# How to solve $\int_{0}^{T}{dt\ } \delta(x-f(t))$ where $\delta$ is the dirac $\delta$-function?

I'm trying to verify this $$$$\int{d^3\vec{r}\ } \langle\rho\rangle_T\; = e,$$$$ where $$\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 $$$$\int_{-\infty}^{\infty}{dx\ } \langle\rho\rangle_T\; = \int_{-a}^{a}{dx\ } \langle\rho\rangle_T,$$$$ but I don't know how to solve the integral that appeared: $$$$\langle\rho\rangle_T\; = \frac{e}{T} \int_{0}^{T}{dt\ } \delta(x-a\sin{\omega t}).$$$$ I have tried to use an elementary $$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 $$$$\langle\rho\rangle_T\; = \frac{e}{a\omega T} \int_{0}^{T}{du\ } \left[1-\left(\frac{u}{a}\right)^2\right]^{-1/2}\delta(x-u).$$$$

Is there a property than I can use? Is the $$u$$-sustitution valid?

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!

Observe \begin{align} \int dx\ \langle \rho\rangle_T =&\ \frac{e}{T}\int dx \int^T_0 dt\ \delta(x-a\sin(\omega t)) = \frac{e}{T}\int^T_0 dt\int dx \ \delta(x-a\sin(\omega t)) \\ =&\ \frac{e}{T}\int^T_0 dt\ 1 = e. \end{align}