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An improper integral (Related to the Fourier transform)

  1. Jul 21, 2011 #1
    How to show this?
    [tex]\int_{-\infty}^{+\infty}e^{-i2\pi xs}ds=\delta(x)[/tex]
    This is a part of a problem of "Bracewell, R. The Fourier Transform and Its Applications, 3rd ed. New York: McGraw-Hill, pp. 100-101, 1999". This isn't a homework, I found it http://mathworld.wolfram.com/FourierTransform1.html" [Broken]. But I'm not sure even if this is true.

    Thanks in advance.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 21, 2011 #2


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    Try showing that

    [tex]\int_{-\infty}^{\infty} dx~\delta(x)f(x) = \int_{-\infty}^\infty dx \int_{-\infty}^\infty ds~e^-{2\pi i s x}f(x) = f(0).[/tex]
  4. Jul 22, 2011 #3
    \int_{-B}^{B}{e^{-i \, 2 \pi \, x \, s} \, ds} =\left.\frac{e^{-i \, 2 \pi \, x \, s}}{-i \, 2 \pi \, x}\right|_{B}^{B} = \frac{\sin{(\pi \, B \, x)}}{\pi \, x}

    Next, consider the function:
    \mathrm{sinc}(x) \equiv \frac{\sin{(\pi \, x)}}{\pi \, x}
    and evaluate the definite integral (by contour integration)
    \int_{-\infty}^{\infty}{\mathrm{sinc}(x) \, dx} = ?
    Then, notice that the integral I evaluated is:
    B \, \mathrm{sinc}(B \, x)
    What happens if you take the limit [itex]B \rightarrow \infty[/itex]?
  5. Jul 22, 2011 #4
    I can show that if f is continuous at 0, then:
    And if f has the Fourier transform, then:
    [tex]\int_{-\infty}^{+\infty}\left [\int_{-\infty}^{+\infty}f(x)e^{-i2\pi xs}dx \right ]ds=f(0)[/tex]
    Does it help?

    Last edited: Jul 22, 2011
  6. Jul 22, 2011 #5
    I can see that if [itex]B \rightarrow \infty[/itex], then:
    \int_{-\infty}^{\infty}{B\mathrm{sinc}(Bx) \, dx} = 1[/tex]
    But [itex]B\mathrm{sinc}(Bx)[/itex] is not equal to 0 for every x≠0, while [itex]\delta (x)[/itex] is.

    Last edited: Jul 22, 2011
  7. Jul 22, 2011 #6
    Last edited by a moderator: Apr 26, 2017
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