Evaluating the $L_2[-1, 1]$ inner product on rescaled Legendre polynomials

Question

Let $z_n(t) = \sqrt{\frac{2n+1}{2}} \frac{1}{2^n n!} \frac {d^n}{dt^n} (t^2-1)^n$, a rescaled Legendre polynomial. As an intermediate step of a larger problem, I need to show that in terms of the $L_2[-1, 1]$ inner product, $<z_n, z_m> = 1$ if $n = m$, $0$ otherwise. To that end I note that:

(1) $(t^2-1)^n = \sum_{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)t^{2n-2k}(-1)^k\text ,$ so $\left(\frac{d^n}{dt^n}(t^2-1)^n\right)$ $= \sum _{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)\left(\prod _{j=0}^{n+1}(2n-2k-j)\right)t^{2n-2k}(-1)^k$ $= \sum _{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)\left(\frac{(2n-2k)!}{(2n-2k-n)!}\right)t^{2n-2k}(-1)^k\text,$ so $\left(\frac{d^{2n}}{dt^{2n}}(t^2-1)^n\right)$ $= \left(\begin{matrix}n\\0\end{matrix}\right)(-1)^0\frac{(2n)!}{(2n-n)!} = \frac{(2n)!}{(2n-n)!}$.

(2) $\left(\frac{d^m}{dt^m}(t^2-1)^n\right)|_{t=1}=\left(\frac{d^m}{dt^m}(t^2-1)^n\right)|_{t=-1}=0$ whenever m < n.

Then I try to actually evaluate the inner product:

$<z_n,z_m>= \int_{-1}^1\left(\sqrt{\frac{2n+1} 2}\frac1{2^nn!}\frac{d^n}{dt^n}(t^2-1)^n\right)\left(\sqrt{\frac{2m+1}2}\frac 1{2^mm!}\frac{d^m}{dt^m}(t^2-1)^m\right)dt$ $=\sqrt{\frac{2n+1} 2}\frac 1{2^nn!}\sqrt{\frac{2m+1} 2}\frac 1{2^mm!}\int_{-1}^1\left(\frac{d^n}{dt^n}(t^2-1)^n\right)\left(\frac{d^m}{dt^m}(t^2-1)^m\right)dt$ $=\sqrt{\frac{(2n+1)(2m+1)}{2^2}}\frac 1{2^{n+m}n!m!}\int_{-1}^1\left(\frac{d^n}{dt^n}(t^2-1)^n\right)\left(\frac{d^m}{dt^m}(t^2-1)^m\right)dt$ $=C\int_{-1}^1\left(\frac{d^n}{dt^n}(t^2-1)^n\right)\left(\frac{d^m}{dt^m}(t^2-1)^m\right)dt$ where $C=\frac{\sqrt{(2n+1)(2m+1)}}{2^{n+m+1}n!m!}$, $=C\left(\left(\left(\frac{d^n}{dt^n}(t^2-1)^n\right)\left(\frac{d^{m-1}}{dt^{m-1}}(t^2-1)^m\right)\right)|_{-1}^1-\left(\int_{-1}^1\left(\frac{d^{n+1}}{dt^{n+1}}(t^2-1)^n\right)\left(\frac{d^{m-1}}{dt^{m-1}}(t^2-1)^m\right)dt \right)\right)$ $=C\left(-\left(\int_{-1}^1\left(\frac{d^{n+1}}{dt^{n+1}}(t^2-1)^n\right)\left(\frac{d^{m-1}}{dt^{m-1}}(t^2-1)^m\right)dt\right)\right)$ $= ...$ $= C(-1)^m \int_{-1}^1 \left(\frac{d^{2n}}{dt^{2n}}(t^2-1)^n\right)\left(\frac{d^{m-n}}{dt^{m-n}}(t^2-1)^m\right)dt$ $=C(-1)^m\left(\frac{d^{2n}}{dt^{2n}}(t^2-1)^n\right)\int _{-1}^1\left(\frac{d^{m-n}}{dt^{m-n}}(t^2-1)^m\right)dt$, with (2) conveniently making the first term of each integration by parts 0 until all that's left is the second term of the very last one.

I know that when m > n, this is (again by (2)) $C(-1)^m\left(\frac{d^{2n}}{dt^{2n}}(t^2-1)^n\right)\left(\frac{d^{m-n-1}}{dt^{m-n-1}}(t^2-1)^m\right)|_{-1}^1 = 0.$ However, trying to evaluate this for m = n is an absolute nightmare, my goal essentially being to show that $\frac{(2n)!}{(2n-n)!}\int _{-1}^1\sum_{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)t^{2n-2k}(-1)^k = C^{-1}$. Actually evaluating the sum I get if I pull everything out of the integration possible and then perform the integration doesn't seem to be humanly possible. Is there an error in my work so far, or am I missing some kind of obvious trick for the last part?

Answer 1

I finally got this to work, so I'd like anyone looking for answeres here to know I made a mistake in (1); it should say: $\left(\frac{d^n}{dt^n}(t^2-1)^n\right)$ $= \sum _{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)\left(\prod _{j=0}^{n-1}(2n-2k-j)\right)t^{2n-2k}(-1)^k$ $= \sum _{k=0}^n\left(\begin{matrix}n\\k\end{matrix}\right)\left(\frac{(2n-2k)!}{(n-2k)!}\right)t^{2n-2k}(-1)^k$ (but with 0 instead of n - 2k whenever n - 2k < 0), and so $\left(\frac{d^{2n}}{dt^{2n}}(t^2-1)^n\right)$ $=(2n)!$.

That in mind, the apparently best way to come up with an inner product of 1 is to convert the integral into one in terms of $t = cos(\theta)$, $dt= sin(\theta)d\theta$, change the bounds as $-1 = cos(\pi)$ and $1 = cos(0)$, and then use an integration by parts trick to get a recursive definition of the integral we actually wanted a value for; ultimately it comes out to $\int _{-1}^1 (t^2 -1)^n = (-1)^n \frac{2^{2n+1} (n!)^2}{(2n+1)!}$, which neatly turns the whole thing into 1 as desired (after the correction to the 2nth-derivative bit in light of the repairs to (1)).