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Fourier Transform.

  1. Feb 29, 2008 #1


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    I need to find the fourie transform of f(x)=N*exp(-ax^2/2).
    (N and a are constants).
    well ofcourse iv'e put into the next integral:
    Iv'e changed variables, that i will get instead of exp(-ax^2/2)exp(-ikx),
    exp(-z^2)exp(-ikz*sqrt(2/a)), but that didn't helped very much, perhaps I need to use here the definition of dirac delta function to simplify this.

    any hints?
    there's also a question to simplify the next terms:
    1.(x^2+3)d(x+5) where d is dirac's delta function.

    as for 2, what i did is: d(2x-8)=d(2(x-4))=d(x-4)/2
    for 3 i think it equals:
    but i didnt use the integral definition, iv'e just used the fact that at x=1 or x=-2 the argument of the functional is zero and thus by definition it diverges there otherwise it equals zero, is this a sound reasoning for this question?
    for 1, not sure, but if we use the integral definition then [tex]\int_{-\infty}^{\infty}f(x)(x^2+3)d(x+5)dx=f(-5)*28=28*\int_{-\infty}^{\infty}f(x)d(x+5)dx[/tex] then (x^2+3)d(x+5)=28*d(x+5), or something like this?

    thanks in advance.
  2. jcsd
  3. Feb 29, 2008 #2
    For the fourier transform I got a solution, however I am not certain whether it is the correct one. Please, use this with care. I rewrote the argument of the exponential function as:

    [tex]\frac{a}{2}x^2+ikx=\frac{a}{2}\left[x^2+\frac{2}{a}ikx\right]= \frac{a}{2}\left[x^2+2\frac{ik}{a}x+\left(\frac{ik}{a}\right)^2\right]+ \frac{k^2}{2a}= \frac{a}{2}\left(x+\frac{ik}{a}\right)^2+\frac{k^2}{2a}[/tex]

    Using then:


    You arrive at:


    Please check this result because the last substitution is a bit awkward and can be regarded as "playing" with real and imaginary numbers. Therefore it might very well be that this is complete and utterly wrong.
  4. Mar 1, 2008 #3


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    not sure myself if your answer is the correct one, i.e obviously when you turned to complex dumby variable you need to use the residue theorem, but my problem is that i don't think that there's a pole there, so the sum of the residues will be zero, i.e you integrate over a straight line of y=k/a in the complex plane, but there are no poles.
  5. Mar 1, 2008 #4
    Why the residue theorem? This is for evaluating the value of a contour integral in the complex plane circling around a few poles (if any). The theorem gives a practical way of determining the value of this integral. I don't think that this has something to do with the integral you asked for. I could be mistaken, then I'm missing something.

    However I had a second look at the result I presented and it is indeed the correct one. I found this in a table in the book "Mathematical Handbook for Scientists and engineers", by G.A. Korn and T.M. Korn. So, now we can safely go to sleep :smile:

    The first one of the dirac questions is (if I remember correctly, it has been a while since I used this) the value of the function that is multiplied with the dirac at the point where the argument of the dirac is zero. This means that it is nothing more than the value of the function in the point x=-5, giving 28. Again, check this please :wink:
  6. Mar 1, 2008 #5


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    so it should be:
    (x^2+3)d(x+5)=28, are you sure?
    then why is my way wrong there, i got: (x^2+3)d(x+5)=28d(x+5)
    ok i can see what you mean if i integrate this function (without entering f(x), i will get 28), but still i don't see how this means what your'e asserting, i mean if (x^2+3)d(x+5)=28, then if i enter 28 in the the integral, i will get an integral which doesn't converge.
  7. Mar 1, 2008 #6
    You are correct. The result is indeed zero everywhere except at the point x=-5 where it is 28. That is mathematically 28d(x+5). That is what I meant, but confusion can arise if one considers the function 28 to be the constant function for all x. Shame on me :redface:
  8. Mar 1, 2008 #7
    For the third dirac question, I made a sketch and I think it is correct as well.
    I have to admit that I can't remember ever to have used this kind of result. The strange thing is that the argument of the function consists of two points where the argument is zero, so it is a kind of "double dirac function". Very nice.

    For the second one I assume by considering the definition of the dirac and applying a transformation gets you the result.
  9. Mar 1, 2008 #8


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    another computation of integrals, after we found g(k) above with the transform, I need to compute the next integrals:
    and [tex]t=\int_{-\infty}^{\infty}g(k)k^2dk[/tex]
    don't see how to do this really, by parts won't work here.
  10. Mar 1, 2008 #9
    Assuming you mean:

    [tex]t=\int_{-\infty}^{\infty}g(k)k^2dk= \int_{-\infty}^{\infty} N\sqrt{\frac{2\pi}{a}}e^{-\frac{k^2}{2a}} k^2dk[/tex]

    It is solvable by partial integration. Consider the following:

    [tex]t=N\sqrt{\frac{2\pi}{a}} \int_{-\infty}^{\infty} e^{-\frac{k^2}{2a}} k^2dk= N\sqrt{\frac{2\pi}{a}} \cdot I[/tex]


    [tex]I=\int_{-\infty}^{\infty} e^{-\frac{k^2}{2a}} k^2dk= (-a)\int_{-\infty}^{\infty} k\cdot e^{-\frac{k^2}{2a}}d\left(-\frac{k^2}{2a}\right)[/tex]

