Show that all derivatives of $f(x) = \frac{\sin(2\pi x)}{\pi x}$ belong to $L^2(\mathbb{R})$ 
I would like to show that all derivatives of $f(x) = \frac{\sin(2\pi x)}{\pi x}$ belong to $L^2(\mathbb{R}).$

One route I have tried is a power series approach. If we expand our function $f$ in a convergent power series, we get
$$f(x) = 2 - \frac{(2)^3}{3!}(\pi x)^2 + \frac{(2\pi)^5}{5!}(\pi x)^4 - \cdots.$$
So we can differentiate term by term to produce the power series for the $k$th order derivative $f^{(k)}$, for all $k \in \mathbb{N}$. Using this perspective, it's clear that we do not have any singularity at $x = 0$ that would cause $\int (f^{(k)})^2$ to blow up there. However, the power series method does not clearly show how each $f^{(k)}$ decays at infinity. 
To show the decay at infinity, I have tried the more brute force approach of just applying the quotient rule. For instance, away from $x = 0$ we have 
$$f'(x)= \frac{2\cos(2\pi x)}{x} - \frac{\sin(2\pi x)}{\pi x^2}.$$
Each term of $(f')^2$ will then have an $x^n$ factor, where $n \ge 2$, so that $f'$ is square integrable away from zero. This idea seems to have promise, but using the quotient rule it is more difficult to establish a formula for each $f^{(k)}$.
I would appreciate any comments about how to combine my thoughts into a finished argument. Alternative solutions are also welcome.
 A: Since $f \in L^2(\mathbb{R}) \iff \hat{f} \in L^{2}(\mathbb{R})$, then it suffices to work in the frequency domain. Using the unitary, ordinary frequency convention for the Fourier transform, we know that
\begin{align}
\mathcal{F}[f^{(n)}](\xi) = (2\pi i\xi)^{n}\hat{f}(\xi)
\end{align}
But since $f$ is a $\mathrm{sinc}$ function, $\hat{f}$ has compact support, and hence $\mathcal{F}[f^{(n)}] \in L^{2}(\mathbb{R})$.
A: Using the power series of $\sin$, one can easily see that $f$ is smooth on $\mathbb R$. So each $f^{(n)}$ exists and is continuous on $\mathbb R$ and to ensure that $f^{(n)}$ belongs to $L^2(\mathbb R)$ all you need to know is its summability at $\pm \infty$.
For $x\neq 0$, one can write $f(x)$ as $f(x)=h(x)g(x)$ with $h(x)=\sin(2\pi x)$ and $g(x)=\dfrac{1}{\pi x}$.
Moreover, $$h^{(k)}(x)=(2\pi)^k\sin\left(2\pi x+\dfrac{k\pi}{2}\right)$$ and 
$$g^{(k)}(x)=\dfrac{(-1)^k k!}{\pi}\dfrac{1}{x^{k+1}} $$
thus, using the general Leibniz rule, we get for any $x\neq 0$ :
$$f^{(n)}(x)=\sum_{k=0}^n \binom {n}{k}\dfrac{(-1)^k k! (2\pi)^{n-k}}{\pi \cdot x^{k+1}}\sin\left( 2\pi x+\dfrac{(n-k)\pi}{2}\right)$$
Each term on the right hand side is a continuous and square integrable function at $\pm \infty$ so $\left(f^{(n)}\right)^2$ is summable at $\pm \infty$ and $f^{(n)}$ belongs to $L^2(\mathbb R)$.
A: The Nick Thompson's answer is perfectly fine, but maybe you want to avoid to consider the Fourier transform. Notice that, since $g(x)=\frac{\sin x}{x}$ satisfies the differential equation:
$$ x g''=xg-2g'\tag{1}$$
by differentiating both sides $n$ times we get:
$$ x g^{(n+2)}+(n+2) g^{(n+1)} = -\left(x g^{(n)} + n g^{(n-1)}\right)\tag{2}$$
and by setting $h_n(x)=x g^{(n)}+n g^{(n-1)}$ the last identity takes the form:
$$ h_{n+2}(x) = -h_{n}(x).\tag{3} $$
Notice that $h_1(x)=\cos x$ and $h_2(x)=-\sin x$, so $|h_n(x)|\leq 1$. Now:
$$ \left|x g^{(n+1)}+(n+1)g^{(n)}\right|\leq 1  \tag{4}$$
leads to:
$$ \left|x^{n+1} g^{(n+1)}+(n+1)x^n g^{(n)}\right|\leq x^n, $$
$$\left| \frac{d}{dx} x^{n+1}g^{(n)}\right|\leq x^n,$$
$$ \left| g^{(n)} \right |\leq \color{blue}{\frac{1}{n+1}}.\tag{5}$$
If now we plug back $(5)$ into $(4)$ we get:
$$ \left| x g^{(n+1)} \right| \leq 2 \tag{6} $$
so:
$$ \int_{1}^{+\infty} \left(g^{n}(x)\right)^2\,dx \leq 4\int_{1}^{+\infty}\frac{dx}{x^2}=\color{red}{4}.\tag{7}$$
Now, we have just to bound the $L^2$ norm on $[0,1]$. By putting together $(5)$ and $(7)$, we get:
$$\int_{0}^{+\infty}\left( g^{(n)}(x)\right)^2\,dx \leq \color{blue}{\frac{1}{(n+1)^2}}+\color{red}{4}.\tag{8}$$
Using $(5)$ and $(6)$, the last bound can be improved up to:
$$\color{green}{\int_{0}^{+\infty}\left( g^{(n)}(x)\right)^2\,dx} \leq\int_{0}^{+\infty}\min\left(\frac{1}{n+1},\frac{2}{x}\right)^2\,dx = \color{green}{\frac{4}{n+1}}.\tag{9} $$
The last line also gives that the $L^2$ norm of $g^{(n)}$ goes to zero as $n\to +\infty$, proving that $\widehat{g}$ is compact-supported, for instance. Also notice that the bound found in $(9)$ is essentially the best one up to a multiplicative constant, since by considering the Fourier transform of $g$ we have:
$$\int_{0}^{+\infty}\left( g^{(n)}(x)\right)^2\,dx=\color{purple}{\frac{\pi}{4n+2}}.\tag{10}$$
A: The functions
$$s_{n,a}(x)=\frac{\sin(x)}{ax^n}\qquad c_{n,a}(x)=\frac{\cos(x)}{ax^n}$$
are $L^2$ on $[1,+\infty)$ (and $(-\infty, -1]$) for every $n\geq 1$ and every $a\in\mathbb{R}\setminus\{0\}$.
We also have that
$$\dfrac{d}{dx}s_{n,a}(x)=c_{n,a}(x)-ns_{n+1,a}(x)$$
$$\dfrac{d}{dx}c_{n,a}(x)=-s_{n,a}(x)+ns_{n+1,a}(x)$$
So if we set
$$V=\mathrm{Span}\{ s_{n,a}(x), c_{n,a}(x)\ \vert\ n\geq 1,\ a\in\mathbb{R}^*\}$$
then every element of $V$ is in $L^2([1,+\infty))$, moreover, the derivative of an element of $V$, as a function in $\mathcal{C}^\infty((0,+\infty))$ is again in $V$.
The function $f(x)=2\sin(x)/x$, which is analytic on $\mathbb{R}$, hence $\mathcal{C}^\infty$, belongs to $V$, as all its derivatives. Therefore $f$ and all its derivatives are in $L^2([1,+\infty))$ (and $L^2((-\infty, -1])$ so, being continuous, they are in $L^2(\mathbb{R})$. Replacing $x$ with $2\pi x$ we obtain the original function.
