Can this integral be calculated using Mathematica?

Question

I have the following integral:

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

Here I wanted to do the $\tau'$ integral first, so I wrote the following

 $Assumptions = c >= 0 && c \[Element] Reals && ϵ > 0 && ϵ \[Element] Reals && δ > 0 && δ \[Element] Reals  && τ \[Element] Reals && t \[Element] Reals;

 I5 = Integrate[E^(-c^2 (τ - t)^2)/(( τ - I ϵ) (t - I δ)), {t, -Infinity, Infinity}]

where I have denoted $\tau'$ using t. However the output turns out to be just the same expression as the input.

Next I tried to use the residue of the integrand about $ \tau' = \mathrm i\epsilon' $ (In the code below, about $ t = \mathrm i\delta $).

Residue[e^(-c^2 (τ - t)^2)/((τ - I ϵ) (t - I δ)), {t, I δ}]

Again the output turns out to be the same expression as the input.

Any suggestions how to compute the integral?

EDIT: It was pointed out in the comments that there were some mistakes in the code, so I edited the same.

Answer 1

C is just a normalization factor, so we set c = 1, make a substitution,

t = (x + y)/2; t1 = (x - y)/2;
Exp[-(t - t1)^2]/(t - A)/(t1 - B) // FullSimplify

Out[]= -((4 E^-y^2)/((2 B - x + y) (-2 A + x + y)))

then the integral with respect to one of the variables is calculated exactly

Integrate[-((
  4 E^-y^2)/((2 B - x + y) (-2 A + x + y))), {y, -Infinity, Infinity}]

Out[]= ConditionalExpression[(
 2 E^(-4 A^2 - 4 B^2 + 4 A x + 4 B x - 
   2 x^2) (E^(-2 B + x)^2 \[Pi] Erfi[2 A - x] + 
    E^(-2 A + x)^2 \[Pi] Erfi[2 B - x] - E^(-2 B + x)^2 Log[2 A - x] -
     E^(-2 A + x)^2 Log[2 B - x] + E^(-2 B + x)^2 Log[-2 A + x] + 
    E^(-2 A + x)^2 Log[-2 B + x]))/(A + B - x), 
 2 Im[A] != Im[x] && 2 Im[B] != Im[x] && 2 A - x \[NotElement] Reals &&
   2 B - x \[NotElement] Reals && -2 A + x \[NotElement] 
   Reals && -2 B + x \[NotElement] Reals]