Question

If Integral of f(x)d(t-a) with limits from (-infinity to + infinity) is f(a) where d is the delta function...what will be the case if limits are finite?

Answer 1

will it be the same and limits will not affect the end result??

Answer 2

\[\int_{-\infty}^{\infty} f(x) \delta(x-a) dx=f(a)\]
?

Answer 3

yup @mukushla ...but for finite limits??

Answer 4

\[\int_{a-\epsilon}^{a+\epsilon}f(t)\delta(t-a)=f(a)\]who says it needs to be infinite? just symmetrical

Answer 5

well then we have
\[\int_{a}^{\infty} f(x) dx=f(a)\]

Answer 6

Max? do u know what he means?

Answer 7

but if limits are like....2 to 8..then??

Answer 8

what is the property of the delta function that will affect the integrand??

Answer 9

2=5-3
8=5+3
epsilon =3
a=5

Answer 10

\[\int_{5-3}^{5+3}f(t)\delta(t-a)dt=f(5)\]

Answer 11

I suppose

Answer 12

yup.....everything is repeating

Answer 13

yes, lag+repeating post=awkward explanation

Answer 14

erm, read above while I reload

Answer 15

the integral has to be symmetrical

Answer 16

let me give an example\[\int\limits_{-2}^{4}(2+t^2)\delta(t+1)dt\]...

Answer 17

i have posted an example above..

Answer 18

not necessary that it should be symmetrical

Answer 19

i think when delta is a fraction such that its integrand denominator = 0 by substituting the limits ,this can also affect the results

Answer 20

didnt understand!!!

Answer 21

I think you were right above actually, the integral does not need to be symmetrical
integration over any region containing the point x=a should give the same result
the symmetry bit is needed to prove various properties

Answer 22

in your integral a=-1
your range is (-2,4) and a=-1 is in that interval
so it is jus t\( f(a)\)

Answer 23

no....@TuringTest ...it will not be the answer.....i am trying to refer some books online.....lets see if i can get it...

Answer 24

I was looking at this for online stuff
http://tutorial.math.lamar.edu/Classes/DE/DiracDeltaFunction.aspx

Answer 25

"Note that the integrals in the second and third property are actually true for any interval containing t=a, provided it’s not one of the endpoints"
so it says

Answer 26

I'm not getting much out of mathworld
http://mathworld.wolfram.com/DeltaFunction.html

Answer 27

The delta function is the limit of some other set of functions(say Gaussians)
that satisfies the criterion of zero width and unit integral in the limt., like
\[\delta(x-a)= \lim_{\epsilon\rightarrow 0}\frac{1}{2\epsilon} if -\epsilon < x-a < \epsilon \]
and 0 otherwise

To use this in an integral that has a as an endpoint, you need to be very careful about how you go about taking the limits.

Answer 28

Any points on either side will do.