1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fourier transform truoble

  1. Apr 26, 2012 #1
    I'm trying to find [itex]\frac{1}{2\pi}\int \limits_{-\infty}^{\infty}e^{-itx}\frac{1}{a^2+x^2}\mathrm{d}x[/itex] where 'a' is a constant.

    First I noticed that there is [itex]\frac {\partial \arctan x}{\partial x}[/itex] in this and using a substitute got [itex]\int \limits_0^{\pi / 2}\cos( t \tan x )\mathrm{d}x[/itex] with some constants in the gaps.
    I then remember that I'm working in complex numbers, factored [itex]a^2+x^2[/itex] and got something essentially along the lines of [itex]\int \frac{e^x}{x}\mathrm{d}x[/itex], or maybe rather [itex]\int \limits_0^{\infty} \frac {\cos tx} {a - ix}\mathrm{d}x[/itex].

    I can't integrate either.
     
    Last edited: Apr 26, 2012
  2. jcsd
  3. Apr 26, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Your integral does not contain "dx" or "dt". Without that we cannot tell what integration you intend. Is the problem
    [tex]\int \frac{e^{-itx}}{a^2+ x^2} dx[/tex]
    or is it
    [tex]\int \frac{e^{-itx}}{a^2+ x^2}dt[/tex]
    ?
     
  4. Apr 26, 2012 #3
    Oh, sorry. It's dx. I'll fix it right away.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fourier transform truoble
  1. Fourier transform (Replies: 4)

  2. Fourier transform (Replies: 1)

  3. Fourier transform (Replies: 1)

  4. Fourier transformation (Replies: 1)

Loading...