The Dirac delta $\delta(z)$ has two poles of the first order, $z=i \varepsilon$, and $z=-i \varepsilon$, which have the residues $(2\pi i\varepsilon)^{-1}$ and $-(2\pi i\varepsilon)^{-1}$, respectively (see below). Now, according to Karl Rottmann's compact encyclopedia of mathematical formulas, these poles may be isolated, in the sense that
\begin{equation} \tag{1}\label{1} \delta(z) = \delta_+(z) + \delta_-(z), \end{equation}
where
\begin{equation} \tag{2}\label{2} \delta_+(z) = \frac{1}{2}\bigg[\delta(z) + \frac{i}{\pi} P\bigg(\frac{1}{z}\bigg)\bigg] \mathrel{\widehat{=}} \frac{1}{2\pi} \int_0^\infty e^{i\omega z} \, d\omega, \end{equation}
and
\begin{equation} \tag{3}\label{3} \delta_-(z) = \frac{1}{2}\bigg[\delta(z) - \frac{i}{\pi} P\bigg(\frac{1}{z}\bigg)\bigg] \mathrel{\widehat{=}} \frac{1}{2\pi} \int_0^\infty e^{-i\omega z} \, d\omega. \end{equation}
The curios-looking $P(z^{-1})$ is defined by
\begin{equation} \tag{4}\label{4} P\bigg(\frac{1}{z}\bigg) \equiv \lim_{\varepsilon\to 0} \, \frac{1}{2} \bigg(\frac{1}{z+i\varepsilon} + \frac{1}{z-i\varepsilon} \bigg) . \end{equation}
Now, my question is twofold: 1) From where does the integral representations of $\delta_+(z)$ and $\delta_-(z)$ [the last part of \eqref{2} and \eqref{3}, respectively] come? And 2): Does this mean that
\begin{equation} \tag{5}\label{5} \frac{1}{\pi} \int_0^\infty \cos \omega z \: d\omega = \frac{1}{2\pi} \int_0^\infty (e^{i\omega z}+e^{-i\omega z}) \, d\omega \mathrel{\widehat{=}} \delta(z) \end{equation}
is a valid statement? My guess for question 2) is yes, but having seen some of the non-intuitive things that may happen when any forms of infinities are involved, I'm vary of drawing conclusions based on intuition and feelings alone.
Remarks/Background
I'm aware that the Dirac delta $\delta(z)$ may be expressed as a hyperfunction,
\begin{equation} \tag{6}\label{6} \delta(z) \equiv \bigg(\frac{1}{2\pi iz}\, ,\,\frac{1}{2\pi iz}\bigg), \end{equation}
whereby the above statements considering the poles and residues of $\delta(z)$ are easy to see. I'm also familiar with the "Fourier" integral representation of $\delta(z)$,
\begin{equation} \tag{7}\label{7} \delta(z) \mathrel{\widehat{=}} \frac{1}{2\pi} \int_{-\infty}^\infty e^{-i\omega z} \, d\omega, \end{equation}
but I struggle to verify that it indeed fulfills the properties of the Dirac delta, i.e. that
\begin{gather} \int_{-\infty}^\infty f(z)\delta(z) \, dz = f(0) \tag{8}\label{8} \\ \int_{-\infty}^\infty \delta(z) \, dz = 1 \tag{9}\label{9}, \end{gather}
with $\delta(z)$ given by \eqref{7} [that \eqref{6} fulfills \eqref{8} and \eqref{9}, however, is easy to verify]. I guess I should also disclose that I'm not very well versed in measure theory – I know that it exists, and I know that $\delta(z)$ is not a function, but I don't know the intricate details.
Any help would be much appreciated!
Edit:
After some more searching on the web, I found that at least on a formal level, \eqref{5} is easy to show, given \eqref{7}: Using that
\begin{equation} \int_0^\infty e^{i\omega z} \, d\omega = - \int_0^{-\infty} e^{-i(-\omega) z} \, d(-\omega) = \int_{-\infty}^0 e^{-i(-\omega) z} \, d(-\omega) \mathrel{\widehat{=}} \int_{-\infty}^0 e^{-i\omega z} \, d\omega, \tag{10}\label{10} \end{equation}
one sees that
\begin{align} \frac{1}{2\pi} \int_0^\infty (e^{i\omega z}+e^{-i\omega z}) \, d\omega &= \frac{1}{2\pi} \bigg(\int_{-\infty}^0 e^{-i\omega z} \, d\omega + \int_0^\infty e^{-i\omega z} \, d\omega \bigg) \\ &= \frac{1}{2\pi} \int_{-\infty}^\infty e^{-i\omega z} \, d\omega \mathrel{\widehat{=}} \delta (z). \tag{11}\label{11} \end{align}
Question 1), however, still remains.
Edit 2:
I have found a way to show \eqref{8} and \eqref{9}. Please ignore this part of the question.