Associated Legendre function/Proofs

This article proves that the Associated Legendre Functions are orthogonal and derives their normalization constant.

Theorem
$$\int\limits_{-1}^{1}P_{l}^{m} \left( x\right) P_{k}^{m} \left( x\right) dx =\frac{2}{2l+1} \frac{\left( l+m\right) !}{\left( l-m\right) !} \delta _{lk}. $$

[Note: This proof uses the more common $$P_{l}^{m} $$ notation, rather than $$P_{l}^{\left( m\right)} $$]

Where: $$P_{l}^{m} \left( x\right) =\frac{\left( -1\right) ^{m} }{2^{l} l!} \left( 1-x^{2} \right) ^{\frac{m}{2} } \frac{d^{l+m} }{dx^{l+m} } \left[ \left( x^{2} -1\right) ^{l} \right], $$ $$0\leq m\leq l.$$

Proof
The Associated Legendre Functions are regular solutions to the general Legendre equation: $$\left( \left[ 1-x^{2} \right] y^{'} \right) ^{'} +\left( l\left[ l+1\right] -\frac{m^{2} }{1-x^{2} } \right) y=0$$ , where $$z^{'} =\frac{dz}{dx}. $$

This equation is an example of a more general class of equations known as the Sturm-Liouville equations. Using Sturm-Liouville theory, one can show that $$K_{kl}^{m} =\int\limits_{-1}^{1}P_{k}^{m} \left( x\right) P_{l}^{m} \left( x\right) dx $$ vanishes when $$k\neq l.$$ However, one can find $$K_{kl}^{m} $$ directly from the above definition, whether or not $$k=l:$$

$$K_{kl}^{m} =\frac{1}{2^{k+l} \left( k!\right) \left( l!\right) } \int\limits_{-1}^{1}\left\{ \left( 1-x^{2} \right) ^{m} \frac{d^{k+m} }{dx^{k+m} } \left[ \left( x^{2} -1\right) ^{k} \right] \right\} \left\{ \frac{d^{l+m} }{dx^{l+m} } \left[ \left( x^{2} -1\right) ^{l} \right] \right\} dx. $$

Since $$k$$ and $$l$$ occur symmetrically, one can without loss of generality assume that $$l\geq k.$$ Integrate by parts $$l+m$$ times, where the curly brackets in the integral indicate the factors, the first being $$u$$ and the second $$v'.$$ For each of the first $$m$$ integrations by parts, $$u$$ in the $$\left. uv\right| _{-1}^{1} $$ term contains the factor $$\left( 1-x^{2} \right) $$; so the term vanishes. For each of the remaining $$l$$ integrations, $$v$$ in that term contains the factor $$\left( x^{2} -1\right) $$; so the term also vanishes. This means:

$$K_{kl}^{m} =\frac{\left( -1\right) ^{l+m} }{2^{k+l} \left( k!\right) \left( l!\right) } \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} \frac{d^{l+m} }{dx^{l+m} } \left[ \left( 1-x^{2} \right) ^{m} \frac{d^{k+m} }{dx^{k+m} } \left[ \left( x^{2} -1\right) ^{k} \right] \right] dx. $$

Expand the second factor using Leibnitz' rule:

$$\frac{d^{l+m} }{dx^{l+m} } \left[ \left( 1-x^{2} \right) ^{m} \frac{d^{k+m} }{dx^{k+m} } \left[ \left( x^{2} -1\right) ^{k} \right] \right] =\sum\limits_{r=0}^{l+m}\frac{\left( l+m\right) !}{r!\left( l+m-r\right) !} \frac{d^{r} }{dx^{r} } \left[ \left( 1-x^{2} \right) ^{m} \right] \frac{d^{l+k+2m-r} }{dx^{l+k+2m-r} } \left[ \left( x^{2} -1\right) ^{k} \right]. $$

