Integral involving Legendre polynomials and simple rational function

Question

I would like to know if there is a way to express the following integral in terms of known functions

$$ I(\ell,a):=\int_{-1}^{1}\frac{P_{\ell}(x)}{x^2+a^2}\mathrm{d}x $$

with $a\in \mathbb{R}$ where $P_{\ell}$ is a Legendre polynomial of order $\ell$. I tried to use Rodrigues' formula, but got nowhere. Obviously, for $\ell$ odd the integral vanishes, so I am interested in even values of $\ell$.

Answer 1

Beside @Mariusz Iwaniuk's result, if we look at the individual values of the integrals $$I(2\ell,a)=\int_{-1}^{+1}\frac{P_{2\ell}(x)}{x^2+a^2}\,dx$$ they all write $$I(2\ell,a)=A_{\ell-1}+B_{\ell}\,\frac{\cot ^{-1}(a)}{a}$$ where $A_n$ and $B_n$ are polynomials of degree $n$ in $a^2$.

Now, the question is : what are these polynomials ?

Using what @Mariusz Iwaniuk provided $$I(2\ell,a)=\frac{2^{2 \ell+1}}{a^2\sqrt{\pi }}\sum_{k=0}^\ell (-1)^k\,\frac{ \Gamma \left(\frac{4 \ell+1-2k}{2}\right) \, _2F_1\left(1,\frac{2\ell+1-2k}{2};\frac{3\ell+3-2k}{2};-\frac{1}{a^2}\right)}{2^{2 k}\,\Gamma (k+1)\, \Gamma (2 \ell+2-2k)}$$

It seems that, for the summation, there is a nasty reccurence relation that I am unable to read.

Answer 2

With Mathematica I have: $$\int_{-1}^1 \frac{P_l(x)}{a^2+x^2} \, dx=\frac{2^{-l} \left(\frac{1}{a^2}\right)^{1+\frac{l}{2}} \sqrt{\pi } \cos \left(\frac{l \pi }{2}\right) \Gamma (1+l) \, _2F_1\left(\frac{1+l}{2},\frac{2+l}{2};\frac{3}{2}+l;-\frac{1}{a^2}\right)}{\Gamma \left(\frac{3}{2}+l\right)}$$

MMA code:

Integrate[LegendreP[l, x]/(a^2 + x^2), {x, -1, 1}] == ( 2^-l (1/a^2)^(1 + l/2) Sqrt[\[Pi]] Cos[(l \[Pi])/2] Gamma[1 + l] Hypergeometric2F1[(1 + l)/2, (2 + l)/ 2, 3/2 + l, -(1/a^2)])/Gamma[3/2 + l]