What is the Fourier Transform of $f(x) = x^2$? What is the Fourier transform of $f(x) = x^2$ ?
I am unsure how the problem is tackled.
 A: Recall that for the function $f_0(x) := x^0 = 1$, we have
$$ \def\F{\mathscr F}\F(f_0) = \sqrt{2\pi} \delta $$
Moreover, if $f_1(x) = x$, then for each $u \in \mathscr S'(\mathbf R)$, we have
$$ \F(f_1 u) = i\F(u)' $$
Hence, for $f = f_1^2f_0$, we have
$$ \F(f_1^2f_0) = i^2\sqrt{2\pi}\delta'' = -\sqrt{2\pi} \delta''$$
A: Using the following definition of the FT
\begin{equation}
\mathcal{F}\{\rho(x)\} \equiv s(k)= \int_{-\infty}^{\infty}\rho(x)e^{-i2\pi kx} dx
\end{equation}
the derivative property of the FT is
\begin{equation}
\mathcal{F}\{\rho'(x)\} = i2\pi k s(k).
\end{equation}
Then, recalling that $\mathcal{F}\{1\} = \delta(k)$ where $\delta(\cdot)$ is a Dirac delta function, we have that
\begin{aligned}
\mathcal{F}\{1\} &= i2\pi k \mathcal{F}\left\{x\right\} \\
\mathcal{F}\left\{x\right\} &= \left(\frac{1}{i2\pi k}\right)\delta(k).
\end{aligned}
Since $\frac{d}{dx} \frac{1}{2}x^2 = x$,
\begin{equation}
\mathcal{F}\{x\} = i2\pi k \mathcal{F}\left\{\frac{1}{2}x^2\right\}.
\end{equation}
Substituting in $\mathcal{F}\left\{x\right\}$ from the previous equation and simplifying, we have
\begin{aligned}
\mathcal{F}\{x^2\} &= 2 \left(\frac{1}{i2\pi k}\right)^2 \delta(k) \\
&= -\frac{1}{2\pi^2 k^2} \delta(k).
\end{aligned}
Repeating the process, it can be shown that
\begin{aligned}
\mathcal{F}\{x^m\} &= m! \left(\frac{1}{i2\pi k}\right)^m \delta(k).
\end{aligned}
