We use the definitions $$F(k)=\mathcal{F}(f(x))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-ikx}f(x) \ dx,$$ where the inverse is defined as $$f(x)=\mathcal{F}^{-1}(F(k))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{ikx}F(k) \ dk.$$

Consider the Fourier transform pair commonly found in tables:

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Given the definition of the Fourier transform and its inverse as already defined, are these transformations symmetric? e.g. is $$\mathcal{F}\left(\frac{\sin(ax)}{x}\right)=\sqrt{\frac{\pi}{2}}H(a-k)H(a+k)$$ the same as $$\mathcal{F}^{-1}\left(\frac{\sin(ak)}{k}\right)=\sqrt{\frac{\pi}{2}}H(a-x)H(a+x)?$$


\begin{align} \mathcal{F}^{-1}_k(F(k)\cos(ckt))&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) \cos(ckt) e^{ikx} \ dk \\ &=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) (e^{ckti}+e^{-ckti}) e^{ikx} \ dk \\ &=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) e^{ik(x+ct)} \ dk+\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^{\infty} F(k) e^{ik(x-ct)} \ dk \\ &=\frac{1}{2}\left(f(x+ct)+f(x-ct)\right). \end{align} (recalling that $f(x)=\mathcal{F}^{-1}(F(k))$)


1 Answer 1


Almost: with your normalization, we have that $$\mathscr{F}\{f\}(x)=\mathscr{F}^{-1}\{f\}(-x)$$ i.e. $$\mathscr{F}^2\{f\}(x)=f(-x)$$ So if $$\mathscr{F}\left\{\frac{\sin(ax)}{x}\right\}(k)=\sqrt{\frac{\pi}{2}}H(a-k)H(a+k)$$ Then $$\mathscr{F}^{-1}\left\{\frac{\sin(ak)}{k}\right\}(x)=\sqrt{\frac{\pi}{2}}H(a+x)H(a-x)$$

  • $\begingroup$ Are you please able to correct my example? I'm still a bit confused. $\endgroup$
    – Steven
    Commented Apr 29, 2019 at 12:55
  • $\begingroup$ @Stuart-JamesBurney Your result was good, because the result of the FT is even. $\endgroup$
    – Botond
    Commented Apr 29, 2019 at 13:00
  • $\begingroup$ Yep, I see that now. I have something that's niggling me though. Consider $$\mathcal{F}\left(f(x)cos(ax)\right)=\frac{1}{2}\left(F(k-a)+F(k+a)\right).$$ By using the definition of the inverse Fourier transform, I have shown that $$\mathcal{F}^{-1}\left(F(k)cos(ckt)\right)=\frac{1}{2}\left(f(x+ct)+f(x-ct))\right.$$ Is this because this too is even? $\endgroup$
    – Steven
    Commented Apr 29, 2019 at 13:05
  • $\begingroup$ @Stuart-JamesBurney is $F(k)=\mathscr{F}\{f(x)\}(k)$? $\endgroup$
    – Botond
    Commented Apr 29, 2019 at 13:13
  • 1
    $\begingroup$ @Stuart-JamesBurney I checked everything multiple times. If $\mathscr{F}\{f\}(k)=\int \mathrm{d}x \, e^{-ikx}f(x)$, then $\mathscr{F}\{f\}(-k)=\int \mathrm{d}x \, e^{ikx}f(x)=\mathscr{F}^{-1}\{f\}(k)$. I checked $\mathscr{F}\{f(x)\cos(ax)\}(k)$ on wikipedia, and it's correct as well. And $\mathscr{F}\{f(x-a)\}(k)=F(k)e^{-ika}$ (from wikipedia), so your inverses are correct as well. $\endgroup$
    – Botond
    Commented Apr 29, 2019 at 14:27

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