1. Not finding help here? Sign up for a free 30min 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

  1. Dec 3, 2012 #1
    In calculating some basic fourier transform I seem stumble on the proble that I don't know how to take the limit in infinity of an exponentialfunction with imaginary exponent. In the attached example it just seems to give zero but I don't know what asserts this property. I would have thought that it would yield something infinite since a cosine or sine does not go to zero at infinity. What is done to arrive at the attached result?
     

    Attached Files:

  2. jcsd
  3. Dec 3, 2012 #2

    fluidistic

    User Avatar
    Gold Member

    Think of ##e^{(ik-a)x}## as equal to ##e^{ikx} \cdot e^{-ax}##. On the right hand side while ##e^{ikx}## is not really defined for when x tends to infinity, it is still bounded (because sine and cosine are bounded) and the term ##e^{-ax}## does tend to 0. So that the product tends to 0 and you get ##\lim _{x \to \infty}e^{(ik-a)x} =0##.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Fourier transform
  1. Fourier transform (Replies: 5)

  2. Fourier transform (Replies: 4)

  3. Fourier transform (Replies: 1)

  4. Fourier transformation (Replies: 1)

Loading...