Associated Legendre function/Proofs

It is demonstrated that the associated Legendre functions are orthogonal in two different ways and their normalization constant for each is derived.

Theorem (orthogonality relation 1)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}, }

where:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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], \quad 0\leq m\leq l.}

Proof

The associated Legendre functions are regular solutions to the associated Legendre differential equation given in the main article. The equation is an example of a more general class of equations known as the Sturm-Liouville equations. Using Sturm-Liouville theory, one can show the orthogonality of functions with the same superscript m and different subscripts:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} =\int\limits_{-1}^{1}P_{k}^{m} \left( x\right) P_{l}^{m} \left( x\right) dx = 0 \quad\hbox{if}\quad k \ne l . }

However, one can find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} } directly from the above definition, whether or not Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=l} . This involves evaluating the overlap integral directly from the definition of the associated Legendre functions given in the main article. Indeed, inserting the definition of the function twice:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} =\frac{1}{2^{k+l}\; k! \; l! } \int\limits_{-1}^{1} \left\{ (1-x^{2})^{m} \frac{d^{k+m} }{dx^{k+m} } \left[ (x^{2} -1)^{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 ≥ k. Use the well-known integration-by-parts equation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-1}^1 u\; v'\; dx = \left. u\,v\right|_{-1}^1 - \int_{-1}^{1} v u' \;dx }

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle uv|_{-1}^1} term contains the factor (1−x²), so the term vanishes. For each of the remaining l integrations, v in that term contains the factor (x²−1) so the term also vanishes. This means:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} =\frac{\left( -1\right) ^{l+m} }{2^{k+l} \; k!\; l! } \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:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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} \binom{l+m}{r} \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 ≤ 2m (remembering that m ≤ l). The other derivative is non-zero only when k + l + 2m − r ≤ 2k, that is, when r ≥ 2m + l − k. Because l ≥ k these two conditions imply that the only non-zero term in the sum occurs when r = 2m and l = k. So:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} =\ (-1)^{l}\ \delta_{kl} \; \frac{(-1)^{l+m} }{2^{2l}\, (l!)^{2}} \binom{l+m}{2m} \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, }

where δ_kl is the Kronecker delta that shows the orthogonality of functions with l ≠ k. The factor Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1)^{l}} at the front of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m}} comes from switching the sign of x²-1 inside (x²-1)^l. To evaluate the differentiated factors, expand (1−x²)^k using the binomial theorem:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( 1-x^{2} \right) ^{k} =\sum\limits_{j=0}^{k} \binom{k}{j} ( -1)^{k-j} x^{2(k-j)}. }

The only term that survives differentiation 2k times is the x^2k term, which after differentiation gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1)^k \, \binom{k}{0}\, 2k! = (-1)^{k}\, (2k)! \, . }

Therefore:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} =\ (-1)^{l}\delta _{kl}\; \; \frac{1}{2^{2l}\; (l!) ^{2} } \frac{(2l)!\,(l+m)!}{(l-m)!} \int\limits_{-1}^{1}(x^{2} -1)^{l} dx \qquad\qquad\qquad\qquad\qquad\qquad (1) }

Evaluate

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-1}^{1}(x^{2} -1)^{l} dx }

by a change of variable:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\cos \theta\; \Longrightarrow\; dx=-\sin \theta d\theta\quad\hbox{and}\quad 1-x^{2} =(\sin \theta)^2. }

Thus,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx =(-1)^{l+1}\int\limits_{\pi}^{0}\left( \sin \theta \right) ^{2l+1} d\theta, }

where we recall that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -1 = \cos\,\pi \quad\hbox{and}\quad 1 = \cos\,0. }

The limits were switched from

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi \rightarrow 0 \; \quad\hbox{and}\quad 0 \rightarrow \pi } ,

which accounts for one minus sign . Integration of

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d(\sin^{n-1}\theta\cos\theta)}{d\theta} = (n-1)\sin^{n-2}\theta - n\sin^{n}\theta }

gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{0}^{\pi }\sin^n \theta\; d\theta =\frac{\left. -\sin^{n-1} \theta \cos \theta \right|_{0}^{\pi } }{n} +\frac{(n-1) }{n} \int\limits_{0}^{\pi }\sin ^{n-2} \theta d\theta. }

Since

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left. -\sin^{n-1} \theta \cos \theta \right| _{0}^{\pi } =0 \quad\hbox{for}\quad n > 1,}

it follows that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 > 1.

Applying this result to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{0}^{\pi }\left( \sin \theta \right) ^{2l+1} d\theta }

and changing the variable back to x yields:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx =\ -\ \frac{2l}{2l+1} \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l-1} dx }

for l ≥ 1. Using this recursively:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx =\ (-1)^{l}\ (\frac{2l }{2l+1} \frac{2\left( l-1\right) }{2l-1} \frac{2\left( l-2\right) }{2l-3} ...\frac{2}{3} )\ \int\limits_{-1}^{1}\ dx }

Noting:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-1}^{1}\ dx\ =\ 2}

and

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2l }{2l+1} \frac{2\left( l-1\right) }{2l-1} \frac{2\left( l-2\right) }{2l-3} ...\frac{2}{3} \ =\ \frac{2^ll! }{(2l+1)(2l-1)(2l-3)...3}\ =\ \frac{2^ll! }{\frac{(2l+1)! }{2^ll!}}\ =\frac{2^{2l} \left( l!\right) ^{2} }{\left( 2l+1\right) !}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-1}^{1}\left( x^{2} -1\right) ^{l} dx = \ (-1)^{l}\ \frac{2^{2l+1} \left( l!\right) ^{2} }{\left( 2l+1\right) !}}

Applying this result to equation (1):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K_{kl}^{m} =\delta _{kl}\; \frac{1}{2^{2l}\; (l!)^{2} } \frac{( 2l)!\;(l+m)!}{(l-m)!}\; \frac{2^{2l+1} \;(l!)^{2} }{( 2l+1)!} = \delta _{kl}\,\frac{2}{2l+1} \frac{( l+m) !}{( l-m) !} \qquad\qquad \mathbf{QED}. }

Clearly, if we define new associated Legendre functions by a constant times the old ones,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bar{P}^m_l(x) \equiv \sqrt{ \frac{2l+1}{2}\; \frac{(l-m)!}{(l+m)!} }\; P^m_l(x) }

then the overlap integral becomes,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K^m_{kl} = \int\limits_{-1}^{1} \bar{P}^m_k(x) \bar{P}^m_l(x) \;dx = \delta_{kl}, }

that is, the new functions are normalized to unity.

Comments

The orthogonality of the associated Legendre functions can be demonstrated in different ways. The proof presented above assumes only that the reader is familiar with basic calculus and it is therefore accessible to the widest possible audience. However, as mentioned, their orthogonality also follows from the fact that the associated Legendre equation belongs to a family known as the Sturm-Liouville equations.

It is possible to demonstrate their orthogonality using principles associated with operator calculus. Let us write

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P^m_{l}(x) = w(x)^{1/2} \; \nabla^m P_l(x) }

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \equiv \frac{d}{dx} \quad \hbox{and}\quad w(x) \equiv (1-x^2)^m. }

Clearly, the case m = 0 is,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P^{0}_l(x) = (1-x^2)^{0/2}\; \nabla^0 P_l(x) = P_l(x). }

The proof given above starts out by implicitly proving the anti-Hermiticity of ∇. Indeed, noting that w(x) is a function with w(1) = w(−1) = 0 for m ≠ 0, it follows from partial integration that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\; w\, g \;|\; \nabla f\; \rangle \equiv \int\limits_{-1}^1\; w(x)\,g(x)\;\big(\nabla f(x)\big) \; dx = \left. w(x)\;g(x)f(x) \right|_{-1}^{1} - \int\limits_{-1}^1 \Big(\nabla w(x)\,g(x)\Big) \, f(x)\; dx = - \langle\; \nabla (w g) \;|\; f\;\rangle }

Hence

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla^\dagger = - \nabla \;\Longrightarrow\; \left(\nabla^\dagger\right)^{m} = (-1)^{m} \;\nabla^{m}. }

To demonstrate orthogonality of the associated Legendre polynomials, we use a result from the theory of orthogonal polynomials. Namely, a Legendre polynomial of order l is orthogonal to any polynomial Π_p of order p lower than l. In bra-ket notation

