# Impulse property - needed mathematical proof

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?
• Integrate by parts. – Jazzmaniac Feb 13 '17 at 12:27
• Shouldn't this be asked in math SE? – Tendero Feb 13 '17 at 13:32
• Which of $f$ and $\phi$ is the impulse? – Dilip Sarwate Feb 13 '17 at 14:23
• This question needs some editing. You didn't explain what $f(t)$ and $\phi(t)$ are. – Matt L. Feb 13 '17 at 17:21

## 1 Answer

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.

• Nice effort! You could also begin with $\int{\phi(t) \delta(t) dt} = \phi(0)$ and then proceed as $\int{\phi(t)' \delta(t) dt} + \int{\phi(t) \delta(t)' dt}= 0$ and simply arrive at the conclusion as $\phi(0)' + \int{\phi(t) \delta(t)' dt}= 0$. for a much simpler layout. Note that when you work with generalized functions such as the $\delta(t)$ all these classical sense operators of differentiation, integration, and limit are just virtually acting on the impulse function. – Fat32 Feb 14 '17 at 21:20
• @Fat32 Thanks! Yes, that may have been simpler. I wanted to start from the very beginning anyway, just in case the OP couldn't work out the missing steps by his own. Apart from that, I must admit I'm not that familiar with distribution theory, so I followed the steps that helped me understand it when I studied this. – Tendero Feb 14 '17 at 21:23