Integration with exponential and inverse power

  1. I confront an integration with the following form:

    [itex] \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)}
    [/itex]
    where [itex]a[/itex] and [itex]b[/itex] are some constants, [itex]\vec{q} = (q_1, q_2)[/itex] and [itex]\vec{k} = (k_1, k_2)[/itex] are two-components vectors.

    In the case of [itex]a\rightarrow \infty [/itex] 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

    [itex]\pi \exp(ab)\left(\Gamma(0,ab)-\Gamma(0,a(\vec{k}^2+b))\right)[/itex]

    where [itex]\Gamma(0,x)=\int_x^\infty t^{-1} e^{-t}\,dt [/itex] 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.
     
    Last edited: Nov 9, 2013
  2. jcsd
  3. 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).
     

  4. Thanks. I just found the solution from another paper. So first one should perform the integration to polar coordinates using the formula:
    [itex] \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[/itex]
    then perform the integration on [itex]p^2[/itex] will yield the above result.
     
    Last edited: Nov 10, 2013
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