Derivative of delta function

Question

I've come across the below. Can you please help me understand the proof for this?

$$\int_{-\infty}^{\infty}x(t)\delta'(t-2.5)dt=-\frac{dx(t)}{dt}{\Big |}_{t=2.5}$$

Answer 1

You can see this by using integration by parts:

$$\begin{align}\int_{-\infty}^{\infty}x(t)\delta'(t-T)dt&=x(t)\delta(t-T){\Big|}_{-\infty}^{\infty}-\int_{-\infty}^{\infty}x'(t)\delta(t-T)dt\\&=x(T)\delta(t-T){\Big|}_{-\infty}^{\infty}-x'(T)\\&=-x'(T)\end{align}$$

where it is assumed that $x(t)$ and $x'(t)$ are continuous at $t=T$.

Answer 2

The proof comes from the Dirac delta function property:

\begin{align} \int\limits_{-\infty}^{\infty} x(t) \delta^{(n)}(t-t_0)dt=(-1)^{n}\frac{d^n}{dt^n}x(t)\bigg\vert_{t=t_0} \end{align}

where $x(t)$ is a continuous function of time with a continuous derivative at $t= t_0 $.

Answer 3

The principal idea working with stuff like delta functions and their derivatives that they aren't actually functions but idealizations with a certain behavior.

The behavior of $\delta(t)$ comes out when integrating its product with another function: then it selects a single function value.

In a similar vein, the behavior of $\delta'(t)$ comes out by doing integration by parts until the behavior of $\delta(t)$ surfaces. That works as long as you can avoid dealing with the product of two such things at a common point of singularity: in that case, stuff tends to fall apart because a number of possible or impossible behaviors may result.