# Dictate Mathematica to use the contour integration method

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:

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