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Integration with exponential and inverse power

by phypar
Tags: exponential, integration, inverse, power
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phypar
#1
Nov8-13, 11:24 PM
P: 11
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.
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dauto
#2
Nov10-13, 09:50 AM
Thanks
P: 1,948
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
#3
Nov10-13, 07:55 PM
P: 11
Quote Quote by dauto View Post
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:
[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.


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