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle\; \Pi_p \;|\; P_l \;\rangle = 0\quad \hbox{if}\quad O\left[\Pi_p\right] \equiv p < l. }

Knowing this,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-1}^{1}P_{l}^{m} \left( x\right) P_{k}^{m} \left( x\right) dx \equiv \langle\; w\, \nabla^m P_k \;|\; \nabla^m P_l\;\rangle = (-1)^m \langle\; \nabla^m \{w\, \nabla^m P_k\} \;|\; P_l\;\rangle . }

The bra is a polynomial of order k, because

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle O\left[\nabla^m P_k\right] = k-m, \quad O\left[w(x)\right] = 2m \; \Longrightarrow\; O\left[w(x)\, \nabla^m P_k(x) \right] = k+m \; \Longrightarrow\; O\left[ \nabla^m \{w(x)\, \nabla^m P_k(x)\}\right] = k, }

where it was used that m times differentiation of a polynomial lowers its order by m and that the order of a product of polynomials is the product of the orders. Since we assumed that k ≤ l, the integral is non-zero only if k = l. Hence it follows readily that the associated Legendre polynomials of equal superscripts and non-equal subscripts are orthogonal. However, the hard work (given above) of computing the normalization for the case k = l remains to be done.

Theorem (orthogonality relation 2)

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-1}^{1}\frac{P_{l}^{m}(x) P_{l}^{n} (x)}{(1-x^{2})} dx = \delta_{mn}\frac{(l+m)!}{m(l-m)!} , m \neq 0 } .

where:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_{l}^{m}(x) = \frac{(-1)^{m}}{(2^{l} l!)} (1-x^{2})^{m/2} \frac{d^{l+m}}{dx^{l+m}} (x^{2}-1)^{l} , 0 \leqq m \leqq l} .

Proof

Let:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{l}^{mn} = \int_{-1}^{1}\frac{P_{l}^{m}(x) P_{l}^{n} (x)}{(1-x^{2})} \ dx = \frac{(-1)^{m+n}}{(2^{l} l!)^2} \int_{-1}^{1} \{(1-x^{2})^{\frac{(m+n)}{2}-1} \frac{d^{l+m}}{dx^{l+m}} (x^{2}-1)^{l}\} \{\frac{d^{l+n}}{dx^{l+n}} (x^{2}-1)^{l}\} dx} .

The functions (x²-1)^k are even functions; so their j^th order derivatives are even or odd functions according as j is even or odd. Therefore, if m or n is even but the other is odd, then one of the two factors in curly braces in the preceding expression is an even function, and the other is an odd function. This makes the entire integrand an odd function. When integrated between the limits that are negatives of each other, it yields 0, as it should, since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta_{mn} = 0} . So we need to consider further only even integrands, for which m and n are either both even or both odd. In this case all exponents in the integrand are non-negative integers (except when m = n = 0).

The integrand can be integrated by parts Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell+n} times, the first factor in curly braces and its derivatives being identified as u and the second factor as v' in the formula:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{-1}^{1} uv' dx = uv \Big|_{-1}^{1} - \int_{-1}^{1} vu' dx} .

Since m and n occur symmetrically, we can assume without loss of generality that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \leqq m} . For the first n-1 integrations by parts, the uv term vanishes at the limits because u includes at least one factor of (1-x²). The same also is true for the n^th integration, unless n = m, in which case the uv term is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{l}^{m} = \frac{(-1)^{m+1}}{(2^{l} l!)^{2}} \frac{d^{m-1}}{dx^{m-1}}\{(1-x^{2})^{m-1} \frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l}\} \{\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l} \} \Big|_{-1}^{1}} .

For the remaining $\ell$ integrations the uv term vanishes at the limits because v includes at least one factor 1-x² (and perhaps u does also). The result is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{l}^{mn} = \delta_{mn}S_{l}^{m} + \frac{(-1)^{l+m}}{(2^{l} l!)^{2}} \int_{-1}^{1} (x^{2}-1)^{l} \frac{d^{l+n}}{dx^{l+n}}\{(1-x^{2})^{\frac{(m+n)}{2}-1} \frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l}\} dx} .

