Continuous extension of $\int_\mathbb{R} dt\, e^{-t^2}/(t-z)$ from $\operatorname{Im} z < 0$ onto $\mathbb R$ I am asked to show that the continuous extension of
$$
F(z) = \int_{-\infty}^{\infty} dt\, \frac{e^{-t^2}}{t-z}, \quad \operatorname{Im} z < 0
$$
onto $\mathbb R$ is given by
$$
\int_{-\infty}^{\infty} dt\, \frac{e^{-t^2}}{t-z} \stackrel{z \to x}{\longrightarrow} \mathcal P \int_{-\infty}^{\infty} dt \frac{e^{-t^2}}{t-x}+i \pi e^{-x^2}, \quad x \in \mathbb R,
$$
where $\mathcal P \int$ denotes a principal value integral. I am unsure of how to approach this.
I can start with the pole at $t=z$ and deform the contour below the real axis so that the point $z$ is above. Then I end up with two terms, one is an integral over a semicircle and the other is the residue of the original function at $t=z$.
The semicircle is just $F(z)$ along $x$, so $\int dt\, e^{t^2}/(t-x)$, and then the second is
$$
2\pi i \operatorname{Res}[e^{-t^2}/(t-z),z=x] = 2\pi i e^{-x^2},
$$
which doesn't equal what I am supposed to show.
 A: The stated result is incorrect. If $z$ is approaching the real axis from $\DeclareMathOperator{\im}{Im} \im z < 0$ then the correct goal is to show that for any fixed $x \in \mathbb R$ we have
$$
\lim_{\substack{z \to x \\ \im z < 0}} \int_{-\infty}^{\infty} \frac{e^{-t^2}}{t-z}\,dt = \mathcal P \int_{-\infty}^{\infty} \frac{e^{-t^2}}{t-x}\,dt - i\pi e^{-x^2}. \tag{$*$}
$$
The main idea behind the proof is that the behavior of the integral as $z \to x$ is controlled by the values of $t$ for which $|t-x| \ll |z-x|$. We will want a little breathing room, so we will instead focus on the slightly larger interval $|t-x| < \sqrt{|z-x|}$. For convenience of notation, let's define
$$
s(z,x) = \sqrt{|z-x|}.
$$
Unwinding the integral expressions in $(*)$ yields
$$
\mathcal P \int_{-\infty}^{\infty} \frac{e^{-t^2}}{t-x}\,dt = \int_{|t-x| < s(z,x)} \frac{e^{-t^2} - e^{-x^2}}{t-x}\,dt + \int_{s(z,x) < |t-x|} \frac{e^{-t^2}}{t-x}\,dt, \tag{1}
$$
and, assuming $\im z < 0$,
$$
\begin{align}
\int_{-\infty}^{\infty} \frac{e^{-t^2}}{t-z}\,dt = &\int_{|t-x| < s(z,x)} \frac{e^{-t^2} - e^{-x^2}}{t-z}\,dt + \int_{s(z,x) < |t-x|}  \frac{e^{-t^2}}{t-z}\,dt - i\pi e^{-x^2} \\
&\qquad - e^{-x^2} \int_{s(z,x) < |t-x|} \left( \frac{1}{t-z} - \frac{1}{t-x} \right) \,dt. \tag{2}
\end{align}
$$
In order to prove $(*)$ we just need to show the following:
$$
\lim_{\substack{z \to x \\ \im z < 0}} \int_{|t-x| < s(z,x)} \left(e^{-t^2} - e^{-x^2}\right) \left(\frac{1}{t-z} - \frac{1}{t-x}\right)dt = 0, \tag{3}
$$
$$
\lim_{\substack{z \to x \\ \im z < 0}} \int_{s(z,x) < |t-x|} e^{-t^2} \left(\frac{1}{t-z} - \frac{1}{t-x}\right)dt = 0, \tag{4}
$$
and
$$
\lim_{\substack{z \to x \\ \im z < 0}} \int_{s(z,x) < |t-x|} \left( \frac{1}{t-z} - \frac{1}{t-x} \right) dt = 0. \tag{5}
$$
Limits $(4)$ and $(5)$ can be shown the same way. If $f(t) = 1$ or $f(t) = e^{-t^2}$ then
$$
\begin{align}
\left| \int_{s(z,x) < |t-x|} f(t) \left( \frac{1}{t-z} - \frac{1}{t-x} \right) dt \right| &\leq \int_{s(z,x) < |t-x|} \left| \frac{1}{t-z} - \frac{1}{t-x} \right| dt \\
&= |z-x| \int_{s(z,x) < |t-x|} \left| \frac{1}{(t-z)(t-x)} \right| dt \\
&= |z-x| \int_{s(z,x) < |t-x|} |t-x|^{-2} \left| \frac{t-x}{t-z} \right| dt.
