Associated Legendre function/Proofs

This addendum 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.

See also

• Associated Legendre function
• Spherical harmonics

References

• Kenneth Franklin Riley, Michael Paul Hobson, Stephen John Bence, "Mathematical methods for physics and engineering", pg. 590, (2006) 3$^{rd}$ Edition, Cambridge University Press, ISBN 0-521-67971-0.