Inverse Fourier Transform of Inverse Square Root Function

[liorda]

Homework Statement

calculate the inverse Fourier transform of $$\left( a^2 + \left( bk \right)^2 \right)^{-1}$$

The Attempt at a Solution

I know that $$FT[e^{-|x|)}](k) = ( \pi (k^2 + 1 ) )^{-1}$$. I've tried to to concatenate the shift FT or the strech FT, but the "+1" in the known FT is in the way.

Sorry for my bad English, it's not my native language.

thanks.

 

[Defennder]

I can't quite read what you're writing here. The FT you've quoted is supposed to be $$F[e^{-\alpha|x|}] = \frac{2\alpha}{\alpha^2+ \omega^2}$$, correct?

Are a and b supposed to be arbitrary constants? If so, then I take $k=\omega$. You must then express the question in terms of the result above.

Note that $$\frac{1}{a^2 + b^2 \omega^2} = \frac{1}{b^2} \ \frac{1}{\frac{a^2}{b^2} + \omega^2}$$.

A factor of 2 is still required. Can you see it now?

 

[liorda]

thanks for answering. You helped me "getting it" :)

I think our normalization factors are different. I know that $$F[e^{-|x|}]=\frac{1}{\pi (\omega^2 +1)}$$, so if a,b>0 (i assume it, since it's the only way to get to the following result) i can write $$F[e^{-\frac{a}{b} |x|}]=\frac{b}{a} \hat{f}(\frac{\omega b}{a}) = \frac{a b e^{i \omega \frac{b}{a}}{\pi (\omega^2 b^2 + a^2)}$$

i'm using $$F[f(x)]=\frac{1}{2\pi}\int_{-\infty}^{\infty} f(x) e^{-i\omega x} dx$$

thanks.