Proving that the delta function is the derivative of the step function.

Question

I want to prove $\frac{\mathrm{d} }{\mathrm{d} x}\Theta =\delta (x)$ using this representation of the delta function: $\delta(x)= \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx}dk $ This should be easy. I just need to integrate $\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx}dk $ with respect to x to get $ \Theta(x) $

This is the first step:

$$\Theta(x) = \int_{-\infty}^\infty \frac{-i}{2 \pi k} e^{ikx} \, dk $$

This doesn't look like the step function to me. So I already suspect that I'm going down the wrong road here.

But, unless I'm being naïve, theoretically this should work.

I tried to do integration by parts, but I think that I'm going down the wrong road because I just have a more complicated mess to deal with. Have I made an error? Is there another way to evaluate this integral?

The Step Function:

$\theta(x) = 1$ if $x>0$

$\theta(x) = 0$ if $x \leq 0$

Answer 1

Informally: $$\int_{-1}^{1} f(x)\Theta'(x) \, dx = \left[\vphantom{\frac11} f(x) \Theta(x) \right]_{-1}^{+1} - \int_{-1}^1 f'(x)\Theta(x)\,dx = f(1) - \left[f(x) \vphantom{\frac11} \right]_0^1 = f(0)$$

which is precisely what you want from $\delta(x)$. So $\delta(x)=\Theta'(x)$.

Answer 2

What you're trying to do technically needs to be understood in the context of distribution theory. But since we can hardly go about introducing the intricate details of distribution theory here, I'll just offer a quick suggestion: show that $$ \int_{-\infty}^\infty \delta(x)\psi(x)~dx = -\int_{-\infty}^\infty \Phi(x)\frac{d}{dx}\psi(x)~dx $$ for all smooth functions $\psi$ with compact support (i.e. vanishes outside of a finite closed interval). Then integrate the right-hand side by parts to see that what you have is $$ \int_{-\infty}^\infty \delta(x)\psi(x)~dx = \int_{-\infty}^\infty \frac{d}{dx}\Phi(x)\psi(x)~dx $$ for all smooth $\psi$ of compact support. Therefore "$\delta(x) = \frac{d}{dx}\Phi(x)$."