For example, we write $$\delta(at + b) = \frac{1}{|a|}\delta\left(t+\frac{b}{a}\right)$$ In a similar way, $$\delta(t^2 + t + 1)=\underline{\hspace{1cm}}$$

Thanks for the help.


  • $\begingroup$ you must be looking for Completing the square and change of variable, right? $\endgroup$ – Gideon Genadi Kogan May 7 '20 at 9:59
  • $\begingroup$ Yeah, but we get complex roots. How do we handle them? $\endgroup$ – user5045 May 7 '20 at 10:41

I shall provide more details if I am correct. IMO if there is not root on the domain of integration, and here I suppose that $t\in \mathbb{R}$, then the argument never vanishes. Then, in an engineer fashion, one could say (but $t\mapsto \delta(t)$ is not a function):

$$\delta(t^2+t+1) \equiv 0\,.$$

In more precise words, I would consider that for any suitable $f(t)$, $t\in \mathbb{R}$:

$$\int_{-\infty}^{\infty}f(t)\delta(t^2+t+1)dt = 0\,.$$

Related links:

  • 1
    $\begingroup$ +1 I certainly agree with your argument. $\endgroup$ – Dilip Sarwate May 7 '20 at 14:27
  • $\begingroup$ And I don't forget that I still have to update the one on $\delta^2$ (long overdue) $\endgroup$ – Laurent Duval May 7 '20 at 14:28

$$\delta (t^2+t+1) = \delta(t + 1/2 - j\cdot \sqrt{3}/2)) + \delta(t + 1/2 + j\cdot \sqrt{3}/2) $$

You get peaks wherever the argument of the delta function is zero. If you constrain t to be real, than the whole expression is simply zero. If t is complex you get two peaks: one at each root.

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    $\begingroup$ The $t$ argument might be missing on the RHS $\endgroup$ – Laurent Duval May 7 '20 at 12:34

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