User:Dan Nessett/Sandboxes/Sandbox 3

The account of this former contributor was not re-activated after the server upgrade of March 2022.

Theorem

$\int _{-1}^{1}{\frac {P_{l}^{m}(x)P_{l}^{n}(x)}{(1-x^{2})}}dx=\delta _{mn}{\frac {(l+m)!}{m(l-m)!}},m\neq 0$.

where:

$P_{l}^{m}(x)={\frac {(-1)^{m}}{(2^{l}l!)}}(1-x^{2})^{m/2}{\frac {d^{l+m}}{dx^{l+m}}}(x^{2}-1)^{l},0\leqq m\leqq l$.

Proof

Let:

$T_{l}^{mn}=\int _{-1}^{1}{\frac {P_{l}^{m}(x)P_{l}^{n}(x)]}{(1-x^{2})}}dx={\frac {(-1)^{m+n}}{(2^{l}l!)^{2}}}\int _{-1}^{1}\{(1-x^{2})^{{\frac {(m+n)}{2}}-1}{\frac {d^{l+m}}{dx^{l+m}}}(x^{2}-1)^{l}\}\{{\frac {d^{l+n}}{dx^{l+n}}}(x^{2}-1)^{l}\}dx$.

The functions (x$^{2}$-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 $\delta _{mn}=0$. 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 l+n times, the first factor in curly braces and its derivatives being identified as u and the second factor as v' in the formula:

$\int _{-1}^{1}uv'dx=uv{\Big |}_{-1}^{1}-\int _{-1}^{1}vu'dx$.

Since m and n occur symmetrically, we can assume without loss of generality that $n\leqq m$. For the first n-1 integrations by parts, the uv term vanishes at the limits because u includes at least one factor of (1-x$^{2}$). The same also is true for the n$^{th}$ integration, unless n = m, in which case the uv term is:

$S_{l}^{m}={\frac {(-1)^{m+1}}{(2^{l}l!)^{2}}}{\frac {d^{m-1}}{dx^{m-1}}}\{(1-x^{2})^{m-1}{\frac {d^{l+m}}{dx^{l+m}}}(x^{2}-1)^{l}\}\{{\frac {d^{l}}{dx^{l}}}(x^{2}-1)^{l}\}{\Big |}_{-1}^{1}$.

For the remaining l integrations the uv term vanishes at the limits because v includes at least one factor (1-x$^{2}$) (and perhaps u does also). The result is:

$T_{l}^{mn}=\delta _{mn}S_{l}^{m}+{\frac {(-1)^{l+m}}{(2^{l}l!)^{2}}}\int _{-1}^{1}(x^{2}-1)^{l}{\frac {d^{l+n}}{dx^{l+n}}}\{(1-x^{2})^{{\frac {(m+n)}{2}}-1}{\frac {d^{l+m}}{dx^{l+m}}}(x^{2}-1)^{l}\}dx$.

The highest power of x in the binomial expansion of (x$^{2}$-1)$^{l}$ is 2l; after l+m derivatives of it are taken, the highest power is l-m. The highest power of x in the expansion of (1-x$^{2}$)$^{(m+n)/2-1}$ is m+n-2; so the highest power in the expression in curly braces is l+n-2. After l+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:

$T_{l}^{mn}=\delta _{mn}S_{l}^{m}=\delta _{mn}{\frac {1}{(2^{l}l!)^{2}}}{\frac {d^{m-1}}{dx^{m-1}}}\{(x^{2}-1)^{m-1}{\frac {d^{l+m}}{dx^{l+m}}}(x^{2}-1)^{l}\}\{{\frac {d^{l}}{dx^{l}}}(x^{2}-1)^{l}\}{\Big |}_{-1}^{1}$.

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 $(x^{2}-1)^{k}=(x-1)^{k}(x+1)^{k}$ 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):

${\frac {d^{l}}{dx^{l}}}(x^{2}-1)^{l}=[{\frac {d^{l}}{dx^{l}}}(x-1)^{l}](x+1)^{l}=l!(x+1)^{l}$ , which equals 2$^{l}$ l! when x = 1:

${\frac {d^{l+m}}{dx^{l+m}}}(x^{2}-1)^{l}={\frac {(l+m)!}{l!m!}}[{\frac {d^{l}}{dx^{l}}}(x-1)^{l}][{\frac {d^{m}}{dx^{m}}}(x+1)^{l}]$

$={\frac {(l+m)!}{l!m!}}[l!][{\frac {l!}{(l-m)!}}(x+1)^{l-m}]$

$=(l+m)!{\frac {l!}{m!(l-m)!}}(x+1)^{l-m}$

${\frac {d^{m-1}}{dx^{m-1}}}\{(x^{2}-1)^{m-1}{\frac {d^{l+m}}{dx^{l+m}}}(x^{2}-1)^{l}\}$

$=(l+m)!{\frac {l!}{m!(l-m)!}}{\frac {d^{m-1}}{dx^{m-1}}}\{(x-1)^{m-1}(x+1)^{l-1}\}$

$=(l+m)!{\frac {l!}{m!(l-m)!}}(m-1)!(x+1)^{l-1}$

$=(l+m)!{\frac {l!}{m(l-m)!}}(x+1)^{l-1}$ , which equals $2^{l-1}(l+m)!{\frac {l!}{m(l-m)!}}$ when x = 1.

$T_{l}^{mn}=2\delta _{mn}{\frac {1}{(2^{l}l!)^{2}}}\{2^{l-1}(l+m)!{\frac {l!}{m(l-m)!}}\}\{2^{l}l!\}=\delta _{mn}{\frac {(l+m)!}{m(l-m)!}}$

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

When n = m = 0:

$T_{l}^{00}=\int _{-1}^{1}{\frac {P_{l}^{0}(x)P_{l}^{0}(x)}{(1-x^{2})}}dx$

$={\frac {1}{(2^{l}l!)^{2}}}\int _{-1}^{1}\{(1-x^{2})^{-1}{\frac {d^{l}}{dx^{l}}}(x^{2}-1)^{l}\}\{{\frac {d^{l}}{dx^{l}}}(x^{2}-1)^{l}\}dx$

$={\frac {1}{(2^{l}l!)^{2}}}\int _{-1}^{1}{\frac {1}{2}}[(x+1)^{-1}-(x-1)^{-1}]\{{\frac {d^{l}}{dx^{l}}}(x^{2}-1)^{l}\}^{2}dx$.

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.