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 $x \leq 0$
Best Answer
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 $\delta(x)=\Theta'(x)$.