Prove Fourier Transformation: $\int_{-\infty}^{\infty} \Theta(t) \sin (w_0 t) e^{- i w t} \,dt =- \frac{w_0}{ w^2-w_0^2 +i \text{sgn}(w)0^{+}}$ Prove:
$$\begin{align}\int_{-\infty}^{\infty} \Theta(t) \sin (w_0 t) e^{- i w t} \,dt &=- \frac{w_0}{ w^2-w_0^2 +i \text{sgn}(w)0^{+}}\\\\
&=- \frac{w_0}{w^2-w_0^2}+ \frac{i \pi }{2}(\delta(w-w_0)-\delta(w+w_0))\end{align}$$
where $\delta(x)$ is the Dirac function, $\Theta(x)$ is the Heaviside step function, $\text{sgn}(x)$ is the Sign function.
I think that the 2nd equation may   need to use Sokhotski–Plemelj theorem, but there is $\text{sgn}(w)$ in the  denominator, so I don't know how to use the identity.
 A: Using the convolution theorem, we have
$$\mathscr{F}\{\sin(\omega_0t)\Theta(t)\}=\frac1{2\pi}\underbrace{\mathscr{F}\{\sin(\omega_0t)\}}_{=i\pi(\delta(\omega +\omega_0)-\delta(\omega- \omega_0))}*\underbrace{\mathscr{F}\{\Theta(t)\}}_{=-\frac{i}{\omega}+\pi \delta(\omega)}$$
Therefore, the Fourier Transform of $\sin(\omega_0t)\Theta(t)$ is given by
$$\begin{align}
\mathscr{F}\{\sin(\omega_0t)\Theta(t)\}&=\frac1{2\pi}\int_{-\infty}^\infty \left(i\pi(\delta(\omega -\omega'+\omega_0)-\delta(\omega-\omega' -\omega_0)) \right)\,\left(-\frac{i}{\omega'}+\pi \delta(\omega') \right)\,d\omega'\\\\
&=\frac{1/2}{\omega+\omega_0}+i\frac{\pi}{2}\delta(\omega+\omega_0)-\frac{1/2}{\omega-\omega_0}-i \frac{\pi}{2}\delta(\omega-\omega ')  \\\\
&=-\frac{\omega_0}{\omega^2-\omega_0^2}+i\frac{\pi}{2} (\delta(\omega+\omega_0)-\delta(\omega-\omega_0))
\end{align}$$
as was to be shown!

In distribution, we have
$$\begin{align}
\lim_{\epsilon\to 0^+}\int_{-\infty }^\infty \sin(\omega_0 t)\Theta(t)e^{-i(\omega-i\epsilon)t}\,dt&=\lim_{\epsilon\to 0^+}\int_0^\infty \sin(\omega_0 t)e^{-i(\omega-i\epsilon)t}\,dt\\\\
&=\frac1{2i}\lim_{\epsilon\to 0^+}\int_0^\infty (e^{i(\omega_0-\omega-i\epsilon)t}-e^{i(-\omega_0-\omega-i\epsilon)t})\,dt\\\\
&=\frac1{2i}\lim_{\epsilon\to 0^+}\left(\frac{1}{\epsilon+i(\omega-\omega_0)}-\frac{1}{\epsilon+i(\omega+\omega_0)}\right)\\\\
&=-\lim_{\epsilon\to 0^+}\frac{\omega_0}{(\omega^2-\omega_0^2)-i(2\omega)\epsilon -\epsilon^2}\\\\
&=-\frac{\omega_0}{(\omega^2-\omega_0^2)-i\text{sgn}(\omega)0^+}
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
&\int_{-\infty}^{\infty}\Theta\pars{t}\sin\pars{\omega_{0}t}
\expo{-\ic\omega t}\,\dd t =
\int_{-\infty}^{\infty}\
\overbrace{\pars{-\int_{-\infty}^{\infty}
{\expo{-\ic\nu t} \over \nu + \ic 0^{+}}\,{\dd\nu \over 2\pi\ic}}}
^{\ds{\Theta\pars{t}}}\ \sin\pars{\omega_{0}t}\expo{-\ic\omega t}\,\dd t
\\[5mm] = &\
-\int_{-\infty}^{\infty}{1 \over \nu + \ic 0^{+}}
\bracks{%
{1 \over 2\ic}\int_{-\infty}^{\infty}
\expo{\ic\pars{-\nu + \omega_{0} - \omega}t}\,\dd t -
{1 \over 2\ic}\int_{-\infty}^{\infty}
\expo{\ic\pars{-\nu - \omega_{0} - \omega}t}\,\dd t}{\dd\nu \over 2\pi\ic}
\\[5mm] = &\
-\int_{-\infty}^{\infty}{1 \over \nu + \ic 0^{+}}
\bracks{%
-\pi\ic\,\delta\pars{-\nu + \omega_{0} - \omega} +
\pi\ic\,\delta\pars{-\nu - \omega_{0} - \omega}}{\dd\nu \over 2\pi\ic}
\\[5mm] = &\
{1 \over 2}\,{1 \over \omega_{0} - \omega + \ic 0^{+}} -
{1 \over 2}\,{1 \over -\omega_{0} - \omega - \ic 0^{+}}
\\[5mm] = &\
\bracks{{1 \over 2}\,\mrm{P.V.}{1 \over \omega_{0} - \omega} -
{1 \over 2}\,\ic\pi\,\delta\pars{\omega_{0} - \omega}} -
\bracks{{1 \over 2}\,\mrm{P.V.}{1 \over -\omega_{0} - \omega} +
{1 \over 2}\,\ic\pi\,\delta\pars{-\omega_{0} - \omega}}
\\[5mm] = &\
\bbx{-\,{1 \over 2}\,\ic\pi\bracks{\vphantom{\Large A}\delta\pars{\omega + \omega_{0}} -
\delta\pars{\omega - \omega_{0}}}}\quad
\mbox{because}\quad\mrm{P.V.}{1 \over \pm\omega_{0} - \omega} = 0
\end{align}

See
  Sokhotski–Plemelj Theorem.

