Integral of unit impulse function?

[Abdulwahab Hajar]

Homework Statement

let's use this symbol to denote the unit impulse function δ
When integrating the unit impulse function (from negative infinity to infinity) ∫δ(t) dt I know that this results in a value of 1 and is only nonzero at the point t = 0.

However for example take this integral into consideration ∫δ(t) e^-jωt
since the delta function is only nonzero at the point zero, we only evaluate this multiplication at the point 0 which yields e⁰ which is 1.

but how can we do that, the integral involves two functions dependant on time shouldn't we integrate from limits for example 0^- to 0⁺ and integrate it by parts or something like that?

Homework Equations

∫δ(t) = 1 at t =0

The Attempt at a Solution

My attempt is attempting to explain it above
Thank you

 

[FactChecker]

There are a few different ways to put the Derac delta function (generalized function, distribution) on a solid theoretical basis (see https://en.wikipedia.org/wiki/Dirac_delta_function). The result and goal of all of them is that ∫δ(t)f(t)dt = f(0).

 

[Abdulwahab Hajar]

There are a few different ways to put the Derac delta function (generalized function, distribution) on a solid theoretical basis (see https://en.wikipedia.org/wiki/Dirac_delta_function). The result and goal of all of them is that ∫δ(t)f(t)dt = f(0).

thank you, you were very helpful...
loving the profile pic btw!