- #1
phypar
- 12
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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.
[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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