    Which gives after partial integration:

    [tex]I=(-a)\left[k \cdot e^{-\frac{k^2}{2a}}\right]_{-\infty}^{\infty} +a\cdot \int_{-\infty}^{\infty} e^{-\frac{k^2}{2a}}dk= (-a)\cdot (0-0)+a\cdot \int_{-\infty}^{\infty} e^{-\frac{k^2}{2a}}dk[/tex]

    Setting now:



    [tex]I=a\cdot \int_{-\infty}^{\infty} \sqrt{2a} \cdot e^{-z^2}dz[/tex]

    Which is:


    The result we are looking for is now:

    [tex]t=N2\pi a[/tex]

    I do not understand the first one. Can you explain this one a little bit more? Can you write the complete integral?
  11. Mar 1, 2008 #10


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    Sorry for the flood of questions.
    I need to calculate the integral:
    by using parseval's equation.
    now parseval's equation is:
    [tex]\int_{-\infty}^{\infty}|f(x)|^2dx=\frac{1}{2\pi}\int_{-\infty}^{\infty}|g(y)|^2dy[/tex] where g(y) is F.T of f(x), obviously here f(x)=(sin(x)/x)^2, so first i need to integrate:
    i tried using residues, but the residue equals zero, and the contour on the big circle goes to zero and so is the small circle around zero (because it equals -pi*Res(z=0), so I don't know what to do here.
    mathematica has a nice term for the integral i need to calculate.
  12. Mar 1, 2008 #11


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    f(x) is the first function i was given in my first post, i.e f(x)=Nexp(-ax^2/2).
    anyhow thanks i understand this part.
  13. Mar 1, 2008 #12
    OK, so the first integral is thus:


    Using the same method as the other one. I will have to look a few things up for the the other questions. "I'll be back" :smile:
  14. Mar 1, 2008 #13


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    I think it's best I will borrow this book by the korns', it could be handy in the open material exam.
  15. Mar 1, 2008 #14
    The FT of (sin(x)/x)^2 seems not so easy. I will do my best, however I don't use contour integration that often, so I will need some time. It seems that it is the appropriate way in this case.

    The book by Korn and Korn is a thick one giving lot's of mathematical material and formulas, but I would use something else for an open exam. For me the one which gave me the best preparation was the one of the schaum series: "Laplace Transforms" by M. Spiegel. Some people find these books rubbish, I find them an excellent source for problems. Not all of them are good, but the ones written by the late prof. Spiegel are very good. It just takes a bit of time to solve the problems.

    [begin edit]
    Another book which can be useful is the one "Complex Variables" also from the schaum series and written by Spiegel.
    [end edit]

    I will further try to find the FT of the function. Just for information, do you have to use Parseval's identity? If I can come up with a different method is that allowed?
    Last edited: Mar 1, 2008
  16. Mar 1, 2008 #15


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    well yes, i need to use parseval's equation.

    thanks for the recommendations, I also think that for this theoretical methods for physicists course schaum's will do, I also thought of browsing mary boas' book.
  17. Mar 1, 2008 #16
    The integral:


    has the solution [tex]\frac{2\pi}{3}[/tex] as you might have obtained from mathematica. I got it from maxima. Anyway it looks as if this is solvable by contour integration, but if you have to use parseval it makes things more complicated. I keep searching for an answer.
  18. Mar 2, 2008 #17
    loop quantum gravity, It seems very difficult or I can't see the solution in a practical way. The problem is that I can't find the FT of the sin^2(x)/x^2 function. I tried several different contours but I end up nowhere. Then I tried to look at it as:

    [tex]\int_{-\infty}^{\infty}\frac{sin^2(x)}{x^2}e^{-ixy}dx= 2 \cdot \int_{0}^{\infty}\frac{sin^2(x)}{x^2}e^{-sx}dx[/tex]

    with the iy replaced by s (another complex variable) and twice the integral of intrest because of the even character of it. It looks now as a Laplace transform of the function. So I found this Laplace tranform in a tables book ("table of laplace transforms, by Roberts and Kaufman) to be the following:

    [tex]L\left\{\frac{sin^2(at)}{t^2}\right\}=a\cdot atan\left(\frac{2a}{s}\right) -\frac{s}{4}log\left(1+\frac{4a^2}{s^2}\right)[/tex]

    Which can be found by considering the property of the Laplace transform that says that if the function is divided by t, you have to integrate the laplace transform. In this case twice. However, if I write this out, replace s by iy, replace the atan of the complex argument, take the absolute value and then the square, I end up with a difficult integral. Maxima gives me a result in the form of a limit.

    So, I don't know what else I could use. Perhaps someone else here can help you. I assume that it must be possible using contour integration, but I can't find the contour.

    My apologies that I can't be of any more assistance.

    If you are looking for a good book with formulas, I can recommend "Mathematical Handbook of Formulas and Tables", written by Spiegel, also one out of the schaum series. It does not only cover basic stuff, but goes into advanced functions as well. I use this one very often.
    Last edited: Mar 2, 2008
  19. Mar 2, 2008 #18


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    thanks coomsat, I appreciate your help nontheless.
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