The integral is, with $\epsilon$ and $\epsilon'$ being small positive constants: (primed variables don't mean a derivative, they are independent variables)

$ \int_{-\infty}^{\infty} \frac{d\tau}{2\pi i} \int_{-\infty}^{\infty} \frac{d\tau'}{2\pi i} \frac{e^{-c^2(\tau - \tau')^2}}{(\tau - i \epsilon)(\tau' - i\epsilon')}.$

I tried to solve for the $\tau'$ integral first where I constructed a contour which goes as a semicircle in the upper half plane but I couldn't get the numerator to die off. Is there any other way to solve the same?

PS. The original integral which I was solving was:

$ \int_{-\infty}^{\infty} \frac{d\tau}{2\pi i} \int_{-\infty}^{\infty} \frac{d\tau'}{2\pi i} \frac{1}{(\tau - i \epsilon)(\tau' - i\epsilon')}.e^{-c^2\tau^2 -c^2\tau'^2 + c^2\tau \tau' \{2+AB\sqrt{2\pi} e^{-A^2B^2/2} -\frac{erf(iAB/{\sqrt{2}})}{i} + i\}}$

where $erf$ is the error function. Since this integral looks ominous to solve, I took the limit of the expression in the curly brackets when $A \to 0$.

$\lim_{A \to 0} {\{2+AB\sqrt{2\pi} e^{-A^2B^2/2} -\frac{erf(iAB/{\sqrt{2}})}{i} + i\}} = 2$

Using this limit I got the expression for the Gaussian integral which is the first equation. Is it possible to solve this case also? Thanks.

  • $\begingroup$ Does $\tau’$ denote the derivative of $\tau$ $\endgroup$ – Henry Lee Oct 17 '18 at 14:17
  • $\begingroup$ @HenryLee No it is an independent variable. $\endgroup$ – Michael Williams Oct 17 '18 at 14:27
  • $\begingroup$ why not just give it a separate symbol then to avoid confusion $\endgroup$ – Henry Lee Oct 17 '18 at 14:27
  • $\begingroup$ @HenryLee I edited to clarify the same, thanks :) $\endgroup$ – Michael Williams Oct 17 '18 at 14:33
  • $\begingroup$ ah good, that is clearer now $\endgroup$ – Henry Lee Oct 17 '18 at 14:34

I assume that the problem is to find $$I(a) = \lim_{(\epsilon, \epsilon') \to (0^+, 0^+)} \frac 1 {(2 \pi i)^2} \iint_{\mathbb R^2} \frac {e^{-c^2 (\tau^2 + 2 a \tau \tau' + (\tau')^2)}} {(\tau - i \epsilon) (\tau' - i \epsilon')} d\tau d\tau'.$$ We have $$I'(a) = \frac {c^2} {2 \pi^2} \iint_{\mathbb R^2} e^{-c^2(\tau^2 + 2 a \tau \tau\ + (\tau')^2)} d\tau d\tau' = \frac 1 {2 \pi \sqrt {1 - a^2}}, \quad -1 < \operatorname{Re} a < 1, \\ \lim_{\epsilon \to 0^+} \int_{\mathbb R} \frac {e^{-c^2 \tau^2}} {\tau - i \epsilon} d\tau = \operatorname{v.\!p.} \int_{\mathbb R} \frac {e^{-c^2 \tau^2}} \tau d\tau + \pi i \operatorname*{Res}_{\tau = 0} \frac {e^{-c^2 \tau^2}} \tau = \pi i, \\ I(0) = \frac 1 4, \\ I(a) = \frac {\arcsin a} {2 \pi} + \frac 1 4, \quad \quad -1 \leq \operatorname{Re} a \leq 1.$$ The double integral is still absolutely convergent for $\operatorname{Re} a = \pm 1$ because absolute convergence in a small sector around the line $\tau' = \mp \tau$ is due to the denominator and elsewhere to the exponential.


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