PV of $\int_{-\infty}^\infty \frac{x^n~ e^{-ax^2} dx}{(x-x_1)(x-x_2)},~ (n=0,1,2,3,...)$ $$\tag{1} I_n = PV \int_{-\infty}^\infty  \frac{x^n ~e^{-ax^2} dx}{(x-x_1)(x-x_2)},~ (n=0,1,2,3,...)~\text{with}~a>0~\text{and}~  x_1~x_2 \in \mathbb{R},$$
Is the principal value of $I_1$ calculable? 
Since $x_1$ and $x_2$ are on the path of integration on the real axis, I was thinking of indenting the contour, passing over them clockwise picking up 
$$\tag{2}-\pi i[\mathrm{Res}(f,x_1)+\mathrm{Res}(f,x_2)]$$
where $f$ is the integrand. To use the residue theorem I close the contour along a semi circle, $\Gamma_\infty$, in the upper half plane. But I suspect that the contribution of the integral 
$$\tag{3}\int_{\Gamma_\infty} dz ~f(z)$$ is non-zero. Since on $\Gamma_\infty$ we have $z = Re^{i\theta}$. 
Assuming this is correct, one finds
$$\tag{4}
I_n = i\pi [\mathrm{Res}(f,x_1)+\mathrm{Res}(f,x_2)]-\int_{\Gamma_\infty} dz ~f(z).
$$
Does all this make sense? And is there a simple estimate/computation of (3) above?
(Note: Mathematica can do the integral (1) in terms of incomplete Gamma functions, but the total expression seems to be indeterminate at integer values of $n$.)
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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Hint:

\begin{align}
&\mbox{With}\ \pars{~a > 0\,;\ x_{1},x_{2} \in \mathbb{R}\ \mbox{and}\
n \in \mathbb{N}_{\geq 0}~}, \,\,\,\mbox{note that}
\\
& I_{n} =
\mrm{P.V.}\int_{-\infty}^{\infty}{x^{n}\expo{-ax^{2}}
\over \pars{x - x_{1}}\pars{x - x_{2}}}\,\dd x =
{1 \over x_{2} - x_{1}}\,\sum_{k = 1}^{2}\pars{-1}^{k}\,
\bbox[#fee,5px]{\ds{\mrm{P.V.}
\int_{-\infty}^{\infty}{x^{n}\expo{-ax^{2}}
\over x - x_{k}}\,\dd x}}\label{1}\tag{1}
\end{align}

\begin{align}
&\bbox[#fee,5px]{\ds{\mrm{P.V.}
\int_{-\infty}^{\infty}{x^{n}\expo{-ax^{2}}
\over x - x_{k}}\,\dd x}} =
\mrm{P.V.}\int_{-\infty}^{\infty}{%
\pars{x + x_{k}}^{n}\expo{\large -a\pars{x + x_{k}}^{2}} \over x}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}{%
\pars{x + x_{k}}^{n}\expo{\large -a\pars{x + x_{k}}^{2}} -
\pars{-x + x_{k}}^{n}\expo{\large -a\pars{-x + x_{k}}^{2}}\over x}\,\dd x
\\[5mm] = &\
\int_{0}^{\infty}{%
\pars{x + x_{k}}^{n}\expo{\large -a\pars{x + x_{k}}^{2}} -
\pars{-1}^{n}\pars{x - x_{k}}^{n}\expo{\large -a\pars{x - x_{k}}^{2}} \over x}\,\dd x
\end{align}
