Computation of an integral depending on the Legendre polynomials

Question

Let $P_l$ be a Legendre polynomial ($l$ is an integer). I want to know why the quantity $$ v_l(k):=(-i)^l\int_{-1}^{+1}\mathrm{e}^{ikx}\,P_l(x)\;\mathrm{d}x $$ is real?

Answer 1

Write $e^{i kx}=\cos(k x)+i\sin(k x)$. So your integral is $$ \nu_l(k)=(-i)^l\left[\int_{-1}^1 dx \cos(kx)P_l(x)+i\int_{-1}^1 dx \sin(kx)P_l(x)\right]\ . $$ But the Legendre polynomial of order $l$ have parity $(-1)^l$. So if $l=2m$ is even, the second integral (with $\sin$) is zero (the integrand is odd and the interval of integration is symmetric) and the final result is the real number [note that $i^{2m}=(-1)^m$] $$ \nu_{2m}(k)=(-1)^m\int_{-1}^1 dx \cos(kx)P_{2m}(x)\ . $$ Similarly, if $l$ is odd, the first integral is now zero, and the $i$ factors cancel.