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Help whith fourier transform

  1. Nov 27, 2008 #1
    1. The problem statement, all variables and given/known data
    Calculate fourier transform of cos(x^2)

    2. Relevant equations

    3. The attempt at a solution

    I want, if it possible, a clue to solve the integral. I dont know how to proceed. I tried integration by parts, but i cant solve it.
    Sorry for my english. How can i use latex?

    Can moderators delete my other topic ? it was a mistake.
    Last edited: Nov 27, 2008
  2. jcsd
  3. Nov 27, 2008 #2
    Think of the cos term and the fact that


    Then using that you should be able to solve the integral.

  4. Nov 27, 2008 #3


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    Homework Helper

    Welcome to PF!

    Hi xoureo! Welcome to PF! :smile:

    For LaTeX, type [noparse][tex] before and [/tex] after …[/noparse]

    for help with the symbols, bookmark http://www.physics.udel.edu/~dubois/lshort2e/node61.html#SECTION008100000000000000000

    and don't worry about the other topic: lots of members do that! :biggrin:
  5. Dec 2, 2008 #4
    I try to do that, but when i get

    [tex]\int_{-\infty}^{\infty} cos(x^2)cos(kx)dx[/tex]

    i get blocked. How should i solve it?, i tried integration by parts ,whithout result. Should i try complex integration? i dont know how to proceed here.

    Thanks for the answers and the welcome :smile:

    PD: sorry for take so much in answering
  6. Dec 2, 2008 #5


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    Hi xoureo!

    No, you haven't tried jeffreydk's :smile: method …

    inegrate ∫e-i(kx + x²)dx separately. :wink:
  7. Dec 2, 2008 #6
    Hi tiny-tim

    Im sorry, i dont understand you. If i use the formula that jeffreydk give, i obtain:

    [tex]\frac{1}{2}\int_{-\infty}^{\infty}e^{i(x^2-kx)}dx +\frac{1}{2}\int_{-\infty}^{\infty}e^{-i(x^2-kx)}dx[/tex]

    The first term is:

    [tex]\frac{1}{2} \left( \int_{-\infty}^{\infty}cos(x^2)e^{-ikx}dx +i\int_{-\infty}^{\infty}sen(x^2)e^{-ikx}dx \right)[/tex]

    and the second is:

    [tex]\frac{1}{2} \left( \int_{-\infty}^{\infty}cos(x^2)e^{-ikx}dx -i\int_{-\infty}^{\infty}sen(x^2)e^{-ikx}dx \right)[/tex]


    [tex] \int_{-\infty}^{\infty}cos(x^2)e^{-ikx}dx= \int_{-\infty}^{\infty}cos(x^2)cos(kx)dx[/tex]

    Which i dont know how to integrate. If i try to integrate

    i get

    [tex]\int_{-\infty}^{\infty} cos(x^2)cos(kx)dx -i\int_{-\infty}^{\infty}sen(x^2)cos(kx)dx[/tex]
    and again i dont know how to integrate this expresions.:confused:
  8. Dec 4, 2008 #7
    Don't go back to sin and cos it's much easier to deal with the exponential terms. Sorry I didn't mention it but it looks like they are Gaussian, you want to get it in the form [itex]\int_{-\infty}^{\infty}e^{-(x-a)^2}[/itex] where this is a Gaussian integral centered at [itex]a[/itex].

    So you're dealing with right now


    Take the first one; if we complete the square in the exponential term,

    [tex]-\alpha x^2+\beta x=-\alpha(x^2-\frac{\beta}{2}x)=-\alpha(x-\frac{\beta}{2\alpha})^2+\frac{\beta^2}{4\alpha}[/tex]

    Now it looks like the Gaussian integral times a constant of [itex]\exp{[\frac{\beta^2}{4\alpha}]}[/itex]. Let's do the first integral now with [itex]\alpha=-i[/itex] and [itex]\beta=-ik[/itex].


    Now try the second one--it should be very similar to the first, then of course include the constants out in front when you're done.
  9. Dec 4, 2008 #8
    @jeffrey: You should be a bit careful with changing variables in the complex plane( you essentially suggest making a variable substitution of something like [tex]x\rightarrow \sqrt{i}(x-k)[/tex] . For doing this you actually use Complex Analysis (especially Cauchy's theorem),you change the path you are integrating, so you would need to construct a closed path in such a way that it contains both integrals and such that the integral vanishes along the other parts of the path. This is actually far from easy and it is more by chance that this works in this case.

    In order to calculate the integrals over exp(ix^2) over the whole real line, to calculate this you use at first symmetry so that you only have to integrate from zero to infinity (and get a factor 2), then you use cauchy's integral theorem to calculate the integral, as a path you choose something which looks like a piece of pie with an angle of pi/4, then you need to argue that the integral over the arc vanishes. Now the integral over the line inclined by pi/4 to the real line is the same as the integral over the real line, then you plug the inclined line in the integral to get the result (here you actually get a gaussian integral).
    Last edited: Dec 4, 2008
  10. Dec 4, 2008 #9
    Oh ok I see what you're getting at--I haven't taken my complex analysis course yet, but that argument is very interesting.
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