The highest power of x in the binomial expansion of (x²-1)^l is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\ell} ; after Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell+m} derivatives of it are taken, the highest power is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell-m} . The highest power of x in the expansion of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (1-x^{2})^{\frac{(m+n)}{2}-1}} is m+n-2; so the highest power in the expression in curly braces is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell+n-2} . After Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell+n-2} derivatives of it are taken, the highest power is 0; that is, the expression is a constant. Taking the remaining 2 derivatives causes the integrand to vanish. Therefore:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{l}^{mn} = \delta_{mn}S_{l}^{m} = \delta_{mn}\frac{1}{(2^{l} l!)^{2}} \frac{d^{m-1}}{dx^{m-1}}\{(x^{2}-1)^{m-1} \frac{d^{l+m}}{dx^{l+m}} (x^{2}-1)^{l}\} \{\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}\} \Big|_{-1}^{1}} .

Since this function is odd, we need evaluate it only at its upper limit x = 1 and double the result.

An expression of the form Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x^{2}-1)^{k} = (x-1)^{k}(x+1)^{k}} can be differentiated using Leibnitz's rule. Only one term in the sum is of interest here, namely the one in which (x-1)^k is differentiated exactly k times so that no factors of (x-1) remain. If it is differentiated fewer times, then it vanishes at x = 1. If it is differentiated more times, then it vanishes everywhere. So, ignoring terms with factors of (x-1), we have (from right to left in the preceding expression):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^{l}}{dx^{l}}(x^{2}-1)^{l} = [\frac{d^{l}}{dx^{l}}(x-1)^{l}](x+1)^{l} = l! (x+1)^{l}} , which equals Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2^{l} \ell!} when x = 1:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l} = \frac{(l+m)!}{l!m!} [\frac{d^{l}}{dx^{l}} (x-1)^{l}] [\frac{d^{m}}{dx^{m}}(x+1)^{l}]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{(l+m)!}{l!m!} [l!] [\frac{l!}{(l-m)!} (x+1)^{l-m}]}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = (l+m)! \frac{l!}{m!(l-m)!} (x+1)^{l-m}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d^{m-1}}{dx^{m-1}}\{(x^{2}-1)^{m-1} \frac{d^{l+m}}{dx^{l+m}}(x^{2}-1)^{l}\}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = (l+m)! \frac{l!}{m!(l-m)!} \frac{d^{m-1}}{dx^{m-1}}\{(x-1)^{m-1}(x+1)^{l-1}\}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = (l+m)! \frac{l!}{m!(l-m)!} (m-1)! (x+1)^{l-1}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = (l+m)! \frac{l!}{m(l-m)!} (x+1)^{l-1}} , which equals Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2^{l-1}(l+m)! \frac{l!}{m(l-m)!} } when x = 1.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{l}^{mn} = 2\delta_{mn}\frac{1}{(2^{l} l!)^{2}} \{2^{l-1}(l+m)! \frac{l!}{m(l-m)!}\} \{2^{l} l!\} = \delta_{mn}\frac{(l+m)!}{m(l-m)!}}

which has been doubled to include the lower limit x = -1. QED.

When n = m = 0:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_{l}^{00} = \int_{-1}^{1} \frac{P_{l}^{0}(x) P_{l}^{0} (x)}{(1-x^{2})} dx}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{1}{(2^{l} l!)^{2}} \int_{-1}^{1} \{(1-x^{2})^{-1} \frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}\} \{\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}\} dx }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{1}{(2^{l} l!)^{2}} \int_{-1}^{1} \frac{1}{2}[(x+1)^{-1} - (x-1)^{-1}] \{\frac{d^{l}}{dx^{l}}(x^{2}-1)^{l}\}^{2} dx} .

When expanded in powers of (x+1) and (x-1) by using Leibnitz's rule, the expression in curly braces includes exactly one term, (x-1)^l, having no factor (x+1) and exactly one term, (x+1)$^{l}$, having no factor (x-1). Therefore in a Laurent series expansion of the integrand about either x = 1 or x = -1 all terms except one have non-negative exponents, and their integral is finite. The one remaining term in the integrand is a constant multiple of 1/(x+1) or 1/(x-1); so its integral logarithmically diverges at the limits, making T$_{l}^{00}$ infinite.