# A simple delta function properties, sifting property

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1. Sep 9, 2015

### hojoon yang

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I don't know why this is possible

To use delta function properties( sifting property)

integral range have to (-inf ,inf)

or at least variable s should be included in [t_0,t_0+T]

but there is no conditions at all (i.e. t_0 < s < t_0+T)

am I wrong?

2. Sep 9, 2015

### blue_leaf77

It is the sameness of the integration limits that eliminates the need for them to be +- infinity. Consider the integral over $t$ only
$$\int_{t_0}^{t_0+T} \delta(t-s) dt$$.
The integrand is a delta function centered at $t=s$, but the range of $s$ is the same as the range of $t$. This means the integrand $\delta(t-s)$ is guaranteed to always be located inside the integration limit of $t$. So,
$$\int_{t_0}^{t_0+T} \delta(t-s) dt = 1$$.

3. Sep 9, 2015

### hojoon yang

u mean delta function property does not care about integral range?

4. Sep 9, 2015

### blue_leaf77

This is only true under the situtaion being considered, namely there is a second integral with respect to $s$ present with the same limits as the first integral.

EDIT: It might have been more justifiable if I had written
$$\int_{t_0}^{t_0+T} \delta(t-s) dt = 1 \hspace{1.2cm} \text{for} \hspace{1.22cm} s=[t_0,t_0+T]$$

Last edited: Sep 9, 2015
5. Sep 13, 2015

### rude man

Taking one integral at a time: ∫ δ(t-s)dt from t0 to t0 + T = 0 unless s is within range of t0 to t0 + T, in which case it = 1. So you have to assume s is within range of t0 to t0 + T, and I agree that should have been specified.

Then the second integral is obviously T so the whole thing is N0/2 times 1 times T = N0T/2.

6. Sep 13, 2015

### Ray Vickson

Do you mean "shifting" or "sifting"? These are both relevant aspects of transforms such as the delta function, but they refer to vastly different properties.