Compact formula for the Fourier transform of $x^n$ exists?

Question

Is it possible to derive a compact formula for the Fourier transform of $f(x)=x^n$ given by $$\tilde{f}(k)=\int\limits_{-\infty}^{+\infty}x^n e^{ikx}dx$$ (where $x$ is a real variable and $n$ is a positive or negative integer or zero) which shows the sigularities explicitly for certain values of $n$?

Answer 1

Yes, the formal computations are indicated in the link by mithusengupta123 Computing Fourier transform of power law

However, as mentioned by Maximilian Janisch , $x^n$ is not an $L^1$ function, so in general you will need the Theory of Distributions to define the Fourier transform on these generalization of functions. A distributions $f$ is defined as linear forms on smooth and compactly supported functions $\varphi ∈ C^\infty_c(\mathbb{R})$, so that instead of looking to $f(x)$ you look at $\langle f,\varphi\rangle$ which if $f$ is a usual fonction will give $$ \langle f,\varphi\rangle := \int_{-\infty}^∞ f(x)\,\varphi(x)\,\mathrm{d}x. $$ Two distributions $f$ and $g$ are equal when for every $\varphi ∈ C^\infty_c(\mathbb{R})$, $$ \langle f,\varphi\rangle = \langle g,\varphi\rangle. $$ (and this is compatible with locally integrable functions: if two locally integrable functions $f$ and $g$ are equal in the above sense, then they are equal almost everywhere).

For example, if $f=x^n$, then $$ \langle f,\varphi\rangle := \int_{-\infty}^∞ x^n\,\varphi(x)\,\mathrm{d}x. $$ But there are other distributions that are not functions, such that the Dirac delta measure $\delta_0$. This distribution is defined by $$ \langle\delta_0,\varphi\rangle = \varphi(0). $$ (In particular $∫\delta_0(\mathrm{d}x) = \langle\delta_0,1\rangle =1$)

With this theory, you get that the Fourier transform of $x^n$ when $n\in\mathbb{N}$ is given by $$ \mathcal{F}(x^n) = 2π i^n \delta^{(n)}_0, $$ where $\delta^{(n)}_0$ is the $n^{th}$ derivative of the Dirac delta measure. When $n<0$, then you will find $$ \mathcal{F}(x^n) = i\pi \frac{(ix)^{|n|-1}}{(|n|-1)!} \mathrm{sign}(x). $$ (This one is in particular a function in the usual sense.)

To prove that, you can start with the case $n=0$ which is given by the Fourier inversion theorem, and then take the derivatives/antiderivatives of the identity obtained to get the others identities (the derivative in the sense of distribution being defined by $\langle f',\varphi\rangle := -\langle f,\varphi'\rangle$).