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Integral/fourier transform

  1. Oct 18, 2013 #1
    I made a few excercises with fourier series, fourier integrals and fourier transforms.

    But i am getting stuck at a few questions,

    most of the time a fourier transform needs to be calculated in part a,
    and than part b ask to solve an intergal with the help of your aswer by part a.

    i made a picture of one of my excersices

    I cannot find this sort of problems in my textbook or internet,
    does someone have a source?



    Attached Files:

  2. jcsd
  3. Oct 18, 2013 #2
    [itex]\sin(ax)\cos(ax) = \frac{1}{2}\sin(2ax)[/itex]

    Let [itex]I[/itex] be the integral we want to solve.

    [itex]I =\int_{0}^{\infty} \frac{\sin(ax)\cos(ax)}{x}dx = \int_{0}^{\infty} \frac{\sin(2ax)}{2x}dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin(2ax)}{2x}dx [/itex]

    [itex]g(x) = \frac{\sin(2ax)}{2x} = f(2x)[/itex]

    [itex]\tilde{g}(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\sin(2ax)}{2x}\cos(kx)dx = \frac{1}{2}\tilde{f}(\frac{k}{2}) [/itex]

    Now let [itex]k = 0[/itex] so the cosine becomes 1.

    [itex]\tilde{g}(0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\frac{\sin(2ax)}{2x}dx = \frac{1}{\sqrt{2\pi}}2I = \frac{1}{2}\tilde{f}(0) = \frac{1}{2}\sqrt{\frac{\pi}{2}} [/itex]

    [itex]\frac{1}{\sqrt{2\pi}}2I = \frac{1}{2}\sqrt{\frac{\pi}{2}} [/itex]

    [itex]I = \frac{1}{4}\sqrt{\frac{\pi}{2}}\sqrt{2\pi} = \frac{\pi}{4}[/itex]

    It seems what you did was evaulate the Fourier transform at [itex]\omega = a[/itex], but your description only gives the Fourier transform for [itex]\left|\omega\right| <a[/itex] and [itex]\left|\omega\right| > a[/itex]. It happens that at [itex]\omega = a[/itex], the value is 1/2 of what it is when [itex]\left|\omega\right| <a[/itex]. That is [itex]\tilde{f}(a) = \frac{1}{2}\tilde{f}(0) = \frac{1}{2}\sqrt{\frac{\pi}{2}}. [/itex] What I wrote above indicates this is correct. This integral can also be performed (tediously) using contour techniques.
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