Why is the delta function not defined as a regular function?

[electron2]

why
$$\int_{0^+}^{t} (x+1)\delta (x)dx=0$$

delta is defined that in 0 it is infinity
and on x differs 0 its value is 0

there heard of some formula
i am not sure if its the right form
$$\int f(x)\delta (x)dx=f(0)$$

why i get 0
??

 

[queenofbabes]

I can make out what you're asking, I think, but in future please express yourself in clearer English.

The delta function is simply defined in the way you described, yes. It has the important property that $$\int_{-\infty}^{\infty} \delta (x)dx = 1$$
, or rather, as long as the point x=0 is within the limits of integration, the result is 1, otherwise it is 0.

Thus looking at your second integral, as the integration is carried out from -infinity, the delta function is 0, except when the integral crosses the point x=0, where delta(x) = 1 and f(x) = f(0), thus the integral "picks out" the value of f(x) at x=0

 

[HallsofIvy]

No, the delta function is NOT defined by "in 0 it is infinity
and on x differs 0 its value is 0" since "infinity" is not an allowable value for functions and, in fact, the delta "function" is not really a function- it is a "distribution" or "functional" and is defined by the property you give: If a< 0< b, then
[tex]\int_a^b f(x)\delta(x) dx= f(0)[/itex]
or, more generally, if a is in the set A then, for any function integrable on A,
[tex]\int_A f(x)\delta(x-a)dx= f(a)[/itex]
while, if a is not in the set,
[tex]\int_A f(x)\delta(x-a)dx= 0[/itex]

 

[g_edgar]

I suppose that $$0^+$$ means do the integral from $$a$$ to $$+\infty$$ with $$a>0$$, then take the limit as $$a$$ goes to $$0$$ from the right. Then your result is $$0$$, but the same thing with $$0^-$$ is $$1$$ .

Anyone have another guess what it means?

 

[Prologue]

The limits of integration do not contain the x value where the delta function explodes. If this x value (exactly zero in this case) is not contained in the limits then the delta function is always zero, so the integral is always zero.

 

[electron2]

No, the delta function is NOT defined by "in 0 it is infinity
and on x differs 0 its value is 0" since "infinity" is not an allowable value for functions and, in fact, the delta "function" is not really a function- it is a "distribution" or "functional" and is defined by the property you give: If a< 0< b, then
[tex]\int_a^b f(x)\delta(x) dx= f(0)[/itex]
or, more generally, if a is in the set A then, for any function integrable on A,
[tex]\int_A f(x)\delta(x-a)dx= f(a)[/itex]
while, if a is not in the set,
[tex]\int_A f(x)\delta(x-a)dx= 0[/itex]

i can't link your formulas with my interval

how to explainm interval in your terms?