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!
Best Answer
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 $f(y)$.