# The three dimensional integral

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.

• Are you sure it's $\delta (1-u_1-u_2-u_3)$ and not $\delta (1-u_1) \delta (1-u_2) \delta (1-u_3)$? Oct 21, 2015 at 11:26
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.
• I tried to integrate the same way. But it does not work. $a$ and $\eta$ are parameters. Oct 21, 2015 at 12:02