\end{align}
$$
If $0 < |z-x| < 1/4$ and $\sqrt{|z-x|} < |t-x|$ then
$$
|t-z| \geq |t-x| - |z-x| > \sqrt{|z-x|} - |z-x| > \frac{1}{2} \sqrt{|z-x|}, \tag{6}
$$
so that
$$
\left|\frac{z-x}{t-x}\right| < 2\sqrt{|z-x|} < 1.
$$
Then from $|t-x| \leq |t-z| + |z-x|$ we get
$$
\left| \frac{t-x}{t-z} \right| \leq 1 + \left| \frac{z-x}{t-z} \right| < 2.
$$
Thus if $|z-x| < 1/4$ we have
$$
\begin{align}
\left| \int_{s(z,x) < |t-x|} f(t) \left( \frac{1}{t-z} - \frac{1}{t-x} \right) dt \right| &< 2 |z-x| \int_{s(z,x) < |t-x|} |t-x|^{-2}\, dt \\
&= 4\sqrt{|z-x|},
\end{align}
$$
from which limits $(4)$ and $(5)$ follow.
For limit $(3)$ we calculate
$$
\begin{align}
&\left| \int_{|t-x| < s(z,x)} \left(e^{-t^2} - e^{-x^2}\right) \left(\frac{1}{t-z} - \frac{1}{t-x}\right)dt \right| \\
&\qquad = |z-x| \int_{|t-x| < s(z,x)} \left| \frac{e^{-t^2} - e^{-x^2}}{t-x} \right| \frac{1}{|t-z|} \,dt \\
&\qquad < 2\sqrt{|z-x|} \int_{|t-x| < s(z,x)} \left| \frac{e^{-t^2} - e^{-x^2}}{t-x} \right| dt \\
&\qquad < 2\sqrt{|z-x|} \int_{|t-x| < 1} \left| \frac{e^{-t^2} - e^{-x^2}}{t-x} \right| dt,
\end{align}
$$
for $|z-x| < 1$, where we used inequality $(6)$ in the second line.  This concludes the proof of $(3)$, and $(*)$ follows.

Proof of (1):
$$
\begin{align}
\mathcal P \int_{-\infty}^{\infty} \frac{e^{-t^2}}{t-x}\,dt &= \lim_{\epsilon \to 0^+} \int_{\epsilon < |t-x|} \frac{e^{-t^2}}{t-x}\,dt \\
&= \lim_{\epsilon \to 0^+} \int_{\epsilon < |t-x| < s(z,x)} \frac{e^{-t^2}}{t-x}\,dt + \int_{s(z,x) < |t-x|} \frac{e^{-t^2}}{t-x}\,dt \\
&= \lim_{\epsilon \to 0^+} \left\{ \int_{\epsilon < |t-x| < s(z,x)} \frac{e^{-t^2} - e^{-x^2}}{t-x}\,dt + e^{-x^2} \int_{\epsilon < |t-x| < s(z,x)} \frac{1}{t-x}\,dt \right\} \\
&\qquad\quad + \int_{s(z,x) < |t-x|} \frac{e^{-t^2}}{t-x}\,dt \\
&= \int_{|t-x| < s(z,x)} \frac{e^{-t^2} - e^{-x^2}}{t-x}\,dt + \int_{s(z,x) < |t-x|} \frac{e^{-t^2}}{t-x}\,dt
\end{align}
$$
since
$$
\int_{\epsilon < |t-x| < s(z,x)} \frac{1}{t-x}\,dt = 0
$$
for all $\epsilon > 0$ and the limit
$$
\lim_{\epsilon \to 0^+} \int_{\epsilon < |t-x| < s(z,x)} \frac{e^{-t^2} - e^{-x^2}}{t-x}\,dt = \int_{|t-x| < s(z,x)} \frac{e^{-t^2} - e^{-x^2}}{t-x}\,dt
$$
exists.