The leftmost derivative in the sum is non-zero only when $$r\leq 2m$$ (remembering that $$m\leq l$$ ). The other derivative is non-zero only when $$k+l+2m-r\leq 2k$$, that is, when $$r\geq 2m+(l-k).$$ Because $$l\geq k$$ these two conditions imply that the only non-zero term in the sum occurs when $$r=2m$$ and $$l=k.$$ So:

$$K_{kl}^{m} =\frac{\left( -1\right) ^{l+m} }{2^{2l} \left( l!\right) ^{2} } \frac{\left( l+m\right) !}{\left( 2m\right) !\left( l-m\right) !} \delta _{kl} \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} \frac{d^{2m} }{dx^{2m} } \left[ \left( 1-x^{2} \right) ^{m} \right] \frac{d^{2l} }{dx^{2l} } \left[ \left( 1-x^{2} \right) ^{l} \right] dx. $$

To evaluate the differentiated factors, expand $$\left( 1-x^{2} \right) ^{k} $$ using the binomial theorem: $$\left( 1-x^{2} \right) ^{k} =\sum\limits_{j=0}^{k}\left( \begin{array}{c} k \\ j \end{array} \right) \left( -1\right) ^{k-j} x^{2\left( k-j\right) }. $$ The only thing that survives differentiation $$2k$$ times is the $$x^{2k} $$ term, which (after differentiation) equals: $$\left( -1\right) ^{k} \left( \begin{array}{c} k \\ 0 \end{array} \right) \left( 2k\right) !=\left( -1\right) ^{k} \left( 2k\right) !$$. Therefore:

$$K_{kl}^{m} =\frac{1}{2^{2l} \left( l!\right) ^{2} } \frac{\left( 2l\right) !\left( l+m\right) !}{\left( l-m\right) !} \delta _{kl} \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx $$ ................................................. (1)

Evaluate $$\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx $$ by a change of variable: $$x=\cos \theta \Rightarrow dx=-\sin \theta d\theta\;and\; 1-x^{2} =\sin \theta. $$ Thus, $$\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l}  dx =\int\limits_{0}^{\pi }\left( \sin \theta \right) ^{2l+1}  d\theta. $$ [To eliminate the negative sign on the second integral, the limits are switched from $$ \pi \rightarrow 0 \; $$ to $$\; 0 \rightarrow \pi $$, recalling that $$ \; -1 = \cos (\pi) \;$$ and $$\; 1 = \cos (0) \; $$].

A table of standard trigonometric integrals shows: $$\int\limits_{0}^{\pi }\sin ^{n} \theta d\theta  =\frac{\left. -\sin \theta \cos \theta \right| _{0}^{\pi } }{n} +\frac{\left( n-1\right) }{n} \int\limits_{0}^{\pi }\sin ^{n-2} \theta d\theta. $$ Since $$\left. -\sin \theta \cos \theta \right| _{0}^{\pi } =0,$$ $$\int\limits_{0}^{\pi }\sin ^{n} \theta d\theta  =\frac{\left( n-1\right) }{n} \int\limits_{0}^{\pi }\sin ^{n-2} \theta  d\theta  $$ for $$n\geq 2.$$ Applying this result to $$\int\limits_{0}^{\pi }\left( \sin \theta \right) ^{2l+1} d\theta  $$ and changing the variable back to $$x$$ yields: $$\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx =\frac{2\left( l+1\right) }{2l+1} \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l-1}  dx $$ for $$l\geq 1.$$ Using this recursively:

$$\int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx =\frac{2\left( l+1\right) }{2l+1} \frac{2\left( l\right) }{2l-1} \frac{2\left( l-1\right) }{2l-3} ...\frac{2\left( 2\right) }{3} \left( 2\right) =\frac{2^{l+1} l!}{\frac{\left( 2l+1\right) !}{2^{l} l!} } =\frac{2^{2l+1} \left( l!\right) ^{2} }{\left( 2l+1\right) !}. $$

Applying this result to (1):

$$K_{kl}^{m} =\frac{1}{2^{2l} \left( l!\right) ^{2} } \frac{\left( 2l\right) !\left( l+m\right) !}{\left( l-m\right) !} \frac{2^{2l+1} \left( l!\right) ^{2} }{\left( 2l+1\right) !} \delta _{kl} =\frac{2}{2l+1} \frac{\left( l+m\right) !}{\left( l-m\right) !} \delta _{kl}. $$ QED.