1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ e should be E and { } should be ( ) when used for grouping terms $\endgroup$ – bill s Oct 17 '18 at 16:58
  • $\begingroup$ @bills I see, thanks. Let me try that out. $\endgroup$ – Michael Williams Oct 17 '18 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 '18 at 17:17
  • 4
    $\begingroup$ The syntax for global assumptions is $Assumptions = c>=0 && c \[Element] Reals && \[Epsilon] > 0 && ... $\endgroup$ – AccidentalFourierTransform Oct 17 '18 at 18:52
  • 1
    $\begingroup$ 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 '18 at 2:53
2
$\begingroup$

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]
$\endgroup$
  • $\begingroup$ 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 '18 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 '18 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.