Integration of $\int_{0}^{\infty} e^{-itw} dt$ We know that $\int_{-\infty}^{\infty}e^{-itw}dt=2\pi\delta(w)$, but how to calculate the half integration $\int_{0}^{\infty}e^{-itw}dt$?
 A: The integral $\int_{-\infty}^\infty e^{-i\omega t}\,dt$ diverges as either a Lebesgue integral or improper Riemann integral.  However, for every "suitable" (smooth with compact support) test function $f(\omega)$, we have
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\left(\int_{-L}^L e^{-i\omega t}\,dt\right)\,d\omega&=\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\,\frac{2\sin(\omega L)}{\omega}\,d\omega\\\\
&=2\pi f(0)
\end{align}$$
Therefore, we can interpret the symbol "$\int_{-\infty}^\infty e^{-i\omega t}\,dt$" to be the regularization of the Dirac Delta
$$\lim_{L\to \infty}\frac{1}{2\pi}\int_{-L}^L e^{-i\omega t}\,dt\sim  \delta(\omega)$$
which is interpreted to mean
$$\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\left(\int_{-L}^L e^{-i\omega t}\,dt\right)\,d\omega =2\pi f(0)$$
Now, applying the regularization $\lim_{L\to \infty}\int_{0}^L e^{-i\omega t}\,dt$ to a suitable test function reveals
$$\begin{align}
\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\left(\int_{0}^L e^{-i\omega t}\,dt\right)\,d\omega&=\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\,\left(\frac{\sin(\omega L)}{\omega}-i\,\frac{1-\cos(\omega L)}{\omega}\right)\,d\omega\\\\
&=\pi f(0)-i\lim_{L\to \infty}\int_{-\infty}^\infty f(\omega)\,\frac{1-\cos(\omega L)}{\omega}\,d\omega \tag 1\\\\
&=\pi f(0)-i\text{PV}\left(\int_{-\infty}^\infty \frac{f(\omega)}{\omega}\,d\omega\right)
\end{align}$$
where $\text{PV}$ denotes that Cauchy-Principal Value. Note that we used the Riemann-Lebesgue Lemma assuming that $\frac{f(\omega)}{\omega}$ is smooth and of compact support for  $|\omega| \ge\epsilon$ for all $\epsilon >0$ to evaluate the integral on the right-hand side of $(1)$.
Therefore, we can write
$$\lim_{L\to \infty}\int_{0}^L e^{-i\omega t}\,dt\sim \pi \delta(\omega)+ \text{PV}\left(\frac{-i}{\omega}\right)$$

A: The integral 
$$
\int_{0}^{\infty}e^{-itw}dt
$$ 
doesn't exist in the normal functional sense. So it's pretty much the same as in the case of $\displaystyle\int_{-\infty}^{\infty}e^{-itw}dt=2\pi\delta(w)$, where you already (maybe implicitly) applied the distributional interpretation. 
However, where you can see 
$$
\int_{-\infty}^{\infty}e^{-itw}dt=2\pi\delta(w)
$$
as the Fourier transform of the constant function $f(x)=1$, you can see 
$$
\int_{0}^{\infty}e^{-itw}dt
$$
as the Fourier transform of the Heaviside function $H(x)=\begin{cases}1,x>0\\0,x<0\end{cases}$. As a result you'll get
$$
\hat{H}(w)=\int_{0}^{\infty}e^{-itw}dt=\pi\delta(w)+\frac 1{i w}
$$
for a detailed derivation and further references I recommend you for example this and of course this answer.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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Hereafter,
  $\ds{\Theta\pars{t} =
\int_{-\infty}^{\infty}{\expo{\ic\nu t} \over \nu - \ic 0^{+}}
\,{\dd\nu \over 2\pi\ic}}$ is the Heaviside Step Function.

\begin{align}
\color{#f00}{\int_{0}^{\infty}\expo{-\ic wt}\,\dd t} & =
\int_{-\infty}^{\infty}\Theta\pars{t}\expo{-\ic wt}\,\dd t =
\int_{-\infty}^{\infty}\bracks{\int_{-\infty}^{\infty}
{\expo{\ic\nu t} \over \nu - \ic 0^{+}}\,{\dd\nu \over 2\pi\ic}}
\expo{-\ic wt}\,\dd t
\\[5mm] & =
-\ic\int_{-\infty}^{\infty}{1 \over \nu - \ic 0^{+}}
\bracks{\int_{-\infty}^{\infty}
\expo{\ic\pars{\nu - w}t}\,{\dd t \over 2\pi}}\,\dd\nu =
-\ic\int_{-\infty}^{\infty}{\delta\pars{\nu - w} \over \nu - \ic 0^{+}}\,\dd\nu
\\[5mm] & =
-\ic\,{1 \over w - \ic 0^{+}} =
\color{#f00}{-\ic\,\mrm{P.V.}\pars{1 \over w} + \pi\delta\pars{w}}
\end{align}
The result must be understood 'under the integral sign'. Namely,
$$
\int_{-\infty}^{\infty}\mrm{f}\pars{w}\int_{0}^{\infty}\expo{-\ic wt}
\,\dd t\,\dd w =
-\ic\,\mrm{P.V.}\int_{-\infty}^{\infty}{\mrm{f}\pars{w} \over w}\,\dd w +
\pi\mrm{f}\pars{0}
$$
A: Hint: maybe you can show symmetry- think of euler's identity
