Does the integral $$C_{n,m}\equiv\int_0^1P_n(x)\widetilde{P}_m(x)dx$$ have any closed answer? $P_n(x)$ is the Legendre polynomial of order $n$. $\widetilde{P}_m(x)\equiv P_m(2x-1)$ is called the shifted Legendre polynomial.
For what $n$ and $m$, $C_{n,m}$ would vanish?
Write $P_n(x)$ in an explicit form 1: $$P_n(x)=\frac{1}{2^n}\sum_{k=0}^{[n/2]}{(-1)^k\binom{n}{k}\binom{2n-2k}{n}x^{n-2k}}$$ Write $\widetilde{P}_m(x)$ in an explicit form 1: $$\widetilde{P}_m(x)=(-1)^m\sum_{j=0}^{l}{(-1)^j\binom{m}{j}\binom{m+j}{j}x^j}$$ Multiply and rearrange: $$P_n(x)\widetilde{P}_m(x)=\frac{(-1)^m}{2^n}\sum_{k=0}^{[n/2]}{(-1)^k\binom{n}{k}\binom{2n-2k}{n}}\sum_{j=0}^{l}{(-1)^j\binom{m}{j}\binom{m+j}{j}x^{n-2k+j}}$$ Integrate: $$C_{n,m}\equiv\int_0^1P_n(x)\widetilde{P}_m(x)dx=\frac{(-1)^m}{2^n}\sum_{k=0}^{[n/2]}{(-1)^k\binom{n}{k}\binom{2n-2k}{n}}\sum_{j=0}^{l}{(-1)^j\binom{m}{j}\binom{m+j}{j}\int_0^1x^{n-2k+j}}dx,$$ $$\int_0^1x^{n-2k+j}dx=\frac{1}{n-2k+j+1}x^{n-2k+j+1}\Big|_0^1=\frac{1}{n-2k+j+1},$$ solution: $$\bbox[border:1px solid black]{ C_{n,m}=\frac{(-1)^m}{2^n}\sum_{k=0}^{[n/2]}{(-1)^k\binom{n}{k}\binom{2n-2k}{n}}\sum_{j=0}^{l}{\frac{(-1)^j}{n-2k+j+1}\binom{m}{j}\binom{m+j}{j}} }$$