Heaviside step function fourier transform and principal values I found the following answer on Math.SE:
Fourier transform of unit step?
However, it is still not clear to me and maybe somebody could explain it clearer.
Problem
I have the following in my notes of a theoretical physics course:
$$
\hat{\Theta}(\omega) = \int_{-\infty}^\infty \Theta (t) e^{i\omega t} \mathrm{d} t 
= \lim_{\varepsilon \to 0} \int_0^\infty e^{i\omega t - \varepsilon t} \mathrm{d}t
= \pi \delta(\omega) + \mathrm{P} \frac{i}{\omega},
$$
where the $\mathrm{P}$ denotes the Cauchy's principal value. 
Question
I understand why I get a delta function in this computation, but I have no idea why I have $\mathrm{P} \frac{i}{\omega}$ instead of just $\frac{i}{\omega}$ in the resulting expression.
 A: $$
\begin{align}
\lim_{\varepsilon \to 0} \int_0^\infty e^{i\omega t - \varepsilon t} \mathrm{d}t
&=\lim_{\varepsilon \to 0} \frac{1}{\varepsilon - i\omega}
\\ 
&=\lim_{\varepsilon \to 0} \frac{1}{\varepsilon + \omega^2/\varepsilon} + \lim_{\varepsilon \to 0} \frac{\omega i}{\varepsilon^2 + \omega^2} 
\\
&=\pi \delta(\omega) + \mathrm{P} \frac{i}{\omega}
\end{align}
$$
where the last step uses the limiting representations of the delta function and the Cauchy principal value.
A: First let us prove that for any particular interval $[-a,a]$, $\def\d{\operatorname{d}\!{}}\displaystyle
\int_{-\infty}^{+\infty}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x$ converges.
As in
\begin{split}
\lim_{c\to+\infty}\int_{0}^{c}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x &= \int_{0}^{+\infty}f(0)\frac{e^{ia\xi}-e^{-ia\xi}}{i\mkern1mu\xi}\d\xi \\
& \quad {} +\lim_{c\to+\infty}\int_a^b\frac{f(x)-f(0)}{ix}\bigl(1-e^{-icx}\bigr)\d{}x,
\end{split}
both of the RHS converge.
For any $f\in\mathcal{S}(\mathbb{R})$, it's obvious that
\begin{align}
\int_{0}^{+\infty}\d\xi\int_{-\infty}^{+\infty}f(x)e^{-ix\xi}\d{}x &= \lim_{a\to+\infty}\int_{0}^{+\infty}\d\xi\int_{-a}^{a}f(x)e^{-ix\xi}\d{}x\\
&= \int_0^{+\infty}f(0)\frac{2\sin a\xi}{\xi}\d\xi + \int_{-\infty}^{+\infty}\frac{f(x)-f(0)}{ix}\d{}x\\
&= \pi{}f(0) + \int_0^{+\infty}\frac{f(x)-f(-x)}{ix}\d{}x
\end{align}

hence

\begin{align}
\langle\hat{H},f\,\rangle &= \langle{H,\hat{f}}\rangle\\
&= \int_{0}^{+\infty}\hat{f}(\xi)\,\mathrm d\xi\\
&= \int_{0}^{+\infty}\mathrm d\xi\int_{-\infty}^{+\infty}f(x)e^{-ix\xi}\,\mathrm dx\\
&= \pi{}f(0) + \int_0^{+\infty}\frac{f(x)-f(-x)}{ix}\d{}x\\
&= \pi{}f(0) + \lim_{x_0\to0^+}\int_{|x|\geqslant x_0}f(x)\frac{1}{ix}\d{}x
\end{align}
we obtain
$$\langle \hat{H},f\,\rangle=\operatorname {V.\!P.}\biggl(\frac{1}{ix}\biggr)[f\,]+\pi f(0)$$
So
$$\hat{H}=\operatorname {V.\!P.}\biggl(\frac{1}{ix}\biggr)+\pi\delta$$
A: While chaohuang's answer gave you basically everything you needed, I'd like to elaborate a little on the last part, and on generalized functions (a.k.a. distributions) in general; I apologize for the unavoidable pun.
I realize that in physics, you might treat generalized functions, like the delta distribution, in a "classical" way similar to regular functions. The truth is, however, that distributions are really only defined in the context of how they act on a set of (nicely behaved) test functions. These test functions are usually either Schwartz space functions, or compactly supported functions. Occasionally we also consider the (much larger) test space of $C^\infty$ functions. So the reason why $\hat{H}(\xi)$ involves $\text{PV}\left(\dfrac{1}{\xi}\right)$, as opposed to just $\dfrac{1}{\xi}$, is because of how it integrates against a test function in Schwartz space.
