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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.

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    $\begingroup$ e should be E and { } should be ( ) when used for grouping terms $\endgroup$
    – bill s
    Oct 17, 2018 at 16:58
  • $\begingroup$ @bills I see, thanks. Let me try that out. $\endgroup$
    – Michael Williams
    Oct 17, 2018 at 17:01
  • $\begingroup$ @bills While the Integrate command is still giving the same, the Residue command is working properly now. Thanks. $\endgroup$
    – Michael Williams
    Oct 17, 2018 at 17:17
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    $\begingroup$ The syntax for global assumptions is $Assumptions = c>=0 && c \[Element] Reals && \[Epsilon] > 0 && ... $\endgroup$
    – AccidentalFourierTransform
    Oct 17, 2018 at 18:52
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    $\begingroup$ You can calculate the real part of the integral using Integrate[E^(-c^2 (τ - t)^2)/((τ - I ϵ) t), {t, -Infinity, Infinity}, Assumptions -> {c > 0 && ϵ > 0 && τ \[Element] Reals}, PrincipalValue -> True]. The imaginary part is trivial to evaluate using $\mathrm{im}\left[\frac{1}{x-i\epsilon}\right]=\pi\delta(x)$. $\endgroup$
    – AccidentalFourierTransform
    Oct 18, 2018 at 2:53

1 Answer 1

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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]
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  • $\begingroup$ Actually your variable substitution works for me if you let c stay c. But I think you need a factor 1/2 in front of the integral since dt = dy/2. Then don't forget to convert x back to the t's. $\endgroup$
    – Bill Watts
    Dec 14, 2018 at 7:37
  • $\begingroup$ @BillWatts Thank you. I think all this should be interesting to the author. But he did not answer. $\endgroup$
    – Alex Trounev
    Dec 14, 2018 at 12:56

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