Fourier Transform of a Polynomial Suppose you are given the polynomial 
\begin{equation}
f(x)=1+x^3
\end{equation}
and the definition of Fourier transform:
\begin{equation}
\hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ikx}f(x)dx, k\in \mathbb{R}
\end{equation}
Obviously, that function has no Fourier transform but its correspoding tempered distribution does. So, in order to find that Fourier transform we do the following:
\begin{equation}
\langle \hat{f}, \phi \rangle =\langle f, \hat{\phi}\rangle=\int_{-\infty}^{\infty}f(x)\hat{\phi}(x)dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(1+x^3)\int_{-\infty}^{\infty}e^{-ixy}\phi(y)dydx \Leftrightarrow \\
\langle \hat{f}, \phi \rangle =\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-ixy}\phi(y)dydx+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}x^3\int_{-\infty}^{\infty}e^{-ixy}\phi(y)dydx
\end{equation}
Now we argue that the inner integral is uniformly convergent to the variable $x$ and therefore by Fubini's theorem we can change the integration order. Moreover, there improper integrals do exist and so do their correspoding Principal Values. So:
\begin{equation}
\langle \hat{f}, \phi \rangle =\frac{1}{\sqrt{2\pi}}\lim_{R\to \infty}\left[ \int_{-\infty}^{\infty}\phi(y)\int_{-R}^{R}e^{-ixy}dxdy+\int_{-\infty}^{\infty}\phi(y)\int_{-R}^{R}x^3e^{-ixy}dxdy \right]
\end{equation}
The first of these two integrals, I can see that is the $\delta(x)$ with a coefficient. Mathematica says that the second one:
\begin{equation}
\frac{1}{\sqrt{2\pi}}\lim_{R\to \infty}\int_{-\infty}^{\infty}\phi(y)\int_{-R}^{R}x^3e^{-ixy}dxdy
\end{equation}
is the $\delta'''(x)$ (again with a coefficient but that is easy to take care of.) 
My question is how to prove that the second integral is the 3rd "derivative" of the Delta distribution. 
Thank you!
 A: Well, derivation under the integral sign gives you directly
$$\delta(y)=\frac{1}{2\pi}\int e^{-ixy}dx$$
$$\delta'(y)=\frac{-i}{2\pi}\int xe^{-ixy}dx$$
$$\delta''(y)=\frac{-1}{2\pi}\int x^2 e^{-ixy}dx$$
$$\delta'''(y)=\frac{i}{2\pi}\int x^3 e^{-ixy}dx$$
Then, by integration by parts, you can verify what it does as a distribution:
$$\int \delta(y)f(y)dy=f(0)$$
$$\int \delta'(y)f(y)dy=\underbrace{\delta(y)f(y)|_{-\infty}^\infty}_0-\int \delta(y)f'(y)dy=-f'(0)$$
$$\int \delta''(y)f(y)dy=f''(0)$$
and so on.
Now you can start with a finite range where the by-parts won't be zero, and you will get the conditions for "convergence".
$$\int_{-A}^A \delta'(y)f(y)dy=-i\int_{-A}^A \int_{-R}^R x e^{-ixy}f(y)dx\,dy=$$
$$=-i\int_{-R}^R e^{-ixy}f(y)|_{-A}^A \, dx+i\int_{-A}^A \int_{-R}^R e^{-ixy}f'(y)dx\,dy$$
$$=-2i \sin AR \frac{f(A)-f(-A)}{A}+i\int_{-A}^A \int_{-R}^R e^{-ixy}f'(y)dx\,dy$$
In the limit, the second integral goes back to the delta function definition, while the first part is supposed to go to $0$ for a well behaved $f(y)$.
