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.
Asked
Active
Viewed 2.5k times
2
-
No such proof can exist, for this “function” isn't a function at all. Also known as Dirac's delta, it belongs to a class called “generalized functions”, or distributions. Its “integral” is 1 by definition. – Harald Hanche-Olsen Dec 28 '13 at 08:51
1 Answers
2
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 $-\infty$.

user44197
- 9,730
-
To expand on what I wrote: Think of the $\delta$ function as performing snap shot. If you want to take a snapshot of $f(x)$ at $x=0$, then multiply it by the $\delta$ function and then integrate it over any interval that contains zero. To take a snapshot at any other point, shift the $\delta$ function to the point where you want to take the snapshot. This property is used to model ideal sampler of signals. $\delta$ function has no meaning outside an integral, it is almost like fish out of water when you take it out of an integral. – user44197 Dec 28 '13 at 08:56