An integral related to the derivative of Legendre polynomials 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. 
