Re: Fourier Transform challenge

John wrote:

MATLAB:

fourier(exp(-x-exp(-x)))

ans =

gamma(1+i*w)


This "solution" is actually meaningless because

    gamma(1+iw) = iw gamma(iw)

and the gamma function is analytic only in the right half plane (real part of z positive).

That said, the original poster's function exp(-x-exp(-x)) is itself integrable- its integal is

    exp(-exp(-x))

as can be verified by differentiation.

Given this result, the Fourier Transform can be integrated "by parts". Start by integrating only over the interval [-T,T]. Then, integrating by parts and taking the limit as T becomes large we get

                   exp(-iwT-exp(-T)) - exp(iwT-exp(T))
    FT{} =  lim   -------------------------------------
            T->oo           1 + iw


Clearly the second term in the numerator converges to zero and so we are left with

                   exp(-iwT-exp(-T))
    FT{} =  lim   -------------------
            T->oo       1 + iw



                   exp(-iwT)
         =  lim   -----------
            T->oo    1 + iw


This limit does not exist in the conventional sense but does exist as a "distribution" or "generalized function" (the Dirac delta function is the usual example- sorry to bring this subject up but I must). In this sense, the limit exists and is zero by the Riemann-Lebesgue Lemma (see Papoulis' book "The Fourier Integral and its Applications").

I suspect the limit being zero explains the excellent roll-off characteristics seen by the original poster.

Mike
.

Relevant Pages

  • Re: Fourier Transform challenge
    ... and the gamma function is analytic only in the right half plane. ... I had also forgotten that the Gamma function is analytic everywhere except at the negative integers. ... Start by integrating only over the interval. ... the limit exists and is zero by the Riemann-Lebesgue Lemma (see ...
    (comp.dsp)
  • Re: Superposition
    ... what does the math reveal? ... I am not trying to define power as negative with respect to zero ... THE INTEGRAL OF A COSINE IS A SINE. ... integrating sine function to cosine...so it has been 40 years since my last ...
    (rec.radio.amateur.antenna)
  • Re: Use of EPSILON () Intrinsic Function
    ... zero for a given parameters combination. ... This upper limit must be specified to perform the integration. ... I'm integrating an oscillating uneven function about the X-axis ... There're 10,000 of such function in the numerical/computer model, ...
    (comp.lang.fortran)
  • Re: Possible proof of Gabriels Theorem?
    ... >> that delta must eventually become zero, however, according to the ... What is the difference between integrating ... >I have spoken to so many students who have passed real analysis courses ...
    (sci.math)
  • Re: Possible proof of Gabriels Theorem?
    ... This means that delta must eventually be ... > that delta must eventually become zero, however, according to the ... What is the difference between integrating ... I have spoken to so many students who have passed real analysis courses ...
    (sci.math)