I tried to evaluate such integral: $$I=\int \frac{du_1du_2du_3u_1^{i\eta}u_2^{-i\eta}\delta(1-u_1-u_2-u_3)\theta(u_1)\theta(u_2)\theta(u_3)}{(u_3+au_1u_2)^2} $$ where $\delta(x)$ is a Dirac $\delta$ - function, $\theta(x)$ is a Heaviside step function and $0<a<1$. The $\theta$ and $\delta$ functions are required to extract necessary area of integration.

This integral can be represented via Gauss Hyper-geometric functions. I spent big amount of time to trying to take this integral. But I haven't done it yet.

Please help me!

  • $\begingroup$ Are you sure it's $\delta (1-u_1-u_2-u_3)$ and not $\delta (1-u_1) \delta (1-u_2) \delta (1-u_3)$? $\endgroup$
    – Yuriy S
    Oct 21, 2015 at 11:26
  • $\begingroup$ Yes. In your case the integrating is trivial. $\endgroup$
    – Peter
    Oct 21, 2015 at 11:59
  • $\begingroup$ Where does this problem come from? $\endgroup$
    – amcalde
    Oct 21, 2015 at 12:04

1 Answer 1


Too long for comment: $$I=\int \frac{du_1du_2du_3u_1^{i\eta}u_2^{-i\eta}\delta(1-u_1-u_2-u_3)\theta(u_1)\theta(u_2)\theta(u_3)}{(u_3+au_1u_2)^2} $$ $$=\int \frac{du_1du_2u_1^{i\eta}u_2^{-i\eta}\theta(u_1)\theta(u_2)\theta(1-u_1-u_2)}{(1-u_1-u_2+au_1u_2)^2} $$ $$=\int_0^\infty du_1u_1^{i\eta}\int_{u_2 = 0}^{1-u_1} \frac{du_2u_2^{-i\eta}}{(1-u_1-u_2+au_1u_2)^2} $$ $$=\int_0^\infty du_1u_1^{i\eta}\int_{u_2 = 0}^{1-u_1} \frac{du_2u_2^{-i\eta}}{(1-u_1-(1-au_1)u_2)^2} $$ Now at this point, it get's uglier. The Hypergeometric functions are not avoidable past here. Is this what you wanted? Do you have a value for $a$? That might help.

  • $\begingroup$ I tried to integrate the same way. But it does not work. $a$ and $\eta$ are parameters. $\endgroup$
    – Peter
    Oct 21, 2015 at 12:02

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