I'm trying to verify this
\begin{equation}
\int{d^3\vec{r}\ } \langle\rho\rangle_T\; = e,
\end{equation}
where $\langle\rho\rangle_T$ is the temporal average of $\rho(x,t) = e\delta(x-a\sin{\omega t})$ and $e$, $a$ and $\omega$ are constants.
If $\rho(x,t) = e\delta(x-a\sin{\omega t})$, the problem is reduced to
\begin{equation}
\int_{-\infty}^{\infty}{dx\ } \langle\rho\rangle_T\; = \int_{-a}^{a}{dx\ } \langle\rho\rangle_T,
\end{equation}
but I don't know how to solve the integral that appeared:
\begin{equation}
\langle\rho\rangle_T\; = \frac{e}{T} \int_{0}^{T}{dt\ } \delta(x-a\sin{\omega t}).
\end{equation}
I have tried to use an elementary $u$-sustitution: $u = a\sin{\omega t}$, then
$$du dt\ a\omega\cos{\omega t} \quad \leftrightarrow
\quad \frac{1}{a\omega} \left[1-\left(\frac{u}{a}\right)^2\right]^{-1/2}\ du = dt.
$$
Or
\begin{equation}
\langle\rho\rangle_T\; = \frac{e}{a\omega T} \int_{0}^{T}{du\ } \left[1-\left(\frac{u}{a}\right)^2\right]^{-1/2}\delta(x-u).
\end{equation}
Is there a property than I can use? Is the $u$-sustitution valid?
Best Answer
Definiton of the Composition of the Dirac Delta $\displaystyle \delta$ with a Function $\displaystyle g$, $\displaystyle \delta \circ g$
The Dirac Delta of a composition with a smooth function $g$, written $\delta \circ g$ is defined such that for any test function $\phi$ we have
$$\begin{align} \langle \delta \circ g, \phi\rangle &=\int_{-\infty}^\infty \delta(g(x))\,\phi(x)\,dx\\\\ &=\sum_{n=1}^N \int_{-\infty}^\infty \frac{\delta(x-x_n)}{|g'(x_n)|}\,\phi((x))\,dx\\\\ &=\sum_{n=1}^N \frac{\phi(x_n)}{|g'(x_n)|} \end{align}$$
where $g$ is assumed to be continuously differentiable, $g'$ is nowhere $0$, and $g$ has $N$ simple roots $x_n$, $1\le n\le N$.
Applying the General Result to the Problem of Interest
Let $\phi(t)=\xi_{[0,T]}(t)$ and let $g(t)=x-a\sin(\omega t)$ so that $g'(t)=-a\omega \cos(\omega t)$. We will assume that $a>0$, $x>0$ and $x/a<1$.
The roots of $g(t)$ are at values of $t_n$ such that $\sin(\omega t_n)=x/a$. These values on $[0,T]$ are at $t_1=\frac1\omega \arcsin(x/a)$ and $t_2=\frac{\pi}\omega-\frac1\omega\arcsin(x/a)$.
At either of the roots of $g(t)$, $|g'(t_n)|=a\omega \sqrt{1-(x/a)^2}$
Putting it together we find that
$$\int_{-\infty}^\infty \xi_{[0,T]}\delta(x-a\sin(\omega t))\,dt=\frac2{a\omega\sqrt{1-\left(\frac xa\right)^2}}$$
whereupon integrating this result over $[-a,a]$ yields
$$\frac eT\int_{-a}^a \frac2{a\omega\sqrt{1-\left(\frac xa\right)^2}}\,dx=2\pi e /(\omega T)=e$$
as was to be shown!