Show that these polynomials are orthonormal in the $L^2$ scalar product 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.
 A: 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).
