Time averaged delta function

Question

Suppose a Dirac delta function shifts horizontally as $\delta(x - a \sin(\omega t))$, could we define a "time average" of it like $$ f(x) = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} \delta(x - a \sin(\omega t)) dt $$ ? What would be the function look like? It is--I think--equivalent to $$ f(x) = \frac{\omega}{2\pi} \int_0^{2\pi/\omega} \delta \left(t-\frac{1}{\omega}\arcsin(x/a) \right) dt $$ It is tempting to argue that if $t=\arcsin(x/a)/\omega$ for a $t$ in $(0, 2\pi/\omega)$, and the integration becomes $\omega/2\pi$. But I think it is wrong. The reason is the integration bound is chosen arbitrarily: any bound with a gap of $2\pi/\omega$ should do.

Answer 1

Dirac delta is a distribution, which means that it is a linear functional acting upon the space of test functions. The same is true for $f$. That said, if you want to recognize $f$, then you can first apply $f$ to test functions.

Now if $\varphi \in \mathcal{D}(\mathbb{R})$ is a test function, then (by abusing notations to denote the distribution-function pairing by the usual integral notation) we find that

\begin{align*} \int_{\mathbb{R}} \varphi(x)f(x) \, dx &=\int_{\mathbb{R}} \varphi(x) \left( \frac{\omega}{2\pi} \int_{0}^{2\pi/\omega} \delta(x - a\sin(\omega t)) \, dt \right) \, dx \\ &= \frac{\omega}{2\pi} \int_{0}^{2\pi/\omega} \left( \int_{\mathbb{R}} \varphi(x) \delta(x - a\sin(\omega t)) \, dx \right) \, dt \\ &= \frac{\omega}{2\pi} \int_{0}^{2\pi/\omega} \varphi(a\sin(\omega t)) \, dt \tag{1} \\ &= \frac{1}{\pi} \int_{-a}^{a} \frac{\varphi(x)}{\sqrt{a^2 - x^2}} \, dx. \tag{2} \end{align*}

The computation $\text{(1)}$ tells that your integral is just a disguise of the following familiar-looking linear functional

$$ \varphi \quad \mapsto \quad \frac{\omega}{2\pi} \int_{0}^{2\pi/\omega} \varphi(a\sin(\omega t)) \, dt. $$

Moreover, $\text{(2)}$ tells that $f$ can be identified with the following function

$$ f(x) = \frac{1}{\pi\sqrt{a^2 - x^2}} \mathbf{1}_{[-a,a]}(x). $$

This is not surprising, since you are building up $f$ by summing up infinitesimal masses.