Eikonal Blog


Fourier transform

Filed under: mathematics — Tags: , — sandokan65 @ 20:31


  • Direct: F(p) = {\bf F}[f](p) :\equiv \int_{-\infty}^{+\infty} \frac{dt}{(2\pi)^{\mu}} e^{ipt} f(t),
  • Inverse: {\bf F}^{-1}[g](t) :\equiv \int_{-\infty}^{+\infty} \frac{dp}{(2\pi)^{1-\mu}} e^{-ipt} g(p).

Here some author use \mu=1, and some authors \mu=\frac12 (e.g. Mitrinović)


  • {\bf F}[f(t+a)](p) = e^{iap} F(p)
  • {\bf F}[e^{iat}f(t)](p) = F(p+a)
  • if \lim_{t\rightarrow \pm\infty} |f(t)| = 0, then {\bf F}[f'(t)](p) = -i p F(p)
  • {\bf F}[\int_{{\Bbb R}}f(t-u)g(u)\frac{du}{(2\pi)^{\mu}}](p) = F(p)  G(p)

Table of Fourier transformations

f(t) F(p)={\bf F}[f](p)
\frac{\sin(at)}{t} \frac{1}{2}\sqrt{\frac{\pi}{2}}(\frac{p+a}{|p+a|} - \frac{p-a}{|p-a|})
exp(-a t^2) \frac{1}{\sqrt{2a}} exp(-\frac{p^2}{4\pi})
\cos(at^2) \frac{1}{\sqrt{2a}} \cos(\frac{p^2}{4a}-\frac{\pi}{4})
\sin(at^2) \frac{1}{\sqrt{2a}} \sin(\frac{p^2}{4a}+\frac{\pi}{4})
|t|^{-a} (0<a<1) $ \sqrt{\frac{2}{\pi}} |p|^{1-a} \Gamma(1-a) \sin(\frac{\pi a}{2})
(1+t^2)^{-1} \sqrt{\frac{\pi}{2}} e^{-|p|}


  • 1) D.S. Mitrinović, J.D. Kečkić: “Jednačine Matematičke Fizike”; 2nd edition; Građevinska Knjiga, Beograd 1978.

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