Integral involving even order Legendre polynomials

Question

Let $a>1$. We want to evaluate the integral \begin{equation*} \int_{-1}^1 \frac{P_{2n}(\xi)\,d\xi}{\sqrt{a^2-\xi^2}} \end{equation*} Mathematica is able to evaluate special cases for various $n$, but does not give an answer for arbitrary $n$. From the special cases, we see that the integral is always of the form \begin{equation*} A_{1n}(a)\sqrt{a^2-1}+A_{2n}(a)\csc^{-1}(a), \end{equation*} where $A_{1n}(a)$ and $A_{2n}(a)$ are polynomials in $a$. So the question is:

Starting from the Rodriguez formula for $P_n(\xi)$ or using any other method, is it possible to determine the polynomials $A_{1n}(a)$ and $A_{2n}(a)$.

Incidentally, the result for this integral for $a=1$ is known, and is given by \begin{equation*} \frac{\pi^2}{\left[\Gamma\left(\frac{1}{2}-n\right)\Gamma(n+1)\right]^2}. \end{equation*}

Answer 1

Actually there's nothing very special about $P_{2n}$, it's just that $P_{2n}$ is an even polynomial.

Let's let

$$ I_k = \int_{-1}^1 \frac{\xi^{2k} d\xi}{\sqrt{a^2 - \xi^2}} = 2\int_{0}^1 \frac{\xi^{2k} d\xi}{\sqrt{a^2 - \xi^2}} $$

Further, let's take $b = 1/a$ and change variables to $t = b\xi$. Now we have

$$ I_k = 2 b^{-2k} \int_0^b \frac{t^{2k}\; dt}{\sqrt{1-t^2}} $$

where $0 \le b \le 1$. It is "well-known" that

$$ \int_0^b \frac{t^{2k}\; dt}{\sqrt{1-t^2}} = \frac{(2k-1)(2k-3) \cdots 1}{(2k)(2k-2)\cdots 2}\arcsin(b) - \sum_{i=1}^k \frac{(2k-1)(2k-3)\cdots (2i+1)}{(2k)(2k-2) \cdots (2i)} b^{2i-1} \sqrt{1-b^2} $$

and then if $P_{2m}(\xi) = \sum_{k=0}^{m} a_k \xi^{2k}$, your integral is $\sum_{k=0}^m a_k I_k$.

Answer 2

Thanks to Robert's answer, I was able to derive the following explicit expression: \begin{equation*} \int_{-1}^1\frac{P_{2n}(\xi)\,d\xi}{\sqrt{a^2-\xi^2}}=\sum_{k=1}^{n+1} \frac{4^na^{2(k-1)}B_{1/a^2}\left(k-\frac{1}{2},\frac{1}{2}\right)\Gamma\left(n+k-\frac{1}{2}\right)} {\Gamma(2k-1)\Gamma\left(k-n-\frac{1}{2}\right)\Gamma(2n-2k+3)}, \end{equation*} where $B_z(a,b)$ denotes the incomplete $\beta$ function. For $a=1$, the above expression reduces to \begin{equation*} \int_{-1}^1\frac{P_{2n}(\xi)\,d\xi}{\sqrt{1-\xi^2}}=\left(\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+1)}\right)^2. \end{equation*}

The above results along with the orthogonality of the Legendre polynomials allows one to express $1/\sqrt{a^2-\xi^2}$ as a Fourier-Legendre series!