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Homework Help: Integral computation over Fourier/Convolution

  1. Jan 27, 2013 #1

    I just tried to solve the following integral:

    \int_{-\infty}^{\infty} \frac{2}{1+(2\pi t)^2} \mathrm{sinc}(2t) dt

    My approach is: Convert both to Fourier domain and the multiplication becomes a convolution. Because of Parsevals theorem, I can either integrate in time or frequency domain. Because of linearity, I can then put the outer integral inside:

    \int_{f=-\infty}^{\infty} \int_{\nu=-\infty}^{\infty} e^{-|\nu|} \frac{1}{2} \Pi\left(\frac{f-\nu}{2}\right) d\nu df \\
    = \int_{\nu=-\infty}^{\infty} e^{-|\nu|} \frac{1}{2} \underbrace{\int_{f=-\infty}^{\infty}\Pi\left(\frac{f-\nu}{2}\right) df}_{2} d\nu

    It can be clearly seen, that only the double-sided exponential stays whose integral is 2.
    So my total solution is 2.

    However, Wolfram alpha gives me (e-1)/e:


    Can anyone explain where I did wrong?


    PS: The sinc is defined as normalized, i.e. [itex]\mathrm{sinc}(t)=\sin(\pi t)/\pi t[/itex] ...
  2. jcsd
  3. Jan 28, 2013 #2


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    This is your fundamental problem. Parseval's theorem would apply if you were integrating the magnitude squared of your function, but you are not doing that.

    In general, if your original function is [itex]x(t)[/itex], then
    $$\int_{-\infty}^{\infty} x(t) dt = \left. \int_{-\infty}^{\infty} x(t) e^{-ift} dt \right|_{f = 0} = \hat{x}(0)$$
    where I have assumed the following definition for the Fourier transform:
    $$\hat{x}(f) = \int_{-\infty}^{\infty} x(t) e^{-ift} dt$$
    Thus, integration in the time domain does not correspond to integration in the frequency domain, but rather to evaluation of the Fourier transform at [itex]f = 0[/itex].
  4. Jan 28, 2013 #3
    Oh, of course, thanks!!

    Do you have a pointer on how to calculate this integral with Fourier transform theorems?

  5. Jan 28, 2013 #4


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    As jbunniii implied, you want to evaluate
    $$ F\left[\frac{2}{1+(2\pi t)^2} \text{sinc }2t\right]_{f = 0}$$ where F[ ] denotes the Fourier transform. What's the Fourier transform of that product?
  6. Jan 28, 2013 #5
    Oh, right, didn't realize that!
    Thanks a lot, got it!
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