What Happens When You Integrate the Square of the Delta Function Over All Space?

In summary, the conversation is about finding the energy of a signal represented by \delta(t) and its integral. The attempt at a solution involves using the equations \int^{\infty}_{-\infty}\delta(t)dt = 1 and \int uv dt = u\int v dt - \int u^{'}(t)\int v dt dt, but the result is 0, which may not be the expected outcome. The person is seeking expert advice on whether the integral was done correctly or if there are other explanations for why it is 0.
  • #1
bishshoy007
7
0

Homework Statement



[tex]\int \delta(t)^{2} [/tex] dt from -infinity to +infinty

Homework Equations



[tex]\int^{\infty}_{-\infty}\delta(t)dt = 1[/tex]

[tex]\int uv dt = u\int v dt - \int u^{'}(t)\int v dt dt[/tex]

[tex] \int^{b}_{a} f(x) dx = F(b) - F(a) [/tex]

[tex]\delta(-\infty) = \delta(\infty) = 0 [/tex]

The Attempt at a Solution



[tex]\int \delta(t) . \delta(t) dt = \delta(t)\int \delta(t) dt - \int \frac{d \delta(t)}{dt} . \int \delta(t) dt dt = \delta(t).1 - \int \frac{d \delta(t)}{dt}dt = \delta(t) - \delta(t) = 0 [/tex]
 
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  • #2
The integral of delta(x)^2 is not defined, it's infinite. Is that really the problem you were given?
 
Last edited:
  • #3
Dick said:
The integral of delta(x)^2 is not defined, it's infinite. Is that really the problem you were given?

Thank you for replying.
[tex]\delta(t)[/tex] is an energy signal. I'm trying to find out the energy of the signal.
I wonder why the integral came out to be zero. I want some expert advice, whether I have done the integral correctly or not, or if there is some other explanations.
 
  • #4
anybody please answer to my problem.
why the integral turns out to be zero.
 

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