Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Featured Insights An Integral Result from Parseval's Theorem - Comments

  1. Jul 12, 2017 #1

    Charles Link

    User Avatar
    Homework Helper

  2. jcsd
  3. Jul 12, 2017 #2

    Charles Link

    User Avatar
    Homework Helper

    Additional comments: The result for this integral is similar to the type of result that would be obtained from residue theory if the integrand gets evaluated at the "pole" at ## x=x_o ##. Alternatively, with the transformations ## y=x-x_o ## and ## y=\pi u ##, in the limit ## T \rightarrow +\infty ##, ## \frac{\sin(\pi u T)}{\pi u }=\delta(u) ##, but this only gives a result for large ## T ## for an evaluation of the integral. So far, the Parseval method is the only way I have of solving it.
     
  4. Jul 14, 2017 #3
    Thanks Charles! Some day I will understand this :)
     
  5. Jul 15, 2017 #4

    Charles Link

    User Avatar
    Homework Helper

    Hi @Greg Bernhardt
    I will be very pleased if someone comes up with an alternative method to evaluate this integral. I am pretty sure I computed it correctly, but I would really enjoy seeing an alternative solution. :) :)
     
  6. Jul 16, 2017 #5

    Charles Link

    User Avatar
    Homework Helper

    I believe I have now also succeeded at evaluating this integral by use of the residue theorem: ## \\ ## The integral can be rewritten as ## I=-\frac{1}{2} \int Re[ \frac{e^{i2z}[1-e^{-i2(z-x_o)}]}{(z-x_o)(z+x_o)}] \, dz ##. ## \\ ## (Using ## sin(x-x_o)sin(x+x_o)=-\frac{1}{2}[cos(2x)-cos(2x_o)] ## and then writing ## cos(2x) ## as the real part of ## e^{i2x} ##). ## \\ ## A Taylor expansion shows the only pole is at ## z=-x_o ## (from the ## (z+x_o) ## term in the denominator. ## \\ ## (Note that ## e^{-i2(z-x_o)}=1-2i(z-x_o)+higher \, order \, terms \, in \, (z-x_o) ##. Thereby, there is no pole at ## z=x_o ##). ## \\ ## The contour will go along the x-axis with an infinitesimal semi-circle loop over the pole at ## z=-x_o ##, and will be closed in the upper half complex plane. (I needed to consult my complex variables book for this next part:). The infinitesimal semi-circle loop over the pole in the clockwise direction results in ## -B_o \pi i ## where ## B_o ## is the residue of the function at ## z=-x_o ##. ## \\ ## Upon evaluating the residue ## B_o ##, this results in ## -\pi \frac{sin(2x_o)}{2x_o} ## for the portion containing the infinitesimal semi-circle loop over the pole (which is not part of the integration of the function along the x-axis, where the entire function is considered to be well-behaved.). Since the contour doesn't enclose any poles, the complete integral around it must be zero, so the functional part along the x-axis must be equal to ## \pi \frac{sin(2x_o)}{2x_o} ##, which is the result that we anticipated.
     
    Last edited: Jul 17, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted