Xrfgtjtk

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
$$P^m_l left({x}right) = left({1 - x^2}right)^{m/2} dfrac {mathrm d^m P_l left({x}right)} {mathrm d x^m}={left({1 - x^2}right)^m frac {mathrm d^{k + m} } {mathrm d x^{k + m} } left({x^2 - 1}right)^k }qquad 0 le m le l$$
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:

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

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

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