Stuck on (simple looking!) discontinuous double integral with gaussian weight and double pole

Question

The question is what is the analytic answer to the limit of the difference of integrals

$(I_{+\epsilon}-I_{-\epsilon})|_{\epsilon \rightarrow 0} =\int_{0}^{\infty}dx \int_{0}^{\infty}dy\left(\frac{e^{-\frac12 x^2-\frac12(y+i \epsilon)^2}}{(x-i (y+i \epsilon))^2}-\frac{e^{-\frac12 x^2-\frac12(y-i \epsilon)^2}}{(x-i (y-i \epsilon))^2}\right)|_{\epsilon \rightarrow 0}$.

This comes from considering the integral

$I_0= \int_{0}^{\infty}dx \int_{0}^{\infty}dy\frac{e^{-\frac12 x^2-\frac12y^2}}{(x-i y)^2}$.

By itself this is not a convergent integral, and playing around with it, it seemed to me that the divergence is logarithmic.

However, by shifting the y variable above and below the real axis there, one can find a slight alteration which should be convergent (I believe), given by the expression at the top of this question.

I'm struggling to get a clean working calculation for this, but the most reasonable ones seemed to give essentially $\pi$ .

Any idea if this is indeed the correct result and how to obtain it cleanly?

Answer 1

Too long for a comment

I don't think I can provide the asimptotics of the form of the integral you presented. The situation is complicated by the fact that the second integral has a singulariry at $x=y$. But if your goal is to get a regular part of $I_0$, it can be done by means of another regularization.

Let's consider $\,\displaystyle I_\epsilon=\int_0^\infty\int_0^\infty\frac{e^{-\frac12 x^2-\frac12y^2}}{(x-i y)^{2-\epsilon}}dxdy, \,\,0<\epsilon\ll1$.

Switching to the spherical system of coordinates, $$I_\epsilon=\int_0^\infty\frac{e^{-r^2/2}}{r^{2-\epsilon}}rdr\int_0^{\pi/2}e^{i(2-\epsilon)\phi}d\phi=2^{\frac\epsilon2-1}\Gamma\left(\frac\epsilon2\right)\frac {1+e^{-\frac{\pi i\epsilon}2}}{2-\epsilon}i$$ $$=\frac i\epsilon+\frac i2(1+\ln2-\gamma)+\frac\pi4+O(\epsilon)$$

Answer 2

Working the original integral.

Using cartesian coordinates, Mathematica returns $$I=\int_0^\infty\frac{e^{-\frac{x^2+y^2}{2}}}{(x-i y)^2}\,dy=\frac i{ \pi x}\,e^{-\frac{x^2}{2}}\,G_{3,2}^{2,3}\left(-\frac{i \sqrt{2}}{x},\frac{1}{2}| \begin{array}{c} 0,\frac{1}{2},1 \\ \frac{1}{2},1 \\ \end{array} \right)$$ which seems impossible to integrate again.

Using a different approach $$I=-\frac{\pi}{2} x \, \text{erfc}\left(\frac{x}{\sqrt{2}}\right)+\frac{x\sqrt{2 \pi } +2 i}{2 x}\,e^{-\frac{ x^2}{2}} -\frac{i}{2} x \,\Gamma \left(0,\frac{x^2}{2}\right)$$ which can be integrated $$\color{blue}{J=\int_\epsilon^\infty I\,dx=-\frac{\sqrt{2 \pi } \epsilon +2 i}{4} e^{-\frac{\epsilon ^2}{2}}+}$$ $$\color{blue}{\frac{1}{4} \left(\pi \left(\epsilon ^2+1\right) \text{erfc}\left(\frac{\epsilon }{\sqrt{2}}\right)+i \left(\epsilon ^2 E_1\left(\frac{\epsilon ^2}{2}\right)-2\, \text{Ei}\left(-\frac{\epsilon ^2}{2}\right)\right)\right)}$$

Expanded as series around $\epsilon=0$ $$\color{green}{J=\left( \frac{\pi }{4}-\sqrt{\frac{\pi }{2}} \epsilon +\frac{\pi }{4}\epsilon ^2-\frac{1}{6} \sqrt{\frac{\pi }{2}} \epsilon ^3+\cdots \right)-}$$

$$\color{green}{\frac i 4 \left(4 \log (\epsilon )+2( \gamma +1-\log (2))+\epsilon ^2 (2 \log (\epsilon )+\gamma -2-\log (2))+\cdots\right)}$$

For $\epsilon=\frac 1{100}$ the above expansion gives $0.772865\, +4.16342\, i$ while the exact value is $0.772943\, +4.16342 \,i$.