The Heaviside function $H$ (seen as a tempered Distribution) is the weak limit
of the function
$$
H(x) = \mathcal{S}'-\lim_{\epsilon \rightarrow 0} H(x) e^{-\epsilon x}.
$$
To see this let $\varphi$ be an arbitrary function of rapid decay. Split
$$
\lim_{\epsilon \rightarrow 0} \int_0^{\infty} dx e^{-\epsilon x} \varphi(x) \,
$$
the integral into positive and negative part and use monotone convergence to
take the limit under the integral.
The Fourier transform of a tempered distribution is continuous and coincides
with the usual Fourier transform on, well anything, but lets say $L^{2}$. Thus
$$
\hat{H} = \mathcal{S}'-\lim_{\epsilon \rightarrow 0}
\int_{0}^{\infty} dx e^{i\,k\,x -\epsilon x}
= \mathcal{S}'-\lim_{\epsilon \rightarrow 0} \frac{i}{k + i \epsilon}
$$
Now for all for all test functions $\varphi$ we have
$$
\hat{H}[\varphi] = \lim_{\epsilon \rightarrow 0}
\int_{-\infty}^{+\infty} dk \frac{i}{k + i \epsilon} \varphi(k)
$$
Let $\eta > 0$ be some small fixed constant and split the Integral
$$
\hat{H}[\varphi] = \left\{
\int_{|k| > \eta} dk + \int_{|k| < \eta} dk
\right\} \frac{i}{k + i \epsilon} \varphi(k)
$$
In the first Integral you can take the limit under the integral, when letting
$\eta \rightarrow 0$ this becomes the principal value term. In the second
Integral write $\varphi(k) = \varphi(0) + O(k)$. The first Term gives the
$\delta$ function:
$$
\int_{|k| < \eta} dk \frac{i}{k + i \epsilon} \varphi(0)
= \varphi(0) (\log (i \epsilon + \eta) - \log(i \epsilon - \eta))
\rightarrow \varphi(0) (\log(\eta) - \log(-\eta)) = -i \pi.
$$
The second term vanishes (left as an exercise :).
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
$$\hat{H}(\xi) = -i\text{PV}\left(\frac{1}{\xi}\right) + \pi\delta$$
Best Answer
Note that
$$\frac{1}{\alpha+i\omega}=\frac{\alpha}{\alpha^2+\omega^2}-\frac{i\omega}{\alpha^2+\omega^2}\tag{1}$$
Taking the limit $\alpha\rightarrow 0^{+}$ of (1) gives
$$\lim_{\alpha\rightarrow 0^+}\frac{1}{\alpha+i\omega}=\lim_{\alpha\rightarrow 0^+}\frac{\alpha}{\alpha^2+\omega^2}+\frac{1}{i\omega}\tag{2}$$
The real part of (2) is a scaled nascent Dirac delta function:
$$\lim_{\alpha\rightarrow 0^+}\frac{\alpha}{\alpha^2+\omega^2}=\pi\delta(\omega)\tag{3}$$
Combining (1), (2) and (3) gives
$$\lim_{\alpha\rightarrow 0^+}\frac{1}{\alpha+i\omega}=\pi\delta(\omega)+\frac{1}{i\omega}\tag{4}$$
So the two expressions in your question are indeed identical if you consider the limit $\alpha\rightarrow 0^+$.