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I was looking for a method to do the following integral: $\int^1_{-1}(1-x^2)\frac{dP_m(x)}{dx}P_n(x)~dx$

I know there should be an explicit representation for the result but I am struggling to work it out.

metamorphy
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Rescy_
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  • For the case $m=n$, the integral is zero. Are you using the definition $$P_{k}(x) = \frac{1}{2^{k}\cdot k!} \frac{d^k}{dx^k}(x^2-1)^k$$? – RobertTheTutor Feb 25 '21 at 12:11

1 Answers1

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By Rodrigues' formula and the Legendre DE, you essentially want to find $$ \int_{-1}^1 [(x^2-1)^m]^{(m-1)}[(x^2-1)^n]^{(n)}\,\mathrm{d}x $$ If $n<m$, integrate by parts $n$ times gives $$ \pm\int_{-1}^1 [(x^2-1)^m]^{(m-n-1)}\underbrace{[(x^2-1)^n]^{(2n)}}_{=(2n)!}\,\mathrm{d}x $$ which is $0$ unless $m=n+1$, in which case you want to calculate $$ \int_{-1}^1(x^2-1)^m\,\mathrm{d}x $$ which with $x=\sin\theta$ and $\int_0^{\pi/2}\cos^{2m+1}\theta\,\mathrm{d}\theta=\frac{(2m)!!}{(2m+1)!!}$ nails it. Similar for the case $n\geq m$.

user10354138
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  • How do we get from line 1 to line 2 formula? Did you use integration by parts? – Rescy_ Feb 25 '21 at 18:54
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    @Rescy_ Legendre DE says $(1-x^2)P_m'(x)$ is an antiderivative of constant times $P_m(x)$. Rodrigues tells you $P_m,P_n$ are constant times differentiating a power of $(x^2-1)$ a number of times. So you get the formula up to constant factor which you should find and keep track. – user10354138 Feb 25 '21 at 18:59