Argument of Dirac delta function is quadratic with complex roots

Question

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.

Answer 1

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:

• Dirac $δ(g(x))$ with complex roots $x_i $
• Regarding $x^2−a^2$ inside the argument of Dirac delta

Answer 2

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

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.