# Associated Legendre Polynomials Orthogonality Proof: $\int_{-1}^1 P_k^m(x) \cdot P_l^m(x) \; \mathrm{d} x = \frac{2(l+m)!}{(2l+1)(l-m)!} \delta_{k,l}$

I have to solve the following equation using associated legendre polynomials,

$$\int_{-1}^1 P_k^m(x) \cdot P_l^m(x) \; \mathrm{d} x = \frac{2(l+m)!}{(2l+1)(l-m)!} \delta_{k,l}$$

Where they are associated Legendre polynomials.

Any hint or help will be great.

Let be $$\mathcal A_{k l}^m = \int_{-1}^1 P_k^m \left({x}\right) P_l^m \left({x}\right) \, \mathrm d x$$ where the associated Legendre functions are given by the well-known Rodrigues formula $$P^m_l(x) = \frac{1}{2^l \, l!} \, \left({1 - x^2}\right)^{m/2} \dfrac {\mathrm d^m P_l \left({x}\right)} {\mathrm d x^m} = {\left({1 - x^2}\right)^m \frac {\mathrm d^{k + m} } {\mathrm d x^{k + m} } \left({x^2 - 1}\right)^k }\qquad 0 \le m \le l$$ Thus we have $$\mathcal A_{k l}^m = \frac 1 {2^{k + l} k! l!} \int_{-1}^1 \left({\left({1 - x^2}\right)^m \frac {\mathrm d^{k + m} } {\mathrm d x^{k + m} } \left({x^2 - 1}\right)^k}\right) \left({\frac{\mathrm d^{l + m} } {\mathrm d x^{l + m} } \left({x^2 - 1}\right)^l }\right) \, \mathrm d x$$ where $$k$$ and $$l$$ occur symmetrically. Let $$l \ge k$$.

We can integrate by parts $$l + m$$ times $$\int_{-1}^1 u v' \mathrm d x = \left.{u v}\right|_{-1}^1 - \int_{-1}^1 v u' \ \mathrm d x$$ where $$u = \left({1 - x^2}\right)^m \frac {\mathrm d^{k + m} } {\mathrm d x^{k + m} } \left({x^2 - 1}\right)^k$$ and $$v' = \frac {\mathrm d^{l + m}} {\mathrm d x^{l + m}} \left({x^2 - 1}\right)^l$$

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 $$\mathcal A_{k l}^m = \frac {\left({-1}\right)^{l + m} } {2^{k + l} k! l!} \int_{-1}^1 \left({x^2 - 1}\right)^l \frac {\mathrm d^{l + m} } {\mathrm d x^{l + m} } \left({\left({1 - x^2}\right)^m \frac {\mathrm d^{k + m} } {\mathrm d x^{k + m} } \left({x^2 - 1}\right)^k }\right) \, \mathrm d x$$

Expanding the second factor using Leibniz's Rule $$\frac {\mathrm d^{l + m} } {\mathrm d x^{l + m} } \left({1 - x^2}\right)^m \frac {\mathrm d^{k + m} } {\mathrm d x^{k + m} } \left({x^2 - 1}\right)^k= \sum_{r \mathop = 0}^{l + m} \binom {l + m} r \frac {\mathrm d^r} {\mathrm d x^r} \left({1 - x^2}\right)^m \frac {\mathrm d^{l + k + 2 m - r} } {\mathrm d x^{l + k + 2 m - r} } \left({x^2 - 1}\right)^k$$ the leftmost derivative in the sum is non-zero only when $$r \le 2 m$$ (remembering that $$m \le l$$) and the other derivative is non-zero only when $$k + l + 2 m - r \le 2 k$$, that is, when $$r \ge 2 m + l - k$$.

Because $$l \ge k$$, these two conditions imply that the only non-zero term in the sum occurs when $$r = 2 m$$ and $$l = k$$.

Thus $$\mathcal A_{k l}^m = \left({-1}\right)^l \delta_{k l} \frac {\left({-1}\right)^{l + m} } {2^{2 l} \left({l!}\right)^2} \binom {l + m} {2 m} \int_{-1}^1 \left({x^2 - 1}\right)^l \frac {\mathrm d^{2 m} } {\mathrm d x^{2 m} } \left({1 - x^2}\right)^m \frac {\mathrm d^{2 l} } {\mathrm d x^{2 l} } \left({1 - x^2}\right)^l \mathrm d x$$ where $$\delta_{k l}$$ is the Kronecker Delta.

The factor $$(-1)^l$$ at the front of $$\mathcal A_{k l}^m$$ comes from switching the sign of $$x^2 - 1$$ inside $$\left({x^2 - 1}\right)^l$$.

