The function in question is $f(x) = H(a - |x|)$, where the Fourier transform is given by $F(k) = (\frac{2}{k}) \sin(ak)$.

Initial attempt:

$F(k) = ( H(a - |x|), e^{-ikx} )$ = $- (\delta(a - |x|), -\frac{1}{ik} e^{-ikx} )$ = $ \frac{1}{ik} \int_{-\infty}^{\infty} \delta(a - |x|) e^{-ikx} dx$. (The first two equalities follow from the distribution theory. Noting that $p_{+} ' = H$, and $p_{+} '' = H' = \delta$).

What I attempted to do was split the integral for $x < 0$ and $x > 0$ leaving me with $\delta(a + x)$ and $\delta(a - x)$, respectively:

$F(k) = \frac{2}{ik}[ \frac{1}{2} (\int_{-\infty}^{0} \delta(a + x) e^{-ikx} dx + \int_{0}^{\infty} \delta(x - a) e^{-ikx} dx )]$

Where $\delta (a - x) = \delta(-(x - a)) = \delta(x - a)$, since the delta function is an even function. By a property of the Fourier transform, the transform of $f(x - a)$ is $e^{-iak} F(k)$. So:

$F(k) = \frac{2}{k} [\frac{e^{iak} + e^{-iak}}{2i}]$, which ifs off from $\frac{2}{k} \sin(ak)$ by the $-$ sign. Is what I did by splitting up the integral not correct?

I also noticed that the Fourier transform $F(k)$ has the function $f(ak) = \sin(ak)$. Can we use the property that for a function $f(ax)$ its Fourier transform is given by $\frac{1}{|a|} F(\frac{k}{a})$? Thanks for the help, I appreciate it.

  • $\begingroup$ $H(a-|x|)$ ? seriously ? you are just trying to compute $\displaystyle\int_{-a}^a e^{-ikx} dx = \frac{e^{-ikx}}{-ik}|_{-a}^a = \frac{e^{-ika}-e^{ika}}{-ik} = \frac{2}{k}\sin(ka)$. and yes you can use $\mathcal{F}[f(ax)](k) = \frac{1}{|a|} \mathcal{F}[f(x)](k/a)$ ... once you proved it ! (a simple change of variable $y = ax$..) $\endgroup$ – reuns Apr 27 '16 at 23:16
  • $\begingroup$ I'm sorry, can you explain a little more clearly why we have the integral $\int_{-a}^{a} e^{-ikx} dx$? $\endgroup$ – Javier Apr 27 '16 at 23:22
  • $\begingroup$ because $H(x) = 1$ whenever $x > 0$, $0$ otherwise, hence $H(a - |x|) = 1$ whenever $|x| < a$, $0$ otherwise, hence $\displaystyle\int_{-\infty}^\infty H(a-|x|) e^{-ikx} dx = \int_{-a}^a e^{-ikx} dx$ $\endgroup$ – reuns Apr 27 '16 at 23:26
  • $\begingroup$ I see now. Thank you so much! I was gravely overthinking this. $\endgroup$ – Javier Apr 27 '16 at 23:28
  • $\begingroup$ but what you wrote isn't useless : see if you can find where the error is, if you can prove each step, and next time you will be good $\endgroup$ – reuns Apr 27 '16 at 23:29

Your approach is perfectly appropriate. There was, however, a flaw in the execution of that way forward.

Note that we have

$$\begin{align} F(k)&=\langle H(a-|x|),e^{-ikx}\rangle\\\\ &=\frac{1}{-ik}\langle H(a-|x|),\frac{de^{-ikx}}{dx}\rangle\\\\ &=\frac{1}{-ik}\langle \text{sgn}(x)H'(a-|x|),e^{-ikx})\\\\ &=-\frac{1}{ik}\langle \delta(a-|x|),\text{sgn}(x)\,e^{-ikx}\rangle\\\\ &=-\frac{1}{ik}\left(\text{sgn}(-a)e^{ika}+\text{sgn}(a)e^{-ika}\right)\\\\ &=\frac{1}{ik}\left(e^{ika}-e^{-ika}\right)\\\\ &=\frac{2\sin(ka)}{ka} \end{align}$$

as was to be shown!


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