What is $\int_{0}^{\infty}e^{ixw}dw$? We know that
$$\int_{-\infty}^{\infty}e^{ixw}dw=\delta(x)$$
More details, see http://en.wikipedia.org/wiki/Dirac_delta_function
Now my question is 
$$\int_{0}^{\infty}e^{ixw}dw=?$$
Be grateful with any hints!
 A: The integral $$\int_0^\infty e^{i x w} dw$$ doesn't exist in the classical sense. However, viewed as the fourier transform of the Heaviside step distribution $$\int_0^\infty e^{i x w}dw=\int_{-\infty}^\infty H(w) e^{i x w} dw= \pi \delta(x)+i \text{p.v.} \left( \frac{1}{x} \right) ,$$
where $\text{p.v.}\left(\frac{1}{x} \right)$ is the principal value distribution. 
A: see http://www.artofproblemsolving.com/Forum/blog.php?u=152939&b=93736
But it seems no relation with the principal value distribution.
A: \begin{align}
\color{#ff0000}{\large\int_{0}^{\infty}{\rm e}^{{\rm i}xw}\,{\rm d}w}
&=
\int_{-\infty}^{\infty}{\rm e}^{{\rm i}xw}\,\Theta\left(w\right)\,{\rm d}w
=
\int_{-\infty}^{\infty}{\rm e}^{{\rm i}xw}
\left[-\int_{-\infty}^{\infty}{{\rm d}k \over 2\pi{\rm i}}\,
{{\rm e}^{-{\rm i}kw} \over k + {\rm i}0^{+}}\right]\,{\rm d}w
\\[3mm]&=
{\rm i}\int_{-\infty}^{\infty}{\rm d}k\,{1 \over k + {\rm i}0^{+}}
\int_{-\infty}^{\infty}{\rm e}^{{\rm i}w\left(x- k\right)}
\,{{\rm d}w \over 2\pi}
=
{\rm i}\int_{-\infty}^{\infty}{\rm d}k\,
{\delta\left(k - x\right) \over k + {\rm i}0^{+}}
=
{{\rm i} \over x + {\rm i}0^{+}}
\\[3mm]&=
{\rm i}\left[{\cal P}\,{1 \over x} - {\rm i}\pi\,\delta\left(x\right)\right]
=
\color{#ff0000}{\large{\cal P}\,{{\rm i} \over x} + \pi\,\delta\left(x\right)}
\end{align}
This is understood 'under the integral sign':
$$
\int_{-\infty}^{\infty}{\rm f}\left(x\right)
\left(\,\int_{0}^{\infty}{\rm e}^{{\rm i}xw}\,{\rm d}w\right)\,{\rm d}x
=
{\rm i}\,{\cal P}\int_{-\infty}^{\infty}{{\rm f}\left(x\right) \over x}\,{\rm d}x
+
\pi\,{\rm f}\left(0\right)
$$
