The Fourier transform below gives a delta function
$$\int_{-\infty}^\infty\!dt ~e^{i\omega t}~=~2\pi\delta(\omega).$$
then what is the value if the integral range becomes $t=0\to\infty$?
$$\int_0^\infty \!dt ~e^{i\omega t} ~=~ \text{?}$$
I find it's related to principal value. What is principal value and how to derive the integral?
First, note that you are taking the Fourier transform of the Heaviside function.
By definition, we see that \begin{align} \langle \hat H, \varphi\rangle :=&\ \langle H, \hat \varphi\rangle = \int^\infty_0 \hat \varphi(\omega)\ d\omega= \int^\infty_0 \int^\infty_{-\infty} e^{i\omega t} \varphi(t)\ dt d\omega\\ =&\ \lim_{\alpha\rightarrow 0^+}\int^\infty_0 \int^\infty_{-\infty} e^{(it-\alpha) \omega} \varphi(t)\ dt d\omega\\ =&\lim_{\alpha \rightarrow 0^+} \int^\infty_{-\infty} \left(\int^\infty_0 e^{(it-\alpha)\omega}d\omega\ \right)\varphi(t) dt\\ =&\ \lim_{\alpha\rightarrow 0^+} \int^\infty_{-\infty} \frac{\varphi(t)}{\alpha-it}dt = \lim_{\alpha\rightarrow 0^+} \int^\infty_{-\infty} \left(\frac{\alpha}{\alpha^2+t^2}+\frac{it}{\alpha^2+t^2}\right)\varphi(t)\ dt. \end{align} Note that \begin{align} \lim_{\alpha\rightarrow 0^+}\int^\infty_{-\infty}\frac{\alpha\varphi(t)}{\alpha^2+t^2}\ dt= \lim_{\alpha\rightarrow 0^+} \int^\infty_{-\infty} \frac{\varphi(\alpha u)}{1+u^2}\ du =\varphi(0) \int^\infty_{-\infty}\frac{du}{1+u^2} = \pi\varphi(0) =\langle \pi\delta, \varphi \rangle \end{align} and \begin{align} \lim_{\alpha\rightarrow 0^+} i\int^\infty_{-\infty} \frac{t\varphi(t)}{\alpha^2+t^2}dt =&\ i\lim_{\alpha\rightarrow 0^+}\lim_{\varepsilon\rightarrow 0^+}\left(\int^\infty_\varepsilon + \int^{-\varepsilon}_{-\infty} \right)\frac{t\varphi(t)}{\alpha^2+t^2}\ dt\\ =&\ i\lim_{\varepsilon\rightarrow 0^+}\left(\int^\infty_\varepsilon + \int^{-\varepsilon}_{-\infty} \right)\frac{\varphi(t)}{t}\ dt\\ =&: \langle \text{P.V.}\left(\frac{i}{t}\right), \varphi\rangle. \end{align}
Since \begin{align} \langle \hat H, \varphi\rangle = \langle \pi\delta+\text{P.V.}\left(\frac{i}{t}\right), \varphi\rangle \end{align} for any smooth $\varphi$, then we see that \begin{align} \hat H = \pi\delta + \text{P.V.}\left(\frac{i}{t}\right). \end{align}
Just to reiterate. Observe that $\displaystyle \text{P.V.}\left(\frac{1}{t}\right)$ acts on functions in the following manner \begin{align} \langle \displaystyle \text{P.V.}\left(\frac{1}{t}\right), \varphi\rangle = \lim_{\varepsilon\rightarrow 0^+}\left(\int^\infty_\varepsilon + \int^{-\varepsilon}_{-\infty} \right)\frac{\varphi(t)}{t}\ dt \end{align} just like how \begin{align} \langle \delta, \varphi\rangle = \varphi(0). \end{align}