# Argument of Dirac delta function is quadratic with complex roots

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.

Purna.

• you must be looking for Completing the square and change of variable, right? – Gideon Genadi Kogan May 7 '20 at 9:59
• Yeah, but we get complex roots. How do we handle them? – 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 $$f(t)$$, $$t\in \mathbb{R}$$:

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

• And I don't forget that I still have to update the one on $\delta^2$ (long overdue) – Laurent Duval May 7 '20 at 14:28
$$\delta (t^2+t+1) = \delta(t + 1/2 - j\cdot \sqrt{3}/2)) + \delta(t + 1/2 + j\cdot \sqrt{3}/2)$$
• The $t$ argument might be missing on the RHS – Laurent Duval May 7 '20 at 12:34