Originally Posted by

**mmzaj** ok , my bad , here is the post again ..

i was trying to come with an easier way of finding the Hilbert transform of a function $\displaystyle \ x(t)$ , and here is what i did :

starting with Fourier transform of x :

$\displaystyle X(f)=\int^{\infty}_{-\infty}\ x(t) \ e^ {\ -i2\pi ft} \ dt$

now

$\displaystyle \ e^ {\ -i2\pi ft} = \frac{1}{2\pi i}\oint\frac{e^ {\ -i2\pi

fz}}{z-t} \ dz$

where the contour encloses t .

"this is done by cauchy's integral formula"

=>

$\displaystyle X(f) = \frac{1}{2\pi i}\int^{\infty}_{-\infty}\ x(t) \oint\frac{e^ {\ -i2\pi fz}}{z-t} \ dz \

dt = \frac{1}{2\pi i} \oint\ e^ {\ -i2\pi fz}\int^{\infty}_{-\infty}\frac{x(t)}{z-t}\ dt \ dz$

now

$\displaystyle \pi\hat{x}(z)=\int^{\infty}_{-\infty}\frac{x(t)}{z-t}\ dt$

where

$\displaystyle \hat{x}(z)$ is the Hilbert transform of x .

=>

$\displaystyle

\ X(f)= \frac{1}{2i}\oint\hat{x}(s)\ e^ {\ -i2\pi fs}\ ds$

"z is replaced with s for conventional reasons"

now the program is to relate the last integral to the inverse laplace (or fourier !!) transform of$\displaystyle \hat{x}(s)$, by choosing the suitable contour(s) , and here is where i'm stuck !! so any help is appreciated .