Inverse Fourier transform of $F(w)=e^{-|w|a}$ where a 0

Question

Inverse Fourier transform of $F(w)=e^{-|w|a}$ where a > 0

richyw · Answer

$$f(x)=\int^\infty_{-\infty}e^{-iwx}e^{-|w|a}dw$$$$f(x)=\int^0_{-\infty}e^{w(a-ix)}dw+\int^\infty_0e^{w(-a-ix)}dw$$The integral on the right is messing me up. I make the substitution $$u=w(a-ix)$$$$du=(a-ix)dw$$

richyw · Answer

oops, that's the substitution for the first integral. I substitute $$v = w(-a-ix)$$$$dv=(-a-ix)dw$$

richyw · Answer

$$\frac{1}{-a-ix}\int_0^{?} e^vdv$$

richyw · Answer

is my top limit of integration now $-\infty$ ? because $v=-w(a+ix)$ If not I would get an infinite answer...

richyw · Answer

ok I think I just missed that detail about the integration limits in the integral. I end up with $$f(x)=\frac{2a}{a^2+x^2}$$

accessdenied · Answer

At the beginning when you take the inverse Fourier transform by integral definition, you have to use the positive exponent right?
  $ \large e^{i \omega x} $ instead of $ \large  e^{-i \omega x} $

accessdenied · Answer

But it seems like I get the same answer with that adjustment.

richyw · Answer

doesn't it depend on how you define the ft and ivt? like one of them has to have the 2π factor and the other one doesn't.

richyw · Answer

attachment

accessdenied · Answer

Hm... I knew there was some uncertainty around the 2 pi factor for definition, but I actually did not know there was room to define the e^(iwx) vs. e^(-iwx) differently as well. :) http://reference.wolfram.com/mathematica/tutorial/IntegralTransformsAndRelatedOperations.html

accessdenied · Answer

So, I figure as long as that is consistent with what you use normally, that will be fine then.
It seems both led to the answer.

richyw · Answer

yup, that page you have even shows the different conventions. in physics you usually see it different. Apparently this is the "pure mathematics" convention. funny because the book is called "applied PDE's" haha. Also thanks for sending that link, because now I know how to change the convention from the default in mathematica!

accessdenied · Answer

yup, I found it looking up the situation more closely. Linked through a stackexchange post about it, top response. http://math.stackexchange.com/questions/574581/fourier-transform-convention-frac1-sqrt2-pi-int-infty-infty-fx?rq=1