To evaluate the differentiated factors, expand $$\left({1 - x^2}\right)^k$$ using the Binomial Theorem $$\left({1 - x^2}\right)^k = \sum_{j \mathop = 0}^k \binom k j \left({-1}\right)^{k-j} x^{2 \left({k-j}\right)}$$

The only term that survives differentiation $$2^k$$ times is the $$x^{2 k}$$ term, which after differentiation gives $$\left({-1}\right)^k \binom k 0 2 k! = \left({-1}\right)^k \left({2k}\right)!$$

Therefore $$\mathcal A_{k l}^m = \left({-1}\right)^l \delta_{k l} \frac 1 {2^{2 l} \left({l!}\right)^2} \frac {\left({2 l}\right)! \left({l + m}\right)!} {\left({l - m}\right)!} \int_{-1}^1 \left({x^2 - 1}\right)^l \ \mathrm d x =\left({-1}\right)^l \delta_{k l} \frac 1 {2^{2 l} \left({l!}\right)^2} \frac {\left({2 l}\right)! \left({l + m}\right)!} {\left({l - m}\right)!} \mathcal B_l$$

The integral $$\mathcal B_l=\int_{-1}^1 \left({x^2 - 1}\right)^l \ \mathrm d x$$ can be evaluated by a change of variable $$x = \cos \theta$$

Thus $$\mathcal B_l= \left({-1}\right)^{l + 1} \int_\pi^0 \left({\sin \theta }\right)^{2 l + 1} \, \mathrm d \theta=\left({-1}\right)^{l} \int_0^\pi \left({\sin \theta }\right)^{2 l + 1} \, \mathrm d \theta$$

Integration of $$\frac {\mathrm d \left({\sin^{n - 1} \theta \cos \theta}\right)} {\mathrm d \theta} = \left({n-1}\right) \sin^{n-2} \theta - n \sin^n \theta$$ gives $$\int_0^\pi \sin^n \theta \, \mathrm d \theta = \frac {\left.{-\sin^{n - 1} \theta \cos \theta}\right|_0^\pi} n + \frac {\left({n - 1}\right)} n \int_0^\pi \sin^{n - 2} \theta \, \mathrm d \theta= \frac {\left({n - 1}\right)} n \int_0^\pi \sin ^{n - 2} \theta \, \mathrm d \theta$$

since $$\displaystyle \left.{-\sin^{n-1} \theta \cos \theta}\right|_0^\pi = 0$$ for $$n > 1$$. Applying this result to $$\int_0^\pi \left({\sin \theta }\right)^{2 l + 1} \, \mathrm d \theta$$ and changing the variable back to $$x$$ yields:

$$\int_{-1}^1 \left({x^2 - 1}\right)^l \, \mathrm d x = - \frac {2 l} {2 l + 1} \int_{-1}^1 \left({x^2 - 1}\right)^{l - 1} \, \mathrm d x\quad\text{for}\;l \ge 1$$

Using this recursively $$\displaystyle \int_{-1}^1 \left({x^2 - 1}\right)^l \, \mathrm d x = \left({-1}\right)^l \left({\frac {2 l} {2 l + 1} \frac {2 \left({l - 1}\right)} {2 l - 1} \frac {2 \left({l - 2}\right)} {2 l - 3} \cdots \frac 2 3}\right) \int_{-1}^1 \, \mathrm d x$$

and noting that

$$\frac {2 l} {2 l + 1} \frac {2 \left({l-1}\right) } {2 l - 1} \frac {2 \left({l - 2}\right)} {2 l - 3} \cdots \frac 2 3= \frac {2^l l!} {\left({2 l + 1}\right) \left({2 l - 1}\right) \left({2 l - 3}\right) \cdots 3} = \frac{2^l l!} {\frac {\left({2 l + 1}\right)!} {2^l l!} } = \frac {2^{2 l} \left({l!}\right)^2} {\left({2 l + 1}\right)!}$$

it follows that $$\mathcal B_l=\int_{-1}^1 \left({x^2 - 1}\right)^l \mathrm d x = \ \left({-1}\right)^l \ \frac{2^{2l+1} \left({l!}\right)^2} {\left({2l+1}\right) !}$$

Therefore we have $$\mathcal A_{k l}^m =\left({-1}\right)^l \delta_{k l} \frac 1 {2^{2 l} \left({l!}\right)^2} \frac {\left({2 l}\right)! \left({l + m}\right)!} {\left({l - m}\right)!} \mathcal B_l= \delta _{k l} \frac 2 {2 l + 1} \frac {\left({l + m}\right) !} {\left({l - m}\right)!}$$ that is

$$\int_{-1}^1 P_k^m \left({x}\right) P_l^m \left({x}\right) \, \mathrm d x = \delta _{k l} \frac 2 {2 l + 1} \frac {\left({l + m}\right) !} {\left({l - m}\right)!}$$

• In your second equation you have a "$k$" which is on the right side but not in the left, and $k$ is not a dummy variable. Did you copy this from some source? What is "k" there? how do you justify changing the power from $m/2$ to $m$? Apr 7, 2017 at 0:35
• It seems that you are using Rodriguez's formula but if so, then you are forgetting the $1/(2^k k!)$ factor.... Apr 7, 2017 at 0:41
• I think I know what happened. You forgot to write the $P_k^m(x)$ factor on the left (and the coefficients of the Rodriguez's formula). Apr 7, 2017 at 0:49
• Herman, I've corrected the missing factor in Rodrigues' formula. Sep 15, 2022 at 6:47