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Messy Fourier Transform Integral

  1. Dec 4, 2009 #1
    I'm teaching myself some basic Fourier analysis from Boas's "Mathematical Methods in the Physical Sciences." I'm a little stuck on Example Problem 1 on p.382. This is a basic example of getting the g(a) Fourier transform of a certain function (pictured on the page) and then plugging it into the formula to get the FT of f(x). The problem has two typos in it according to the online errata, which I've taken into account.

    Anyway, you get the g(a) function which is g(a)=sin(a)/(pi*a) (I'm using "a" for the alpha that appears in the book). I understand this.

    Then you want to get your Fourier representation of f(x) so you plug g(a) into the integral from - to + infinity of g(a)*e^(iax) da. This I understand.

    Now for this integral she expands e^(iax) using Euler to get cos(ax)+i*sin(ax). If you multiply this by g(a), i.e. by sin(a)/(pi*a), you get two terms in your integral, i.e. the integral is of (sin(a)(cos ax))/(pi*a) + (sin(a)(i*sin(ax))/(pi*a) da.

    But she does not show these two terms of the multiplication. Rather, she says that the integral is equal to two times the integral of the first term, because the function "sin(a)/a is an even function." So the complex second term is dropped. This I don't understand. The second term is a function of a and x, times i. I can't really tell if its even or odd. If it's odd, I can see dropping it from the integral since the integral is symmetric around the origin so the integral of the term will be = 0. But the fact that its got an "i" in it is throwing me, too.

    Could somebody please help me with this second term in the integral? Thanks.
  2. jcsd
  3. Dec 4, 2009 #2
    A function of two variables has two parity properties. For example y*cos(x) is even wrt x but odd wrt y. So when you integrate over the function wrt to only one variable, it's the parity wrt to that variable which counts.
    For example sin(a*x) is odd wrt a (even wrt x) so integrating over a symmetric interval wrt a will result in zero.
  4. Dec 4, 2009 #3
    Okay I think I get it. The integral is of i*(sin(a)*sin(ax))/a. We're integrating wrt a. Now sin(a)/a is even. But sin(ax) is odd if we're integrating wrt a. So its an even function times an odd function which gives an odd, and so its integral is zero over the symmetric interval. And so the "i" nicely goes away.

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