PV of $\int_{-\infty}^\infty \frac{x^n~ e^{-ax^2} dx}{(x-x_1)(x-x_2)},~ (n=0,1,2,3,...)$

Question

$$\tag{1} I_n = PV \int_{-\infty}^\infty \frac{x^n ~e^{-ax^2} dx}{(x-x_1)(x-x_2)},~ (n=0,1,2,3,...)~\text{with}~a>0~\text{and}~ x_1~x_2 \in \mathbb{R},$$ 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 $n$.)

Answer 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}