Integration with exponential and inverse power

[phypar]

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<br /> \vec{q}}{((\vec q-\vec k)^{2})(\vec{q}^{2}+b)}<br />
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.

 

[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).

 

[phypar]

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&lt;1,~~n\geq0
then perform the integration on p^2 will yield the above result.