What is the Fourier Transform of $f(x) = x^2$?

Question

What is the Fourier transform of $f(x) = x^2$ ?

I am unsure how the problem is tackled.

Answer 1

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 $$ \F(f_1^2f_0) = i^2\sqrt{2\pi}\delta'' = -\sqrt{2\pi} \delta''$$

Answer 2

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}