How to prove $P \int_{-\infty}^\infty \frac{e^{-ax^2-bx}}{x} dx = -i\pi$? For $a,b>0$, Mathematica says that
$$P \int_{-\infty}^\infty \frac{e^{-ax^2-bx}}{x} dx = -i\pi,$$
where $P$ denotes the principal-value integral. How can I derive the above?
Due to the factor $e^{-ax^2}$, the integrand grows exponentialy as $x \to \pm i \infty$. Hence, contour method with respect to the semicircle contour is not applicable.
 A: My Mathematica 13 claims the answer is $i\pi$:

This is clearly wrong, though, since the principal value of $\int_{-\infty}^{\infty} f(x) \, \mathrm{d}x$ of a real-valued function $f$ on $\mathbb{R}$ cannot be imaginary.
1. What is the correct answer? We have
\begin{align*}
\mathrm{PV}\!\int_{-\infty}^{\infty} \frac{e^{-ax^2-bx}}{x} \, \mathrm{d}x
&= \frac{1}{2} \int_{-\infty}^{\infty} \frac{e^{-ax^2-bx} - e^{-ax^2+bx}}{x} \, \mathrm{d}x \\
&= -\frac{1}{2} \int_{-\infty}^{\infty} e^{-ax^2} \left( \int_{-b}^{b} e^{xs} \, \mathrm{d}s \right) \, \mathrm{d}x \\
&= -\frac{1}{2} \int_{-b}^{b} \left( \int_{-\infty}^{\infty} e^{-ax^2+sx} \, \mathrm{d}x \right) \, \mathrm{d}s \\
&= -\frac{1}{2} \int_{-b}^{b} \left( \sqrt{\frac{\pi}{a}} e^{s^2/4a}  \right) \, \mathrm{d}s\\
&= -\pi \operatorname{erfi}\left(\frac{b}{2 \sqrt{a}}\right).
\end{align*}
Here, $\operatorname{erfi}(\cdot)$ stands for the imaginary error function.
Addendum. The key idea in this solution is in fact an integral version of Feynman's trick. This means that we can adapt this logic to give a solution using Feynman's trick. Indeed, set
$$ f(b) = \mathrm{PV}\!\int_{-\infty}^{\infty} \frac{e^{-ax^2-bx}}{x} \, \mathrm{d}x. $$
Then $f(0) = 0$ since the integrand is an odd function. Moreover,
$$ f'(b)
= - \mathrm{PV}\!\int_{-\infty}^{\infty} e^{-ax^2-bx} \, \mathrm{d}x
= - \sqrt{\frac{\pi}{a}} e^{b^2/4a}. $$
Therefore,
$$ f(b)
= f(0) + \int_{0}^{b} f'(s) \, \mathrm{d}s
= - \int_{0}^{b} \left( \sqrt{\frac{\pi}{a}} e^{s^2/4a}  \right) \, \mathrm{d}s
= -\pi \operatorname{erfi}\left(\frac{b}{2 \sqrt{a}}\right).
$$
2. What might have gone wrong? I suspect that what Mathematica computed is
\begin{align*}
\int_{-\infty}^{\infty} \frac{e^{-ax^2-bx}}{x} \, \mathrm{d}x
&= \lim_{\substack{\varepsilon & \to 0^+ \\ R & \to \infty}} \left( \int_{-R}^{-\varepsilon} \frac{e^{-az^2-bz}}{z} \, \mathrm{d}z + \int_{\varepsilon}^{R} \frac{e^{-az^2-bz}}{z} \, \mathrm{d}z \right) \\
&= \lim_{\substack{\varepsilon & \to 0^+ \\ R & \to \infty}} \left( \int_{\gamma_{\varepsilon}} \frac{e^{-az^2-bz}}{z} \, \mathrm{d}z - \int_{\gamma_{R}} \frac{e^{-az^2-bz}}{z} \, \mathrm{d}z \right),
\end{align*}
where $\gamma_r$ denotes the counter-clockwise oriented, upper semicircular contour centered at $0$. Then
$$ \lim_{\varepsilon \to 0^+} \int_{\gamma_{\varepsilon}} \frac{e^{-az^2-bz}}{z} \, \mathrm{d}z
= i\pi \, \underset{z=0}{\mathrm{Res}} \, \frac{e^{-az^2-bz}}{z}
= i\pi, $$
but I suspect that Mathematica incorrectly assumed
$$ \color{red}{ \int_{\gamma_{R}} \frac{e^{-az^2-bz}}{z} \, \mathrm{d}z \quad \xrightarrow[?]{R\to\infty} \quad 0 }, $$
which is a fairly bold claim (and is indeed false in light of the above computation.)
A: The principal value integral can be calculated using the method of residues. The integral can be written as the sum of the residues at the poles of the integrand inside the contour. To find the poles, we need to find the zeroes of the denominator of the integrand, which is x^2 + (b/a). These zeroes occur at x = ±i√(b/a). Since the integral is symmetric about the x-axis, we only need to consider one pole, say x = i√(b/a). To calculate the residue, we need to evaluate the limit of (x - i√(b/a))*e^(-ax^2 - bx) as x approaches i√(b/a). This limit can be calculated using L'Hopital's rule. Finally, we need to multiply the residue by -iπ to find the principal value integral.
