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?
  • $\begingroup$ Integrate by parts. $\endgroup$ – Jazzmaniac Feb 13 '17 at 12:27
  • 3
    $\begingroup$ Shouldn't this be asked in math SE? $\endgroup$ – Tendero Feb 13 '17 at 13:32
  • $\begingroup$ Which of $f$ and $\phi$ is the impulse? $\endgroup$ – Dilip Sarwate Feb 13 '17 at 14:23
  • 4
    $\begingroup$ This question needs some editing. You didn't explain what $f(t)$ and $\phi(t)$ are. $\endgroup$ – Matt L. Feb 13 '17 at 17:21

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


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.

  • $\begingroup$ 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. $\endgroup$ – Fat32 Feb 14 '17 at 21:20
  • $\begingroup$ @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. $\endgroup$ – Tendero Feb 14 '17 at 21:23

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