I try to show, that the polynomials $$\varphi_k(x) = \dfrac{k!}{(2k)!} \dfrac{d^k}{dx^k} (x^2-1)^k$$ are orthonormal regarding to the $L^2$-scalar product over $[-1,1]$.
I think i have to show, that $$\int_{-1}^{1}\varphi_k(x) \varphi_j(x) =\left\{\begin{matrix} 1 ,\ \text{if} \ j=k& \\ 0, \ \text{if} \ j \ne k& \end{matrix}\right.$$
Is this correct?
I tried to do this. but didn't succeed. I thought i have to do partial integration but, i don't get it. Can anyone help me?
Thank you very much.
The polynomials are orthogonal but NOT orthoNORMAL
It seems that the polynomials as stated in your question are orthogonal but not orthonormal. First of all $\varphi_0\equiv 1$ has norm $\sqrt2\neq 1$. $k=1$ doesn't work either: $$ \varphi_1(x) = \frac12\frac\partial{\partial x}(x^2-1) = x $$ Then $$ \int_{-1}^1 \varphi_1(x) \varphi_1(x) \,{\rm d}x = \int_{-1}^1 x^2 \,{\rm d}x = \frac23 \neq 1 $$ and so on for all the other $k\in\mathbb N$.
I would rather state that $$\tag{Th} \varphi_k(x) ~=~ \frac{\sqrt{k+1/2}}{2^kk!} \frac{{\rm d}^k}{{\rm d}x^k} (x^2-1)^k $$ is the correct normalization.
A deeper look
First of all, for the sake of simplicity of notation, let us write $\partial_k$ in place of $\frac{\partial^k}{\partial x^k}$.
We want to normalize the functions $\psi_k=\partial_k(x^2-1)^k$ as $$ \varphi_k(x) = \alpha_k\psi_k(x) \qquad \exists\alpha_k\in\mathbb R \text{ to be determined} $$ so that the polynomials $\{\varphi_k\}_{k\in\mathbb N}$ are orthonormal.
Now, the interval $[-1,+1]$ is symmetric, and therefore the integral of odd functions over $[-1,+1]$ is $0$. Besides, if $f(x)$ is odd, then $f'(x)$ is even, and similarly if $f(x)$ is even, then $f'(x)$ is odd.
It will be useful to introduce the following notation: let $$ \sigma(f) = \begin{cases} +1 & \text{if $f$ is even}\\ -1 & \text{if $f$ is odd} \end{cases} $$ (the cases $f\equiv 0$ or $f$ neither even nor odd are not relevant, so let us not be fussy about it). Note that $$ \sigma(fg)=\sigma(f)\sigma(g) \quad\text{and}\quad \sigma(\partial_k f)=(-1)^k\sigma(f) $$ Now, $(x^2-1)^k$ is clearly an even function. For the above property, $$\tag{$\sigma$} \sigma(\psi_k) = (-1)^k \quad\text{and therefore}\quad \sigma(\psi_k\psi_j) = (-1)^{k+j} $$ It follows that if one between $k$ and $j$ is even and the other is odd, then $\langle\psi_k,\psi_j\rangle=0$.
It remains to check the cases $k,j$ both even and $k,j$ both odd. Assume wlog $1\leq j\leq k$ (the case when $j=0$ is trivial); integrating by parts, \begin{align} \langle\psi_k,\psi_j\rangle = & \int_{-1}^1 \partial_k(x^2-1)^k\,\, \partial_j(x^2-1)^j {\rm d}x \\ = & \partial_{k-1}(x^2-1)^k \partial_j(x^2-1)^j \Bigg|_{-1}^1 - \int_{-1}^1 \partial_{k-1}(x^2-1)^k\,\, \partial_{j+1}(x^2-1)^j {\rm d}x \end{align} Iterating $j$ times, we get \begin{align}\tag{1} \langle\psi_k,\psi_j\rangle = & \sum_{\ell=1}^{j} (-1)^{\ell+1} \partial_{k-\ell}(x^2-1)^k \partial_{j+\ell-1}(x^2-1)^j \Bigg|_{-1}^1 \\ & +\,\, (-1)^j \int_{-1}^1 \partial_{k-j}(x^2-1)^k\,\, \partial_{2j}(x^2-1)^j {\rm d}x \end{align} Note that $$ (x^2-1)^j = x^{2j} + \text{(polyn. of degree $2j-2$)} $$ therefore $$\tag{2} \partial_{2j}(x^2-1)^j=(2j)! $$ Besides, $(x^2-1)^k=(x-1)^k(x+1)^k$ has two roots $\pm1$ both of multiplicity $k$. Therefore $\pm1$ are both roots of the $\ell$-th derivatives $\partial_\ell(x^2-1)^k$, for $0\leq\ell<k$, that is, $$\tag{3} \partial_\ell\,(x^2-1)^k\big|_{x=\pm 1} ~=~ 0 \quad 0\leq\ell<k $$ It follows that every addend of the sum in $(1)$ is $0$.
From this fact and from $(2)$ we get $$\tag{4} \langle\psi_k,\psi_j\rangle = (-1)^j (2j)! \int_{-1}^1 \partial_{k-j}(x^2-1)^k {\rm d}x $$ If $k\neq j$, that is, if $k>j$ for our assumption, then $k-j-1\geq 0$ yielding $$ \langle\psi_k,\psi_j\rangle = (2j)! \partial_{k-j-1}(x^2-1)^k \Bigg|_{-1}^1 $$ Since $0\leq k-j-1<k$, from $(3)$ it follows that $$ \langle\psi_k,\psi_j\rangle = 0 \qquad j\neq k $$ So far we proved the orthogonality.
Let us focus on the case $j=k$. In light of $(4)$ (with $j=k$), $$ \langle\psi_k,\psi_j\rangle ~=~ (-1)^k(2k)!\int_{-1}^1(x^2-1)^k\,{\rm d}x $$ Iterating integration by parts, you can easily check that $$ \int_{-1}^1(x^2-1)^k = (-1)^k2^{2k+1}\frac{k!^2}{(2k+1)!} $$ so that $$ \langle\psi_k,\psi_j\rangle ~=~ (-1)^k2^{2k+1}\frac{k!^2}{(2k+1)!}(2k)!(-1)^k ~=~ 2^{2k}\frac{k!^2}{k+1/2} $$ Then, being $\varphi_k=\alpha_k\psi_k$ we have $$ \langle\varphi_k,\varphi_k\rangle ~=~ 2^{2k}\frac{k!^2}{k+1/2}\alpha_k^2 $$ In order to be orthonormal, the above quantity has to equal $1$, that is, $$ \alpha_k = \frac{\sqrt{k+1/2}}{2^kk!} $$ which proves my thesis (Th).