# Re: Fourier Transform; thanks, and...

[ Follow Ups ] [ Post Followup ] [ Calculus Message Board ] [ FAQ ]

Posted by Tim and Alex on September 19, 2002 at 20:53:11:

In Reply to: Re: Fourier Transform posted by Brad Paul on September 18, 2002 at 17:25:53:

: : Given g(k) = N/(k^2 + a^2) where N, a are parameters

: : and
: : inf
: : /\
: : |
: : f(x) = (1/((2*pi)^0.5))*| g(k)exp(ikx) dk
: : |
: : \/
: : -inf

: : i.e. f(x) <= F.T. => g(k)

: : what is f(x)?

: : We get as far as

: : I(k) = (N/((2*pi)^0.5)) . ((exp(ikx).arctan(x/a))/a)

: : f(x) = I(inf) - I(-inf);

: : and we cannot understand the significance.
: : Any assistance would be greatly appreciated.

: Tim and Alex,

: I'm sorry I can't be of more help. But all I have for you is the result.

: The Fourier Transform of your g(k) function is:

: G(x)=F-1{g(k)}=N (pi/2)1/2 e-a |x| /a

: If you find a slick way of getting this result do post it.

Thanks a lot for your time. The quickest way to the solution is complex analysis (getting the residuals of poles at +/- a on the imaginary axis as roots from splitting (k^2)+(a^2) into (k+ia)(k-ia)). It had been a while since we did the whole integration in the complex plane thingy but it turns out that integrating around the +a gives you your solution without the absolute, and around the -a gives the same solution (thus the absolute on the x in the exponent). It's a pretty ubiquitous feature of maths that the right way to attacking a problem will give you the answer with absolutely no fuss. Another way suggested was integration by parts to get an infinte series of integrals, but this would be a real pain. Has anyone heard of any software that does these kind of Fourier transforms?

Thanks again.

Name:
E-Mail:

Subject: