Integral of Gaussian function with first order pole 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.
 A: 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.
