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

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

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

$\displaystyle \int_{-\infty}^{\infty}[\frac{\sin \omega}{\pi \omega}]=1$

?

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

?