Prove that Unit Impulse Function Integral is equal to one?

Question

Unit impulse function is one of the special functions which is widely used in the field of signal processing. It has nice properties that helps in some situations specially its sifting property. But This depends on the fact of its integral is equal to one. As you know the continuous impulse function S(t) is equal to inf at zero and equal to zero elsewhere. So the boundaries of the integral of it is reduced to the period [0,0]. I need a proof of how did we calculate its integral to get 1.

Answer 1

You can't prove it using standard notions of integrals since the $\delta$ function is not a well defined function. In fact, $\delta$ function is defined by the integral property

If $f(x)$ is continuous at $x=0$ then $$ \int_{-\infty}^{\infty} f(x) \delta(x) dx = f(0)$$

You just have to accept three things: (a) such a function "exists" in some sense, (b) such a function is unique and (c) such a function is incredibly useful!

That said, set $f(x) = 1$ the constant function to get $$ \int_{-\infty}^{\infty} \delta(x) dx = 1$$

By the way, don't be trapped into thinking $\delta$ function is the limiting case where it blows up at zero. There are examples where the limit is constant or even $-\infty$.