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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!

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  • $\begingroup$ Check if this helps you in any way. $\endgroup$
    – creative
    Jun 29, 2015 at 9:13
  • $\begingroup$ @Loophole, Uhm, I did that and I know the result. The thing is that I do know how to reach there by applying the definition of the Fourier transform for a tempered function.. $\endgroup$
    – Bazinga
    Jun 29, 2015 at 9:17
  • $\begingroup$ correction to my above comment: I meant "do not know how to reach...." $\endgroup$
    – Bazinga
    Jun 29, 2015 at 10:09
  • $\begingroup$ Try integration by parts. Transfer the derivative from the delta function to the other term. Repeatedly. $\endgroup$
    – orion
    Jun 29, 2015 at 10:33
  • $\begingroup$ @orion I did that but in the end you do not end up with $\delta'''$.. $\endgroup$
    – Bazinga
    Jun 29, 2015 at 11:37

1 Answer 1

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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 $f(y)$.

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  • $\begingroup$ That looks good! So in the first all I have to do is compute the Fourier transform of the delta then take the 3rd derivative and it is enough to replace the expression on the integral of my function with the one of $\delta '''$ right? $\endgroup$
    – Bazinga
    Jun 29, 2015 at 14:44
  • $\begingroup$ Exactly. All you need is the first 4 lines. The rest is just a sketch of a proof (to show that this derivative of the delta function does what it's supposed to do for well-behaved functions). $\endgroup$
    – orion
    Jun 29, 2015 at 14:56
  • $\begingroup$ Yeap I see! Thank you! $\endgroup$
    – Bazinga
    Jun 29, 2015 at 15:50
  • $\begingroup$ note that the Fourier transform of $\delta'''$ is obviously $h'''(0)$ with $h(x) = e^{-2 i \pi \xi x}$ i.e. $h'''(0) = (-2i\pi)^3 \xi^3$. all you have to do next is proving the Fourier inversion theorem for the tempered distributions, not so hard with mollifiers : if $T$ is tempered then $T \ast \varphi$ is $C^\infty$ and tempered with $\varphi$ a Schwartz function , and $\phi . (T \ast \varphi)$ is Schwartz, hence its Fourier transform is $\hat{\phi} \ast (\hat{T}. \hat{\varphi})$. $\endgroup$
    – reuns
    Jun 1, 2016 at 0:04
  • $\begingroup$ finally as usual consider $\varphi_\epsilon(x) = e^{-\epsilon^2 x^2}$ and $\phi_\epsilon(x) = \frac{1}{|\epsilon| \sqrt{\pi}} e^{-x^2/\epsilon^2}$, and let $\epsilon \to 0$. since $\hat{\varphi_\epsilon} = C\phi_{\epsilon/\sqrt{\pi}}$ for some constant $C$, everything is smooth ! @Mitscaype $\endgroup$
    – reuns
    Jun 1, 2016 at 0:09

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