Integral involving Legendre polynomial

Question

In a physics problem the following integral shows up $$\int\limits_0^{2\pi}P_m(\cos{(\theta-\alpha)})\,\cos^{m+2}{(n\alpha)}\;d\alpha,$$ where $P_m$ is the Legendre polynomial and $n,m$ are integer numbers. How this integral can be evaluated?

Answer 1

By periodicity of trigonometric functions we have $$ I(m,n)=\int\limits_0^{2\pi}P_m(\cos{\alpha})\,\cos^{m+2}{(n\alpha+n\theta)}\;d\alpha.\tag{1} $$ Moreover, by the symmetry of Legendre polynomials: $$ I(m,n)=\left(1+(-1)^{mn+m}\right)\int\limits_0^{\pi}P_m(\cos{\alpha})\,\cos^{m+2}{(n\alpha+n\theta)}\;d\alpha. $$ This last formula shows that the integral is $0$ when $m$ is odd, and $n$ is even: $$ I(2M-1,2N)=0. $$ Otherwise it is not useful.

Now in (1) we use the trigonometric expansion (Andrews, Askey, & Roy, 6.4.11): $$ {P}_{m}\left(\cos \alpha\right)= \sum_{k=0}^{m}\frac{(1/2)_k(1/2)_{m-k}}{k!(m-k)!}\cos(m-2k)\alpha,\quad \alpha\in\mathbb{R}. $$ We also have $$ (\cos x)^{2M+1}=\frac{1}{2^{2 M+1}}\sum_{j=-M}^{M+1}\binom{2 M+1}{j+M}\cos(2j-1)x, $$ and $$ (\cos x)^{2M}=\frac{1}{2^{2 M}}\sum_{j=-M}^{M}\binom{2 M}{j+M}\cos(2jx) . $$ Using orthogonality of $\cos x$ on $(0,2\pi)$ we find:

$$ I(2M-1,2N+1)=\frac{\pi }{2^{2 M}}\sum_{j=-M}^{M+1}\frac{\left(\frac{1}{2}\right)_{M+N-2Nj-j} \left(\frac{1}{2}\right)_{M-N-1+2Nj+j}}{(M+N-2Nj-j)! (M-N-1+2Nj+j)!}\binom{2 M+1}{j+M}\cos\left[(2N+1)(2j-1)\theta\right]. $$

This formula has been checked numerically.

This formula seems like a major simplification compared with the double sums. The other cases are left as an exercise.

Answer 2

Here is an explicit formula for the value of this integral:

\begin{split} &\frac{\pi}2 \sum_{s=1+\lceil\frac{m(n-1)}{2n}\rceil}^{m+2} \binom{m+2}s \cos(n(m+2-2s)\theta) \sum_{j=n(m+2-2s)}^{\lfloor \frac{m(n+1)}2\rfloor -n(s-1)} \frac1{2^{2j-n(m+2-2s)}} \\ &\times\binom{2j-n(m+2-2s)}j \binom{m}{2j-n(m+2-2s)}\binom{j+n(s-1)-\frac{m(n-1)+1}2}m. \end{split}