Integration by parts with Legendre Functions

Question

I need help deriving

$\int_{-l}^l [P_l^m(x)]^2 = \frac{2}{2l+1} \frac{(l+m)!}{(l-m)!}$

for the associated Legendre functions

I am supposed to use

$P_l^m(x) = (-1)^{-m}\int_{-l}^l \frac{(1-x^2)^{\frac{m}{2}}}{2^ll!}\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l$

and

$P_l^m(x) = \frac{1}{2^ll!}^{-m}\int_{-l}^l(1-x^2)^{\frac{m}{2}}\frac{d^{l-m}}{dx^{l-m}}(x^2-1)^l$

I tried to multiply them together, ignoring the terms without x and got

$\int_{-l}^l\frac{d^{l+m}}{dx^{l+m}}(x^2-1)^l\frac{d^{l-m}}{dx^{l-m}}(x^2-1)^l$

I'm supposed to integrate by parts m-times to get both derivates to be $\frac{d^{l}}{dx^{l}}$

This is where I think I'm making my mistake, after doing the integration by parts and getting the same number of derivatives for each term, $\frac{d^{l}}{dx^{l}}$

I get

$(-1)^m\int_{-l}^l(\frac{d^{l}}{dx^{l}}(x^2-1)^l)^2$

I guess the (-1)^m will cancel with part of the original equations but I don't see how to get to the final conclusion

$\frac{2}{2l+1} \frac{(l+m)!}{(l-m)!}$

Any guidance or help would be greatly appreciated, Thanks a lot!

Answer 1

The Rodrigues' formulae you supplied have gone somewhat astray; they should be

$P_n^m(x)=\frac{1}{2^n n!} (x^2-1)^{m⁄2} [(x^2-1)^n ]^{(n+m)}$

and

$P_n^{-m}(x)=\frac{1}{2^n n!} (x^2-1)^{-m⁄2} [(x^2-1)^n ]^{(n-m)}$

Let us derive the value of $‖P_n^m‖^2$ for positive m:

$\newcommand{\partial}[1]{\left[#1\right]}$ $\newcommand{\bracket}[1]{\left(#1\right)}$ \begin{equation} \begin{split} ‖P_n^m‖^2&=\int_{-1}^{+1} [P_n^m]^2 dx \\ &=\frac{1}{2^{2n} n!^2} \int_{-1}^{+1}(x^2-1)^m\cdot [(x^2-1)^n ]^{(n+m)}\cdot [(x^2-1)^n ]^{(n+m)}dx \end{split} \end{equation}

Making use of

\begin{equation} \begin{split} [(x^2-1)^n]^{(n-m)}=\frac{(n-m)!}{(n+m)!} (-1)^m (1-x^2)^m [(x^2-1)^n]^{(n+m)} \end{split} \end{equation}

we get

\begin{equation} \begin{split} ‖P_n^m‖^2&=\frac{1}{2^{2n} n!^2} \int_{-1}^{+1}(-1)^m\frac{(n+m)!}{(n-m)!}[(x^2-1)^n ]^{(n+m)}\cdot [(x^2-1)^n]^{(n-m)}dx \end{split} \end{equation}

Partial integration $n-m$ times then gives

\begin{equation} \begin{split} ‖P_n^m‖^2&=\frac{1}{2^{2n} n!^2}(-1)^m\frac{(n+m)!}{(n-m)!} \cdot (-1)^{n-m}\int_{-1}^{+1}[(x^2-1)^n ]^{(2n)}\cdot [(x^2-1)^n]^{(0)}dx \\ &=\frac{1}{2^{2n} n!^2}(-1)^n\frac{(n+m)!}{(n-m)!} \int_{-1}^{+1}(2n)!(x^2-1)^ndx \\ &=\frac{1}{2^{2n} n!^2}(-1)^n\frac{(n+m)!}{(n-m)!}(2n)!(-1)^n\frac{n!^2}{(2n)!}\cdot \frac{2^{2n+1}}{2n+1} \\ &=\frac{(n+m)!}{(n-m)!} \frac{2}{2n+1} \end{split} \end{equation}

And it is easy (using $P_n^{-m}(x) = (-1)^m \frac{(n-m)!}{(n+m)!} P_n^m(x)$) to derive that

\begin{equation} \begin{split} ‖P_n^{-m}‖^2&=\frac{(n-m)!^2}{(n+m)!^2} ‖P_n^m‖^2 \\ &=\frac{(n-m)!}{(n+m)!} \frac{2}{2n+1} \end{split} \end{equation}

Whence

\begin{equation} \begin{split} ‖P_n^{m}‖^2&=\frac{(n+|m|)!}{(n-|m|)!}\frac{2}{2n+1} \end{split} \end{equation}

For negative and positive $m$, $-n \le m \le n$.