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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.

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    $\begingroup$ Where did you see the formula? Why do you need to know how to prove it? $\endgroup$
    – Yemon Choi
    Commented Mar 13, 2019 at 16:40
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    $\begingroup$ 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?) $\endgroup$
    – Carlo Beenakker
    Commented Mar 13, 2019 at 16:56
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    $\begingroup$ 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. $\endgroup$
    – K K
    Commented Mar 13, 2019 at 16:59
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    $\begingroup$ After the comment by OP this looks a normal question. $\endgroup$
    – Fedor Petrov
    Commented Mar 13, 2019 at 22:12

2 Answers 2

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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.

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Not a derivation (yet), but at least a reduction to a more familiar form:

$$\int_{-1}^{1}\sqrt{\frac{1-x}{2}} P_n(x) \text{d}{x}=4\int_0^1 z^2\,P_n(1-2z^2)\,dz$$

which is a special case, $\mu=3/2$, of a formula from Gradshteyn & Ryzhik (7.233):

$$\int_0^1 x^{2\mu-1}P_n(1-2x^2)\,dx=\frac{(-1)^n\Gamma^2(\mu)}{2\Gamma(\mu+n+1)\Gamma(\mu-n)},\;\;{\rm Re}\,\mu>0.$$

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