# Integration with exponential and inverse power

by phypar
Tags: exponential, integration, inverse, power
 P: 9 I confront an integration with the following form: $\int d^2{\vec q} \exp(-a \vec{q}^{2}) \frac{\vec{k}^{2}-\vec{k}\cdot \vec{q}}{((\vec q-\vec k)^{2})(\vec{q}^{2}+b)}$ where $a$ and $b$ are some constants, $\vec{q} = (q_1, q_2)$ and $\vec{k} = (k_1, k_2)$ are two-components vectors. In the case of $a\rightarrow \infty$ in which the exponential becomes 1, I can perform the integration using Feynman parameterization. In the general case I have now idea to calculate it. I know the answer is $\pi \exp(ab)\left(\Gamma(0,ab)-\Gamma(0,a(\vec{k}^2+b))\right)$ where $\Gamma(0,x)=\int_x^\infty t^{-1} e^{-t}\,dt$ is the incomplete gamma function. But i don't know how to arrive at this result. can someone give any clue to perform this kind of integration? thanks a lot.
 P: 830 I would try using Feynman parametrization anyways, and than convert the integration to polar coordinates (The fact that you end up with an incomplete gamma function is a clue that polar coordinates were used).
P: 9
 Quote by dauto I would try using Feynman parametrization anyways, and than convert the integration to polar coordinates (The fact that you end up with an incomplete gamma function is a clue that polar coordinates were used).

Thanks. I just found the solution from another paper. So first one should perform the integration to polar coordinates using the formula:
$\int_0^\pi d\theta \cos(n\theta)/( 1+a\cos(\theta))=\left(\pi/\sqrt{1-a^2}\right)\left((\sqrt{1-a^2}-1)/ a\right)^n,~~~a^2<1,~~n\geq0$
then perform the integration on $p^2$ will yield the above result.

 Related Discussions Calculus & Beyond Homework 4 Calculus 4 Precalculus Mathematics Homework 6 Calculus & Beyond Homework 3 Precalculus Mathematics Homework 6