# Eikonal Blog

## 2010.01.04

### Fourier transform

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

Definition:

• 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)

Properties:

• ${\bf F}[f(t+a)](p) = e^{iap} F(p)$
• ${\bf F}[e^{iat}f(t)](p) = F(p+a)$
• 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) \$ $\sqrt{\frac{2}{\pi}} |p|^{1-a} \Gamma(1-a) \sin(\frac{\pi a}{2})$
$(1+t^2)^{-1}$ $\sqrt{\frac{\pi}{2}} e^{-|p|}$

## Sources

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