So I am given this simple example, where $T \in \mathcal{S}(\mathbb{R})$:

\begin{equation} T=(\mu +\lambda x+\beta x^2)H(x) \end{equation} where $H(x)\in \mathcal{S}(\mathbb{R})$ (also notated as the Heaviside $\theta$). Also I use the following definition for the Fourier transform: \begin{equation} \hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx, k\in \mathbb{R} \end{equation}

Now I would like to derive the Fourier transform of the above,so I make use of the properties of the Fourier transform for tempered distributions. For example, since I know the Fourier transform of $H(x)$ to be equal to: \begin{equation} \hat{H}(k)=-\frac{i}{k\sqrt{2\pi}}+\sqrt{\frac{\pi}{2}}\delta(k) \end{equation} I can use the relation:

\begin{equation} \mathcal{F}[-ixH(x)]=\hat{H}'(k) \Leftrightarrow \mathcal{F}[xH(x)]=i\hat{H}'(k) \end{equation} and derive the Fourier transform for $\lambda(xH(x))$. I guess I can do that also for the expression $\beta (x^2H(x))$, right? In the end I found my result to be: \begin{equation} \hat{T}(k)=\frac{1}{\sqrt{2\pi}}\left( \frac{2i\beta -\lambda k-\mu ik^2}{k^3}\right)+\sqrt{\frac{\pi}{2}}\left( \delta(k)+i\delta'(k)-\delta''(k)\right) \end{equation} and Mathematica gives: \begin{equation} \hat{T}(k)=-\sqrt{\frac{2}{\pi}}\frac{i \beta }{k^3}+\sqrt{\frac{\pi}{2}} \mu \delta(k)-\frac{\lambda}{\sqrt{2\pi}k^2}+\frac{i \mu }{\sqrt{2\pi}k} \end{equation} So I have a couple of $\delta$ derivatives which should not be there.. Why am I so wrong?

Thank you!


1 Answer 1


Your solution is correct, apart from the fact that you forgot the scale factors for the Dirac terms. It should be

$$\hat{T}(k)=\frac{1}{\sqrt{2\pi}}\left( \frac{2i\beta -\lambda k-\mu ik^2}{k^3}\right)+\sqrt{\frac{\pi}{2}}\left( \mu\delta(k)+i\lambda\delta'(k)-\beta\delta''(k)\right)$$

It looks like Mathematica can't be trusted here. First of all, it gets the sign of the imaginary part wrong, at least according to the definition of the Fourier transform that you use. You should check how Mathematica defines the Fourier transform. Probably it uses a different sign in the exponent, which would explain the different sign of the imaginary part. If this is the case, then - according to its definition of the Fourier transform - Mathematica is correct concerning the non-distributional part of the answer. However, your answer including the derivatives of the Dirac delta impulse is definitely correct, and it seems that Mathematica does not handle the derivatives correctly here.

Note that if you use Wolfram Alpha to compute the Fourier transform of the different terms separately, you do get derivatives of the Dirac delta: link1, link2

Just when computing the Fourier transform of the expression as a whole, the result misses the derivatives of the Dirac impulse: link3.

As a final sanity check, you can use Wolfram Alpha to compute the inverse Fourier transform of its Fourier transform result, which shows that you do not get back the original function: link4

  • $\begingroup$ Thank you, I was really not sure if my result was correct, since I am on an introductory level as far as the Fourier transform of the tempered distributions are concerned. $\endgroup$
    – Bazinga
    Commented Jun 30, 2015 at 16:24

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