Prove that $\int f(\theta(x))\delta(x)dx = \int f(y)dy$ ??

Question

Show that the above is true. where $\delta(x)$ is the delta and $\theta(x)$ is the Heaviside Step function. We know that $$\frac{d\theta(x)}{dx} = \delta(x)$$

I am stuck on this. Please help me? Thanks

Answer 1

It would be hard to give a proof, since writing things like $\frac{d\theta}{dx}=\delta$ already puts you on dubious logical ground. In any case, if we assume that then your integral is just $$\int dx \theta'(x)f(\theta(x)) = \int dyf(y)$$ just by making the substitution $y=\theta(x)$.

Answer 2

Even for very nice choices of $f$, I fail to see how this is meaningful beyond a nice symbolic "equality" via u-substitution as per Daniel Littlewood's answer. The reason being that $f\circ \theta$ is in most instances not well-defined as a test function at $0$ so integration against $\delta$ is ill-defined. (The one exception being $f\equiv 0$.)

Taking $f(x) = \exp(-x^2)$ for instance (arguably one of the nicest functions in many senses of the word), the integral on the right hand side is $\sqrt{\pi}$. The integral on the left hand side should pick out the value of $f\circ\theta$ at $0$, however at $0$, $f\circ\theta$ has a jump discontinuity so the Dirac delta averages the left and right hand limits of $f\circ\theta$. These limits evaluate to $(f\circ\theta)(0^-) = f(0) = 1$ and $(f\circ\theta)(0^+) = f(1) = e^{-1}$. Their average is definitively not $\sqrt{\pi}$.

Answer 3

It is useful to review the subsitution theorem which tells us what constitutes a valid subsitution:

Let $\phi$ be a real function which has a derivative on the closed interval $[a,b]$. Let $I$ be an open interval which contains the image of $[a,b]$ under $\phi$. Let $f$ be a real function which is continuous on $I$. Then $$\int_a^b f(\phi(t))\phi'(t)\,{\rm d}t = \int_{\phi(a)}^{\phi(b)}f(u)\,{\rm d}u$$

The function $\theta(t)$ you are trying to substitute does not have a derivative at $x=0$ and the image of $\theta(t)$ under $\mathbb{R}$ is two single points!

We can however apply substitution by $\theta$ on any interval that does not contain $0$ to get

$$\int_a^b f(\theta(t))\cdot 0\,{\rm d}t = \int_{0}^{0}f(u)\,{\rm d}u$$

or

$$\int_a^b f(\theta(t))\cdot 0\,{\rm d}t = \int_{1}^{1}f(u)\,{\rm d}u$$

which both are trivially true.