1. ## opposite furier transfom

$F^{-1}[\frac{\sin \omega}{\pi \omega}]=\frac{h(x^+)+h(x^-)}{2}$
$F^{-1}[\frac{\sin \omega}{\pi \omega}][0]=1$

first of all the fulf tansform is 1 when its |x|<1 0 other wise
so why they chose x=0
??
from this oppisite transform how can they say that
$\int_{-\infty}^{\infty}[\frac{\sin \omega}{\pi \omega}]=1$
?

its only a tranform how can they say that the integral gives the same result straigh away
?

2. Originally Posted by transgalactic
$F^{-1}[\frac{\sin \omega}{\pi \omega}]=\frac{h(x^+)+h(x^-)}{2}$
$F^{-1}[\frac{\sin \omega}{\pi \omega}][0]=1$

first of all the fulf tansform is 1 when its |x|<1 0 other wise
so why they chose x=0
??
from this oppisite transform how can they say that
$\int_{-\infty}^{\infty}[\frac{\sin \omega}{\pi \omega}]=1$
?

its only a tranform how can they say that the integral gives the same result straigh away
?
I'm sorry but most of your post is incomprehensible.

But to see why the final integral is 1 from:

$\mathfrak{F}^{-1}[\frac{\sin \omega}{\pi \omega}][0]=1$

just write out the integral for this inverse FT.

CB