How can $$\Theta(x) = \int_{-\infty}^\infty \frac{-i}{2 \pi k} e^{ikx} \, dk $$ possibly be the Heaviside Step Function?
What I'm looking for is a direct visualization or maybe an approximate numerical calculation that confirms that this weird thing is the unit step function.
To give some context:
I tried 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 was the first step:
$$\Theta(x) = \int_{-\infty}^\infty \frac{-i}{2 \pi k} e^{ikx} \, dk $$
Then I gave up because this did't look like the step function to me.
I tried to do integration by parts, but I just have a more complicated mess. A few people showed me an alternative method of proof. Today, I was reading The Six Core Theories of Modern Physics by Charles Stevens and I saw that, according to Stevens, what I originally wrote is the step function. How could this be? All he says is "This must be the unit step function because its derivative is the dirac delta.
The Step Function:
$\theta(x) = 1$ if $x>0$
$\theta(x) = 0$ if $x \leq 0$
Hint: Split it into real and imaginary parts. You'll need to evaluate $\int_{-\infty}^\infty \frac{\cos(u)}{u}du$ which is 0 because you're integrating an odd function, though the integral is divergent. You'll also need to evaluate $\int_{-\infty}^\infty \frac{\sin u}{u}du$ which is twice the Dirichlet integral, and is therefore equal to $\pi$.
It won't be exactly those, but armed with that you should be able to prove it.
Correction: $\int_{-\infty}^\infty \frac{\cos(u)}{u}du$ is actually divergent, so you'll need to take the principle value.