# 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 $n$.)

• Not the Residue, it should be semi-circles around that two poles, with angle goes from $\pi$ to $0$. – JRen Jul 5 at 20:00


# 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}

• The pv integral in red, if one indent the contour over the pole at $x_k$, it looks finite and the epsilon semi circle goes to zero (like $\epsilon ^n$). While the final version not so much as it looks like $1/\epsilon$ – Your Majesty Apr 18 '18 at 23:07
• Also at large radi $R$ it goes like $e^{-R^2}$ so it should die off fast enough right? – Your Majesty Apr 20 '18 at 13:29
• @JustAsk Yes, that's true. The exponential term provides the convergence as $x \to \infty$. – Felix Marin Apr 20 '18 at 18:17
• Actually I'm not so sure that the arc (at infinity) contribution is so trivial. On the arc, the Gaussian $e^{-z^2}$ becomes proportional to $e^{-i R^2 \sin(2\phi)}$ also. For instance, $\lim_{\epsilon\rightarrow 0 }\int_{-\infty}^\infty dx\, \frac{e^{-x^2}}{x-1+i \epsilon} = -\frac{\pi (erfi(1)+i)}{e}$ while jumping over the pole ($-i \pi \mathrm{Res}(x=1-i \epsilon)$) and taking the $\epsilon \rightarrow 0$ limit gives $-\frac{i \pi }{e}$. So something's missing. And I suspect the rest comes from the arc at infinity. Any thoughts? Note: Erfi(z) = erf(i z)/i – Your Majesty Apr 23 '18 at 0:08
• See for instance math.stackexchange.com/q/2749560/110467 – Your Majesty Apr 23 '18 at 1:23