# Prove: $\int_{-1}^1 x^n P_n(x) dx = \frac{2^{n+1}n!^2}{(2n+1)!}$

I'm trying to prove: $$\int_{-1}^1 x^n P_n(x) dx = \frac{2^{n+1}n!^2}{(2n+1)!}$$

My attempt consisted in applying the Rodrigues formula: $$\int_{-1}^1 x^n P_n(x)dx = \dfrac{1}{2^n n!} \int_{-1}^1 x^n \dfrac{d^n}{dx^n}(x^2-1)^ndx$$

and integrating by parts n times:

$$\int_{-1}^1 x^nP_n(x)dx=\dfrac{1}{2^n n!}\left(x^n \dfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^n \bigg|_{-1}^1 + (-1)^{n} n!\int_{-1}^1 (x^2-1)^ndx\right)$$

The first term vanishes at the extremes. And the remaining integral I solved through substitution:

$$t = x^{2} \rightarrow dx = \dfrac{dt}{2\sqrt{t}}$$

$$\dfrac{1}{2}\int^{1}_{0}(t - 1)^{n}\dfrac{dt}{2\sqrt{t}} = \dfrac{1}{2}\int^{1}_{0}(t - 1)^{n} t^{-1/2} dt$$

Using the beta function: $$(-1)^{n}\dfrac{1}{2}\int^{1}_{0}(1 - t)^{n} t^{-1/2} dt =\dfrac{(-1)^{n}}{2}\beta(1/2, n + 1) =\dfrac{(-1)^{n}}{2}\dfrac{\Gamma(1/2)\Gamma(n + 1)}{\Gamma(n + 3/2)}$$

I rewrote the gamma functions as

$$\Gamma(1/2) = \sqrt{\pi}$$

$$\Gamma(n + 1) = n!$$

and $$\Gamma(n + 3/2) = \dfrac{(n+1)(2n)!\sqrt{\pi}}{2^{2n}n!}$$

Finally I reached:

$$(-1)^{n} n!\int_{-1}^1 x^{n-1} \dfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx = \dfrac{(-1)^{n}n!2}{2^{n}n!}\dfrac{(-1)^{n}}{2}\dfrac{\sqrt{\pi}2^{2n}n!n!}{(n+1)(2n)!\sqrt{\pi}} = \dfrac{2^{n}n!n!}{(n+1)(2n)!}$$

I tried simplifying furter, but I didn't get anywhere

• I'm afraid your formula $\,\Gamma(n + 3/2) = \dfrac{(n+1)(2n)!\sqrt{\pi}}{2^{2n}n!}$ is not correct. Taking, for instance, $n=0$ you get $\sqrt\pi$. But, on the other hand, $\,\Gamma(3/2)=\frac{1}{2}\Gamma(1/2)=\frac{\sqrt \pi}{2}$. For $n=1$ you get $\sqrt\pi$ and $\,\Gamma(5/2)=\frac{3}{2}\Gamma(3/2)=\frac{3\sqrt \pi}{4}$ correspondingly Nov 1, 2022 at 20:05
• $\Gamma(n+3/2)=(2n+1)!\sqrt{\pi}/2^{2n+1}n!$ Nov 1, 2022 at 20:25
• You're completely right, I wrote the gamma function as an integral and was able to get to this result, thank you! Nov 1, 2022 at 22:55
• $P_n(x)$ is the Legendre P function? Nov 1, 2022 at 23:13

A method based upon the series for $$P_{n}(x)$$ is seen by the following.

Using $$P_{n}(x) = \sum_{k=0}^{\lfloor{n/2}\rfloor} \frac{(-1)^k \, \left(\frac{1}{2}\right)_{n-k} \, (2 x)^{n-2 k}}{k! \, (n-2 k)!}$$ then \begin{align} I &= \int_{-1}^{1} x^n \, P_{n}(x) \, dx \\ &= \sum_{k} a_{k}^{n} \, 2^{n-2 k} \, \int_{-1}^{1} x^{2 (n-k)} \, dx \\ &= \sum_{k} a_{k}^{n} \, 2^{n-2k} \frac{1}{2n -2 k + 1} \, \left( 1^{2n -2k +1} - (-1)^{2n -2k +1} \right) \\ &= \sum_{k} \frac{(-1)^k \, \left(\frac{1}{2}\right)_{n-k} \, 2^{n -2k +1}}{k! \, (n -2k)! \, (2n -2k +1) } \\ &= \frac{2^{n+1} \, \left(\frac{1}{2}\right)_{n}}{n! \, (2n+1)} \, {}_{3}F_{2}\left(-\frac{n}{2}, \frac{1-n}{2}, -n-\frac{1}{2}; \frac{1}{2}-n, \frac{1}{2}-n; 1 \right) \\ &= \frac{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \, {}_{3}F_{2}\left(-\frac{n}{2}, \frac{1-n}{2}, -n-\frac{1}{2}; \frac{1}{2}-n, \frac{1}{2}-n; 1 \right) \\ &= \frac{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \, \frac{n!}{\binom{2n}{n} \, \left(\frac{1}{2}\right)_{n}} \\ &= \frac{n!}{2^{n-1} \, \left(\frac{3}{2}\right)_{n}} \\ &= \frac{2^{n+1}}{\binom{2n}{n} \, (2n+1)}. \end{align}

• $\displaystyle +1$. Pretty fine. Nov 2, 2022 at 0:11

We'll use the following:

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

2. The leading coefficient of $$P_n(x)$$ ( coefficient of $$x^n$$) equals $$\frac{\binom{2n}{n}}{2^n}$$

3. $$\int_{-1}^1 x^k P_n(x) d x = 0$$ for all $$0 \le k \le n-1$$.

From here we conclude

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