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.
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.