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.

Integral transforms

Filed under: mathematics — Tags: — sandokan65 @ 20:29

General linear integral transformation:

    F(p) = {\bf T}[f](p) :\equiv \int_a^b K(p,t) f(t) dt

Here we assume that this integral exist, K is a fixed complex function, p \in {\Bbb C} and a,b\in{\Bbb R}.

Specific examples:

  • Laplace transformation: {\bf L} defined by a=0, b=+\infty, K(p,t)=e^{pt}.
  • Fourier transformation: {\bf F} defined by a=-\infty, b=+\infty, K(p,t)=\frac1{\sqrt{2\pi}}e^{ipt}.
  • Mellin transformation: {\bf M} defined by a=0, b=+\infty, K(p,t)=t^{p-1}.
  • Hankel transformation: {\bf H} defined by a=0, b=+\infty, K(p,t)=t \ J_n(pt) (a Bessel function of order n).

Trinomial equation related to Mellin transform

Filed under: mathematics — Tags: , , , — sandokan65 @ 20:08

The algebraic equation y^n + x y^p -1 = 0 for (n>p) has following series solutions:

    y_0(x) = \frac1{n} \sum_{r=0}^\infty \frac{(-)^r}{r!}  \frac{\Gamma(\frac{1+pr}{n}) x^r}{\Gamma(\frac{1+pr}{n}+1-r)}.

Other n-1 solutions are given by the complex rotations as y_k(x) = \omega^k y_0(\omega^{pk} x) where \omega:\equiv e^{i\frac{2\pi}{n}}, k=\overline{0,n-1}.

The solutions satisfy following ODE:

    (-d_x)^n y(x) = p (-x d_x -n) y(x).

Source: “The Functions of Mathematical Physics” by Harry Hochstadt

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