Taking the derivative of the limit representation of the Heaviside function (I should preface this by saying that my background is in physics-forgive me if I'm doing/asking something obvious!) 
I am given this limit representation of the Heaviside step function
$$
\theta(t) = \lim_{\epsilon \rightarrow 0^+} \frac{1}{2\pi i} \int_{-\infty}^{\infty} d\omega \frac{e^{i\omega t}}{\omega - i \epsilon}
$$
to show that 
$$
\frac{d}{d t}\theta(t) = \delta(t)
$$
My first instinct to simplify this problem was to move the limit inside the integral, like so:
\begin{equation*}
\begin{aligned}
\theta(t) &= \lim_{\epsilon \rightarrow 0^+} \frac{1}{2\pi i} \int_{-\infty}^{\infty} d\omega \frac{e^{i\omega t}}{\omega - i \epsilon} \\
&= \lim_{\epsilon \rightarrow 0^+} \int_{-\infty}^{\infty} \frac{1}{2\pi}  \frac{e^{i\omega t}}{i\omega - i^2 \epsilon} d\omega \\
&= \int_{-\infty}^{\infty} \frac{1}{2\pi} \lim_{\epsilon \rightarrow 0^+} \frac{e^{i\omega t}}{i\omega - i^2 \underbrace{\epsilon}_\text{= 0}} d\omega \\
&= \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega \frac{e^{i\omega t}}{i\omega}
\end{aligned}
\end{equation*}
However, when I googled moving limits inside integrals, I was told by the internet not to do so unless the integral is constant. I know that the Heaviside is not constant and has a discontinuity. But then I also thought "if we are doing the limit from the positive side of zero, wouldn't that be avoiding the discontinuity?" I am not sure this reasoning works, so that would be my question 1.
From this point I took a derivative with respect to time of the Heaviside
\begin{equation*}
\begin{aligned}
\frac{d}{dt}[\theta(t)] &= \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega \frac{\frac{d}{dt}[e^{i\omega t}]}{i\omega} \\
&= \frac{1}{2\pi} \int_{-\infty}^{\infty} d\omega \frac{i\omega e^{i\omega t}}{i\omega} = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega t} d\omega \\
&= \frac{1}{\sqrt{2\pi}}[\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{i\omega t} d\omega]
\end{aligned}
\end{equation*}
It was at this point I thought the integral reminded me of the Fourier transform. I know that the Fourier transform of the delta function is 
\begin{equation}
\mathcal{F}_x[\delta(x-x_0)](k) = \int_{-\infty}^{\infty} e^{-2\pi i k x}\delta(x-x_0)dx = e^{-2\pi i k x_0}
\end{equation}
and also
\begin{equation*}
\begin{aligned}
\mathcal{F}[\delta(x)](k) &= 1 \\
\mathcal{F}[\theta(x)](k) &= \frac{1}{ik} + \pi\delta(k)
\end{aligned}
\end{equation*}
I then took the Fourier transform of $\delta(x-0) = \delta(x)$:
\begin{equation*}
\begin{aligned}
\mathcal{F}[\delta(x)](\omega) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-i\omega x}\delta(x)dx \\
&= \frac{1}{\sqrt{2\pi}} e^{-i\omega (0)} = \frac{1}{\sqrt{2\pi}}
\end{aligned}
\end{equation*}
Since I know that $\mathcal{F}^{-1}[1](x) = \delta(x)$ I figured that could apply a similar method of performing the inverse Fourier transform on $\frac{1}{\sqrt{2\pi}}$. 
\begin{equation*}
\begin{aligned}
\mathcal{F}^{-1}[\frac{1}{\sqrt{2\pi}}](x) &= \frac{1}{\sqrt{2\pi}}\delta(x) = \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{i\omega x}dx 
\end{aligned}
\end{equation*}
I returned to the time derivative of the Heaviside function 
\begin{equation*}
\begin{aligned}
\frac{d}{dt}[\theta(t)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{i\omega t} d\omega 
\end{aligned}
\end{equation*}
The RHS of $\frac{d}{dt}[\theta(t)]$ looks incredibly similar to the RHS of $\mathcal{F}^{-1}[\frac{1}{\sqrt{2\pi}}](x)$, except for a change in variables. Thus I said that
\begin{equation*}
\begin{aligned}
\frac{d}{dt}[\theta(t)] = \mathcal{F}^{-1}[\frac{1}{\sqrt{2\pi}}](\omega) = \underbrace{\frac{1}{2\pi}\delta(\omega) = \frac{1}{2\pi}\delta(t)}_\text{change variables}
\end{aligned}
\end{equation*}
However! This is not the statement 
$$
\frac{d}{d t}\theta(t) = \delta(t)
$$
but instead that
$$
\frac{d}{d t}\theta(t) = \frac{1}{\sqrt{2\pi}}\delta(t)
$$
which means I obviously did something wrong while taking my Fourier transforms, but I'm not entirely sure what.
tl;dr: 1.) Can I take the limit inside of the integral in the expression $\theta(t) = \lim_{\epsilon \rightarrow 0^+} \frac{1}{2\pi i} \int_{-\infty}^{\infty} d\omega \frac{e^{i\omega t}}{\omega - i \epsilon}$? and 2.) what am I not understanding about Fourier transforms that resulted in this extra factor of $\frac{1}{\sqrt{2\pi}}$ in the final answer?
 A: The answers to your question are respectively


*

*you can't exchange the limit and the integral because otherwise the integral does not converge, and

*you mixed up your conventions for the Fourier transform.


The "right" way to show this equality invokes the theory of distributions.
For $\epsilon >0$, define the nascent step function by
$$\theta_{\epsilon}(t)\stackrel{\text{def}}{=}\int_{-\infty}^{\infty}\frac{\mathrm{d}\omega}{2\pi\mathrm{i}}\frac{\mathrm{e}^{\mathrm{i}\omega t}}{\omega-\mathrm{i}\epsilon}\text{.}$$
We're tasked with showing
$$\frac{\mathrm{d}}{\mathrm{d}t}\lim_{\epsilon\to 0^+}\theta_{\epsilon}(t)\stackrel{?}{=}\delta(t)\text{.}$$
This is an equality of distributions: by definition, it amounts to showing that for all test functions $\phi$,
$$\langle(\lim_{\epsilon\to 0^+}\theta_{\epsilon})',\phi\rangle\stackrel{?}{=}\langle \delta,\phi\rangle\stackrel{\text{def}}{=}\phi(0)\text{.}$$
But by the definitions of $(\quad)'$ and $\lim$ for distributions we have
$$\begin{split}\langle(\lim_{\epsilon\to 0^+}\theta_{\epsilon})',\phi\rangle &=-\langle\lim_{\epsilon\to 0^+}\theta_{\epsilon},\phi'\rangle \\
&=-\lim_{\epsilon\to 0^+}\langle \theta_{\epsilon},\phi'\rangle\text{.}
\end{split}$$
Using the calculus of complex variables (or Fourier theory),
$$\theta_{\epsilon}(t)=\begin{cases}0 & t<0 \\ \tfrac{1}{2} & t=0 \\ \mathrm{e}^{-\epsilon t} & t>0\end{cases}$$
so
$$\begin{split}-\langle \theta_{\epsilon},\phi'\rangle&=-\int_0^{\infty}\mathrm{e}^{-\epsilon t}\phi'(t)\mathrm{d}t\\
&=\phi(0)-\epsilon\int_0^{\infty}\mathrm{e}^{-\epsilon t}\phi(t)\mathrm{d}t\end{split}$$
for any test function $\phi$. Taking the limit of both sides as $\epsilon \to 0^+$ gives the result.
