I am trying to show that,

\begin{align} I = \int_{-1}^1 x^nP_n(x)\,\mathrm{d}x = \frac{2^{n+1}n!n!}{(2n+1)!} \end{align}

So far I have done the following. Rodrigues formula is as follows:

\begin{align*} P_n(x) = \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} x^{n-2k} \end{align*}


\begin{align*} \begin{aligned} N&=n/2, && \text{if} \quad n=\text{even} \\ N&=(n-1)/2, && \text{if} \quad n=\text{odd} \end{aligned} \end{align*}

Substitute Rodrigues formula,

\begin{align*} I &= \int_{-1}^1 x^n \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} x^{n-2k} \,\mathrm{d}x \\ &= \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \int_{-1}^1 x^{2n-2k} \,\mathrm{d}x \\ &= \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \left. \frac{x^{2n-2k+1}}{2n-2k+1}\right\rvert_{-1}^1 \\ &= \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \frac{1 - (-1)^{2n-2k+1}}{2n-2k+1} \\ &= \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \frac{1 + (-1)^{2n-2k}}{2n-2k+1} \end{align*}

Since $2n-2k$ is even,

\begin{align*} I &= \sum_{k=0}^N \frac{(-1)^k (2n-2k)!}{2^nk!(n-k)!(n-2k)!} \frac{2}{2n-2k+1} \\ &= \sum_{k=0}^N \frac{2^{1-n}(-1)^k (2n-2k)!}{k!(n-k)!(n-2k)!(2n-2k+1)} \end{align*}

Can someone give me a hint about how to proceed or do I need to prove with another way?

  • $\begingroup$ Try Bonnet’s recursion formula $(n+1)P_{n+1}(x)=(2n+1)xP_{n}(x)-nP_{n-1}(x)$. $\endgroup$ – marty cohen Jun 14 '16 at 20:06
  • $\begingroup$ I tried. After using recursion formula I inserted Rodrigues formula again for $P_{n+1}$ and $P_{n-1}$ but got stuck again. $\endgroup$ – Shibli Jun 14 '16 at 20:21
  • $\begingroup$ @Shibli: just keep the $\frac{d^n}{dx^n}$ in Rodrigues' formula and apply integration by parts $n$ times. $\endgroup$ – Jack D'Aurizio Jun 14 '16 at 20:46

It is faster to exploit the generating function that comes from Rodrigues' formula or Bonnet's recursion formula: $$\frac{1}{\sqrt{1-2xt+t^2}} = \sum_{n=0}^\infty P_n(x) t^n\tag{1} $$ By replacing $t$ with $xt$ we have: $$ \int_{-1}^{1}\frac{dx}{\sqrt{1-2x^2 t+x^2 t^2}}=\sum_{n\geq 0}\left(\int_{-1}^{-1}x^n P_n(x)\,dx\right)t^n\tag{2} $$ so our integrals can be computed through the Taylor series of the LHS, regarded as a function of $t$.
The LHS of $(2)$ is: $$ \frac{2\arcsin\!\sqrt{t(2-t)}}{\sqrt{t(2-t)}}=\frac{4\arcsin\sqrt{\frac{t}{2}}}{\sqrt{t(2-t)}} \tag{3} $$ and your initial claim turns out to be equivalent to a well-known identity.

Truth to be told, it is also very interesting sideways: some series related with the $\Gamma^2$ function can be easily computed by exploiting the mentioned property of Legendre polynomials.

Using Rodrigues' formula we have: $$\int_{-1}^{1}x^n P_n(x)\,dx = \frac{1}{2^n n!}\int_{-1}^{1}x^n\cdot\frac{d^n}{dx^n}(x^2-1)^n\,dx \stackrel{RIBP}{=}\frac{1}{2^n}\int_{-1}^{1}(1-x^2)^n\,dx\tag{4}$$ where $\text{RIBP}$ stands for repeated integration by parts.
But the RHS of $(4)$ is a value of the Euler's beta function: $$\frac{1}{2^n}\int_{-1}^{1}(1-x^2)^n\,dx=\frac{1}{2^n}\int_{0}^{1}z^{-1/2}(1-z)^n\,dz = \frac{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}{2^n\,\Gamma\left(n+\frac{3}{2}\right)}\tag{5}$$ and we are equally done, since: $$\frac{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}{2^n\,\Gamma\left(n+\frac{3}{2}\right)} = \frac{2\cdot n!}{(2n+1)!!} = \frac{2^{n+1} n!^2}{(2n+1)!}.\tag{6}$$

  • $\begingroup$ In the link you gave, there are two $\arcsin$ but here only one. I could not figure out the relation (sorry I am bad in math). $\endgroup$ – Shibli Jun 14 '16 at 20:41
  • $\begingroup$ @Shibli: It does not matter, please see my updated answer (the direct approach through Rodrigues' formula). $\endgroup$ – Jack D'Aurizio Jun 14 '16 at 20:44

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