Proof of (2):
For $\im z \neq 0$ we have
$$
\begin{align*}
\int_{-\infty}^{\infty} \frac{e^{-t^2}}{t-z}\,dt &= \left[ \int_{|t-x| < s(z,x)} + \int_{s(z,x) < |t-x|} \right] \frac{e^{-t^2}}{t-z}\,dt \\
&= \int_{|t-x| < s(z,x)} \frac{e^{-t^2} - e^{-x^2}}{t-z}\,dt + e^{-x^2} \int_{|t-x| < s(z,x)} \frac{1}{t-z}\,dt \\
&\qquad \quad + \int_{s(z,x) < |t-x|}  \frac{e^{-t^2}}{t-z}\,dt. \tag{7}
\end{align*}
$$
Now
$$
\begin{align}
\int_{|t-x| < s(z,x)} \frac{1}{t-z}\,dt &= \int_{|t-x| < R} \frac{1}{t-z}\,dt - \int_{s(z,x) < |t-x| < R} \frac{1}{t-z}\,dt \\
&= \int_{|t-x| < R} \frac{1}{t-z}\,dt - \int_{s(z,x) < |t-x| < R} \left( \frac{1}{t-z} - \frac{1}{t-x} \right) dt \\
&= \int_{|u| < R} \frac{1}{u+x-z}\,du - \int_{s(z,x) < |t-x| < R} \left( \frac{1}{t-z} - \frac{1}{t-x} \right) dt
\end{align}
$$
for $R > s(z,x)$. Sending $R \to \infty$ we get
$$
\begin{align}
\int_{|t-x| < s(z,x)} \frac{1}{t-z}\,dt &= \mathcal P \int_{-\infty}^{\infty} \frac{1}{u+x-z}\,du - \int_{s(z,x) < |t-x|} \left( \frac{1}{t-z} - \frac{1}{t-x} \right) dt \\
&= -i\pi - \int_{s(z,x) < |t-x|} \left( \frac{1}{t-z} - \frac{1}{t-x} \right) dt, \tag{8}
\end{align}
$$
where we have used the facts (i): $\im(z-x) < 0$ and (ii):
$$
\mathcal P \int_{-\infty}^{\infty} \frac{1}{u-\zeta}\,du =
       \begin{cases}
           i\pi & \text{if } \im \zeta > 0, \\
           -i\pi & \text{if } \im \zeta < 0.
       \end{cases}
$$
Equation $(2)$ follows from inserting $(8)$ into $(7)$.

Remarks.
One important aspect of this approach is that it is not specific to the integrand $e^{-t^2}$. It can be used to show that
$$
\lim_{\substack{z \to x \\ \im z < 0}} \int_{-\infty}^{\infty} \frac{\varphi(t)}{t-z}\,dt = \mathcal P \int_{-\infty}^{\infty} \frac{\varphi(t)}{t-x}\,dt - i\pi \varphi(x)
$$
for any bounded, locally Hölder continuous function $\varphi \colon \mathbb R \to \mathbb C$ for which
$$
\int_{\delta < |t|} \frac{\varphi(t)}{t}\,dt \qquad \text{exists} \tag{$**$}
$$
for $\delta > 0$. The proof can be modified to remove the need for condition $(**)$.
This result, together with the analogous limit for $\im z > 0$, make up the Sokhotski-Plemelj theorem on the real line.