To make 
\begin{align}
\hat{H}(\xi) &= \int_{-\infty}^{\infty}H(x)e^{-ix\xi}\;dx\\
&= \lim_{\epsilon\to0^+}\;\int_{0}^{\infty}e^{-\epsilon x}e^{-ix\xi}\;dx\\
&= \lim_{\epsilon\to0^+}\;(\epsilon+i\xi)^{-1}\\
&= \lim_{\epsilon\to0^+}\;-i(\xi-i\epsilon)^{-1}\\
&= -i(\xi-i0)^{-1}
\end{align}
rigorous, you need to see how it acts as a linear functional on Schwartz space:
\begin{align}
\hat{H}[\varphi] &= \int_{-\infty}^{\infty}\hat{H}(\xi)\varphi(\xi)\;d\xi\\
&= \lim_{\epsilon\to0^+}\;\int_{-\infty}^{\infty}-i(\xi-i\epsilon)^{-1}\varphi(\xi)\;d\xi\\
&= -i\lim_{\epsilon\to0^+}\;\int_{-\infty}^{\infty}\left(\frac{\xi+i\epsilon}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi\\
&= -i\left(\lim_{\epsilon\to0^+}\;\int_{-\infty}^{\infty}\left(\frac{\xi}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi + i\lim_{\epsilon\to0^+}\;\int_{-\infty}^{\infty}\left(\frac{\epsilon}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi\right)\\
&=-i\lim_{\epsilon\to0^+}A(\epsilon)+\lim_{\epsilon\to0^+}B(\epsilon)
\end{align}
where
\begin{align}
A(\epsilon) &= \int_{-\infty}^{\infty}\left(\frac{\xi}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi\\
&= \int_{-\infty}^{\infty}\frac{d}{d\xi}\left(\frac{1}{2}\ln(\xi^2+\epsilon^2)\right)\varphi(\xi)\;d\xi\\
&= -\frac{1}{2}\int_{-\infty}^{\infty}\ln(\xi^2+\epsilon^2)\varphi'(\xi)\;d\xi\\
\end{align}
and
\begin{align}
B(\epsilon) &= \int_{-\infty}^{\infty}\left(\frac{\epsilon}{\xi^2+\epsilon^2}\right)\varphi(\xi)\;d\xi\\
&= \int_{-\infty}^{\infty}\frac{d}{d\xi}\left(\arctan\left(\frac{\xi}{\epsilon}\right)\right)\varphi(\xi)\;d\xi\\
&= -\int_{-\infty}^{\infty}\arctan\left(\frac{\xi}{\epsilon}\right)\varphi'(\xi)\;d\xi\\
\end{align}
with
$$\lim_{\epsilon\to0^+}\;A(\epsilon) = -\int_{-\infty}^{\infty}\ln(|\xi|)\varphi'(\xi)\;d\xi$$
and
$$\lim_{\epsilon\to0^+}\;B(\epsilon) = -\frac{\pi}{2}\int_{-\infty}^{\infty}\text{sgn}(\xi) \varphi'(\xi)\;d\xi$$
by various Lebesgue integration theorems, and the fact that $\varphi$ and its derivatives are rapidly decaying. In particular, whereas the area under $1/\xi$ is infinite to either side of the singularity (making the integration of $\varphi/\xi$ ill-defined around $\xi=0$), the function $\ln|\xi|$ has finite integral near $\xi=0$, and so $\lim_{\epsilon\to0^+}\;A(\epsilon)$ is well-defined.
An easy computation then gives us that
$$\lim_{\epsilon\to0^+}\;A(\epsilon) = \lim_{\epsilon\to0^+}\; \int_{\mathbb{R}\setminus(-\epsilon,\epsilon)}\frac{1}{\xi}\varphi(\xi)\;d\xi =: \text{PV}\left(\frac{1}{\xi}\right)[\varphi]$$
by definition of the Cauchy principal value distribution. Similarly, we find that
$$\lim_{\epsilon\to0^+}\;B(\epsilon) = 2\left(\frac{\pi}{2}\varphi(0)\right):= \pi\delta[\varphi]$$
by definition of the delta distribution.
So in conclusion, we have finally arrived at the fact that
$$\hat{H}(\xi) = -i\text{PV}\left(\frac{1}{\xi}\right) + \pi\delta$$
A: Because in distribution theory (implied by the $\delta$ distribution) we want the functions to be locally integrable while $\omega \mapsto \frac 1{\omega}$ isn't.
So $\displaystyle\int_a^b \frac {dx}x$ is not well defined while $\quad P\displaystyle\int_a^b \frac {dx}x$ will be defined for different signs of $a$ and $b$.
