Integration with exponential and inverse powerby phypar Tags: exponential, integration, inverse, power 

#1
Nov813, 11:24 PM

P: 9

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 twocomponents 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. 



#2
Nov1013, 09:50 AM

P: 1,273

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).




#3
Nov1013, 07:55 PM

P: 9

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{1a^2}\right)\left((\sqrt{1a^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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