I want to calculate the integral $$ I=\int_{-1}^{1} \Big(\frac{\mathrm{d}P_{n+1}(t)}{\mathrm{d}t}\Big) \Big(\frac{\mathrm{d}P_{m+1}(t)}{\mathrm{d}t}\Big) \mathrm{d}t $$ where $P_n(t)$ is Legendre polynomials. By virtue of the identity $$ \frac{\mathrm{d}P_{n+1}(t)}{\mathrm{d}t}=2\sum_{k=0}^{[n/2]}\frac{P_{n-2k}(t)}{\vert\vert P_{n-2k}(t)\vert\vert^2} $$ where ${\vert\vert P_{n}(t)\vert\vert}=\sqrt{\frac{2}{2n+1}}$, and Using the property of Double Series, then $$ I=4\int_{-1}^{1} \Big(\sum_{k_1=0}^{[n/2]}\frac{P_{n-2k_1}(t)}{\vert\vert P_{n-2k_1}(t)\vert\vert^2}\Big) \Big(\sum_{k_2=0}^{[m/2]}\frac{P_{m-2k_2}(t)}{\vert\vert P_{m-2k_2}(t)\vert\vert^2}\Big) \mathrm{d}t\\ =4\sum_{k_1=0}^{[n/2]}\sum_{k_2=0}^{[m/2]} \frac{1}{\vert\vert P_{n-2k_1}(t)\vert\vert^2} \frac{1}{\vert\vert P_{m-2k_2}(t)\vert\vert^2} \int_{-1}^{1}{P_{n-2k_1}(t)}{P_{m-2k_2}(t)}\mathrm{d}t\\ =4\sum_{k_1=0}^{[n/2]}\sum_{k_2=0}^{[m/2]} \frac{1}{\vert\vert P_{n-2k_1}(t)\vert\vert^2} \delta_{n-2k_1,m-2k_2}\\ $$
If $n=m$, then $$ I=4\sum_{k=0}^{[n/2]}\frac{1}{\vert\vert P_{n-2k}(t)\vert\vert^2} =2\sum_{k=0}^{[n/2]}\big(2(n-2k)+1\big)\\ =2([n/2]+1)\big((2n+1)-2[n/2]\big) =(n+1)(n+2) $$
But if $n\neq m$, How can I move on?
Legendre polynomials are orthogonal.here So it may work to calculate that integral by using integrating by parts twices.