Impulse property - needed mathematical proof

Question

Prove: $$ \left. \int_{-\infty}^{\infty}\frac{d\delta(t)}{dt}\phi(t)dt=\frac{-d\phi(t)}{dt}\right|_{t=0} $$

$\delta(t)$ is impulse signal and other one is any general signal.

• Can anybody do the proof for me?
• Or at-least how to proceed further?

Answer 1

For any pair of functions $u(t)$ and $v(t)$ we have that

$$\frac{d}{dt}(u(t)v(t))=v(t)\frac{du(t)}{dt}+u(t)\frac{dv(t)}{dt}$$

If we integrate on both sides of the equation we get

$$\int\frac{d}{dt}(u(t)v(t)) \ dt=\int v(t)\frac{du(t)}{dt} \ dt+\int u(t)\frac{dv(t)}{dt} \ dt$$

In this case, $u(t)=\delta(t)$ and $v(t)=\phi(t)$, so

$$\int\frac{d}{dt}(\delta(t)\phi(t)) \ dt=\int \phi(t)\frac{d\delta(t)}{dt} \ dt+\int \delta(t)\frac{d\phi(t)}{dt} \ dt$$

The left side of the equation can be solved easily as

$$\int\frac{d}{dt}(\delta(t)\phi(t)) \ dt = \frac{d}{dt}\int\delta(t)\phi(t) \ dt=\frac{d}{dt}\phi(0)=0$$

as $\phi(0)$ is a constant. We also know that the delta function has the following property

$$\int \delta(t)\frac{d\phi(t)}{dt} \ dt = \left.\frac{d\phi(t)}{dt}\right|_{t=0}$$

So going back to the first equation and replacing with these results:

$$0=\int \phi(t)\frac{d\delta(t)}{dt} \ dt+\left.\frac{d\phi(t)}{dt}\right|_{t=0}$$

And that's the proof.