2-Dimensional FOURIER TRANSFORM

Question

How can I do to calculate the Inverse Fourier Transform of:

$$G(w,y)=e^{-|w|y}$$ where w is real (w is the transform of x). I want to have $g(x,y)$, where $G$ is the Fourier Transform of $g$

Thanks

Answer 1

What we wrote with kryomaxim is correct:$$\frac{1}{2\pi}\intop_{w=-\infty}^{+\infty}e^{-\left|w\right|y}e^{iwx}dw=\frac{1}{2\pi}\intop_{w=-\infty}^{0}e^{wy}e^{iwx}dw+\frac{1}{2\pi}\intop_{w=0}^{+\infty}e^{-wy}e^{iwx}dw$$ $$=\frac{1}{2\pi}\intop_{w=-\infty}^{0}e^{w\left(y+ix\right)}dw+\frac{1}{2\pi}\intop_{w=0}^{+\infty}e^{w\left(ix-y\right)}dw=\frac{1}{2\pi}\left[\frac{e^{w\left(y+ix\right)}}{y+ix}\right]_{w=-\infty}^{w=0}+\frac{1}{2\pi}\left[\frac{e^{w\left(ix-y\right)}}{ix-y}\right]_{w=0}^{w=+\infty}$$ $$=\frac{1}{2\pi}\left(\frac{1}{y+ix}-\frac{1}{ix-y}\right)=\frac{1}{2\pi}\left(\frac{ix-y}{\left(y+ix\right)\left(ix-y\right)}-\frac{y+ix}{\left(y+ix\right)\left(ix-y\right)}\right)=\frac{1}{2\pi}\frac{-2y}{-x^{2}-y^{2}}=\frac{1}{\pi}\frac{y}{x^{2}+y^{2}}$$ where I used the fact that$$\lim_{\xi\rightarrow+\infty}\left|\frac{e^{\xi\left(ix-y\right)}}{ix-y}\right|=\lim_{\xi\rightarrow+\infty}\frac{e^{-\xi y}}{\sqrt{x^{2}+y^{2}}}=0 $$ to compute the second integral, provided that $y>0$ (same for the first). Please check your answer before saying that we said something wrong.

Answer 2

Very usefull in physics, these Fourier transforms!

If you have defined the Fourier transform as$$\hat{f}\left(w\right)=\textrm{TF }\left[f\right]\left(w\right):=\intop_{\mathbb{R}}f\left(x\right)e^{-iwx}dx$$ then you have the inversion formula$$\textrm{TF }^{-1}\left[\hat{f}\right]\left(x\right)=\frac{1}{2\pi}\intop_{\mathbb{R}}\hat{f}\left(x\right)e^{iwx}dw.$$ Here, $f\left(w\right)=f_{y}\left(w\right)=G\left(w,y\right)$ (for a fixed $y$ ) and thus$$g\left(x,y\right)=\textrm{TF }^{-1}\left[G\right]\left(x,y\right)=\frac{1}{2\pi}\intop_{\mathbb{R}}e^{-\left|w\right|y}e^{iwx}dw.$$ It is a one-dimensional Fourier transform, although there are two variables (you only integrate the $x$ dependence).

Answer 3

$g(x,y) = \frac{1}{2 \pi}\int_{-\infty}^\infty G(w,y) e^{iwx} dw = \frac{1}{2 \pi}\int_{-\infty}^\infty e^{-|w|y+iwx} dw = \frac{1}{2 \pi}( \int_{- \infty}^0 e^{iw(-iy+x)} dw +\int_{0}^\infty e^{iw(iy+x)}dw)$. In the last step I have split the integral into a positive part and a negative part of $w$ such that $|w|=-w$ for negative numbers of $w$.