# Integral formula involving Legendre polynomial

I would like a proof of the following equation below, where $$(P_n)$$ denotes Legendre polynomials. I guess this formula to hold; I checked the first several values. $$\begin{equation} \int_{-1}^{1}\sqrt{\frac{1-x}{2}} P_n(x) \text{d}{x}=\frac{-4}{(2n-1)(2n+1)(2n+3)}. \end{equation}$$

This formula is motivated by research about the random walk on the sphere.

• Where did you see the formula? Why do you need to know how to prove it? Mar 13 '19 at 16:40
• the formula is correct, but for a helpful response here you want to provide more motivation for the question (in particular, the "simple proof" request, seems to imply you already have a "complicated" proof?) Mar 13 '19 at 16:56
• I guessed this formula. It could be right because I tried the first several values. This formula can be used in research about the random walk on the sphere.
– K K
Mar 13 '19 at 16:59
• After the comment by OP this looks a normal question. Mar 13 '19 at 22:12

Using the relation $$P_n(x)=\frac{1}{2n+1}\left(P'_{n+1}(x)-P'_{n-1}(x)\right)$$ and integration by parts, we get for your integral the expression $$\frac{1}{2n+1}\left(-\int_{-1}^1 \frac{P_{n+1}(x)}{2\sqrt{2-2x}}dx+\int_{-1}^1 \frac{P_{n-1}(x)}{2\sqrt{2-2x}}dx\right).$$
But $$\int_{-1}^1 \frac{P_{n}(x)}{\sqrt{2-2x}}dx=\frac{2}{2n+1},$$ which can be proven by using the well known expansion $$\frac{1}{\sqrt{1-2xt+t^2}}=\sum_{n}P_n(x)t^n$$ and the orthogonality relation.
$$\int_{-1}^{1}\sqrt{\frac{1-x}{2}} P_n(x) \text{d}{x}=4\int_0^1 z^2\,P_n(1-2z^2)\,dz$$
which is a special case, $$\mu=3/2$$, of a formula from Gradshteyn & Ryzhik (7.233):
$$\int_0^1 x^{2\mu-1}P_n(1-2x^2)\,dx=\frac{(-1)^n\Gamma^2(\mu)}{2\Gamma(\mu+n+1)\Gamma(\mu-n)},\;\;{\rm Re}\,\mu>0.$$