Continuous-time Fourier transform of $t^n$

Question

I was trying to calculate the Continuous-time Fourier Transform of $t^n, n=0,1,2,...$.

What I did is:
$$x(t)=t^n,\;x^{(1)}(t)=nt^{n-1},\;x^{(2)}(t)=n(n-1)t^{n-2},\;...,\;x^{(n)}(t)=n!$$
and then because
$$\mathcal{F}[x^{(n)}(t)](λ) = (iλ)^n\mathcal{F}[x(t)](λ)$$
we have
$$\mathcal{F}[t^n](λ) = (iλ)^{-n}\mathcal{F}[n!](λ) = (iλ)^{-n}n!\mathcal{F}[1](λ) = (iλ)^{-n}\,n!\,(2πδ(λ))$$

However, I found in Wolfram Alpha that $\mathcal{F}[t^n](λ) = 2πe^{-3/2iπn}δ^{(n)}(λ)$.

The problem is I cannot understand what I am doing wrong, can anyone help?
Thanks a lot.

Answer 1

To see your mistake, take a look at the case $n=1$. We have $x(t)=t$ and $x'(t)=1$. This results in

$$2\pi\delta(\lambda)=i\lambda X(\lambda)\tag{1}$$

Note that the LHS of $(1)$ is zero for $\lambda\neq 0$, and, consequently, $X(\lambda)$ must also be zero for $\lambda\neq 0$. So you can't just divide by $i\lambda$ because we're only interested in $\lambda=0$.

However, Eq. $(1)$ can be solved by noticing that

$$F(\lambda)\delta'(\lambda)=F(0)\delta'(\lambda)-F'(0)\delta(\lambda)\tag{2}$$

assuming that $F(\lambda)$ is continuous at $\lambda=0$. With $F(\lambda)=\lambda$ we get from $(2)$

$$\lambda\delta'(\lambda)=-\delta(\lambda)\tag{3}$$

Consequently, $X(\lambda)=2\pi i\delta'(\lambda)$ solves $(1)$.

In general, computing the Fourier transform of $x(t)=t^n$ is most straightforward by using the differentiation property and the duality of the Fourier transform:

$$(i\lambda)^{n}\Longleftrightarrow\delta^{(n)}(t)\tag{4}$$

and, from duality,

$$(-it)^n\Longleftrightarrow 2\pi\delta^{(n)}(\lambda)\tag{5}$$

from which we finally get

$$t^n\Longleftrightarrow 2\pi i^n\delta^{(n)}(\lambda)\tag{6}$$

Note that WolframAlpha's default scaling and sign conventions for the Fourier transform are different from the ones you're using, that's why I get a factor of $\sqrt{2\pi}$ when I click on your link (and instead of that exponential factor I directly get $(-i)^n$).