd.

There is a way of evaluating this without this long and tedious terms that dont cancel out?

Bernard
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Vinicius Fuckner Linhares
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1 Answers1

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I will at least offer a solution for the even case. If $n=2k$ is even, then $P_n$ is an even function and $x^2P_n(x)$ is even as well. Thus, we would have

$$\int_0^1x^2P_{2k}(x)\,dx=\frac{1}{2}\int_{-1}^1x^2P_{2k}(x)\,dx$$

We can explicitly write the $x^2$ in terms of Legendre polynomials, since

$$P_0(x)=1,\;\;\;\;P_2(x)=\frac{3x^2-1}{2}$$

Thus,

$$x^2=\frac{2}{3}P_2(x)+\frac{1}{3}P_0(x)$$

Thus,

$$\int_0^1x^2P_{2k}(x)\,dx=\frac{1}{6}\int_{-1}^1\left(2P_2(x)+P_0(x)\right)P_{2k}(x)\,dx$$

For $k\neq 0$ and $k\neq 1$, by the orthogonality of the polynomials this integral must be zero. Otherwise, we get some value, using the normalization constant of the polynomial, which is

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

Thus, we have

$$\int_0^1x^2P_{2k}(x)\,dx=\left\{\begin{array}{ccc} \frac{1}{3} & : & k=0 \\ \frac{2}{15} & : & k=1 \\ 0 & : & \text{otherwise} \end{array}\right.$$

For odd values of $n$, I can't see an obvious way of doing this without resorting to ugly formulae for the polynomials.

Josh B.
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