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

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Posted by Brad Paul on September 20, 2002 at 08:13:09:

In Reply to: Re: Fourier Transform; thanks, and... posted by Tim and Alex on September 19, 2002 at 20:53:11:

: : : 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.

Good work!

I used Mathematica to do the Fourier Transform. I also looked it up in
a table to double check it.

But be warned Mathematica is a good thing and a bad thing. I had I never
used Mathematica I might have thought on my own to use complex analysis.

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