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

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Homework Statement



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

Homework Equations



\int^{\infty}_{-\infty}\delta(t)dt = 1

\int uv dt = u\int v dt - \int u^{'}(t)\int v dt dt

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

\delta(-\infty) = \delta(\infty) = 0

The Attempt at a Solution



\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
 
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The integral of delta(x)^2 is not defined, it's infinite. Is that really the problem you were given?
 
Last edited:
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.
\delta(t) 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.
 
anybody please answer to my problem.
why the integral turns out to be zero.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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