Limit of sinc and dirac delta

Question

If I have a gate function $x(t) = \left\{ \begin{array}{ll} x_0 & \mbox{if $-T < x < T$};\\ 0 & \mbox{elsewhere}.\end{array} \right. $

Then:

$$\lim_{T\to \infty} X(\omega) = \lim_{T\to \infty} x_0\int_{-T}^{T}e^{-i\omega t}dt = \lim_{T\to \infty} 2x_0T sinc(\omega T) = \left\{ \begin{array}{ll} \infty & \mbox{if $\omega = 0$}\\ 0 & \mbox{elsewhere}.\end{array} \right. =x_02\pi \delta(w)$$

I don't understand the last two steps (the limit and the delta):

$$\lim_{T\to \infty} 2x_0T sinc(\omega T) = \left\{ \begin{array}{ll} \infty & \mbox{if $\omega = 0$}\\ 0 & \mbox{elsewhere}.\end{array} \right. =x_02\pi \delta(w)$$

Could someone explain this ?

EDIT:

Actually, I get that :

$$\lim_{T\to \infty} 2x_0T sinc(\omega T) = \left\{ \begin{array}{ll} \infty & \mbox{if $\omega = 0$}\\ 0 & \mbox{if $\omega \to \infty$}.\end{array} \right.$$

but I don't see how it's zero anywhere else ... so how can this be a dirac delta ?

Answer 1

The T inside the argument of the sinc squeezes the function towards the origin. Most of the area of a sinc function is near the origin already so the contribution the rest of the function has gets more and more insignificant. T outside of the argument makes sure that the total area remains constant.

$$\int ^\infty _{-\infty}Tf(Tx)dx=\int ^\infty _{-\infty}f(x)dx$$

So while the total area remains constant, increasing T gathers the area near the origin

Some illustration from wolfram: