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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?

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  2. jcsd
  3. Dec 3, 2012 #2


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    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##.
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