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I want to evaluate an integral, which can be in principle computed exactly. For $a>0$ and $k_1, k_2 >0$, define $G(x) = (x+ia)^3$. The integral that I have is $$I = \int_{\mathbb R^4} dy_1dy_2 dz_1 dz_2 e^{-ik_2 z_1} e^{-ik_1 z_2} e^{ik_2 y_1} e^{ik_1 y_2} \frac{G(z_2-y_1) G(z_1-y_2)}{G(z_2-z_1) G(z_2-y_2) G(z_1-y_1) G(y_1-y_2)}.$$

Using the contour integration in the complex plane, although very tedious, $I$ can be computed exactly. (The function $e^{\pm ikx}$ dictates whether integral is performed for upper or lower half plane, and the residue can be in principle calculated.)

However, Mathematica cannot solve this integral. Can I force Mathematica to use the contour integration method, or is there any other method to compute $I$?

I used the following code, which does not give the answer within 1 hour:

$Assumptions = {k1 > 0, k2 > 0, a > 0, m > 0};
G[x_] = (x + I a)^3;
Integrate[ 
 Exp[-I k2 z1] Exp[-I k1 z2] Exp[I k2 y1] Exp[I k1 y2] (
  G[z2 - y1] G[z1 - y2])/(
  G[z2 - z1] G[z2 - y2] G[z1 - y1] G[y1 - y2]), {y1, -Infinity, 
  Infinity}, {y2, -Infinity, Infinity}, {z1, -Infinity, 
  Infinity}, {z2, -Infinity, Infinity}]
$\endgroup$
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    $\begingroup$ You should add the Mathematica code you tried. That way people can see exactly what you've done, and also try playing with it themselves instead of typing it from scratch. $\endgroup$
    – theorist
    May 4 at 4:16
  • $\begingroup$ @theorist I added the code. $\endgroup$
    – eigenvalue
    May 4 at 10:57
  • $\begingroup$ Your statement "Using the contour integration in the complex plane, although very tedious, $I$ can be computed exactly" is empty words. $\endgroup$
    – user64494
    May 4 at 15:24
  • $\begingroup$ @user64494 Could you please rephrase your request for clarification? Expressions like "empty words" and "built on sand" sound very harsh to me. $\endgroup$
    – MarcoB
    May 4 at 16:38
  • $\begingroup$ @eigenvalue: Can you please post the exact (or numeric) value of the integral? $\endgroup$
    – josh
    May 4 at 21:19

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