Wednesday, 17 January 2018

integration - Integrating w.r.t. two different variables



I can't really follow what's going on here to get from line 2 to line 3:
0d2p(2π)2Λ2(p2+Λ2)(p2+m2)=10dzd2p(2π)2Λ2(p2+zΛ2+(1z)m2)2=10dz14πΛ2zΛ2+(1z)m2=14πΛ2Λ2m2lnΛ2m2




There are two different integration variables and then one of them vanishes. I've tried evaluating the integral w.r.t p and then again w.r.t. p because it's d2p, but that didn't work (just using wolfram alpha). From line one to line two I can follow, it's a Feynman parametrization, but what do you do to get from line 2 to line 3?


Answer



The integral over momentum space is missing in Line 2 of the OP. That is to say that the expression d2pΛ2(p2+zΛ2+(1z)m2)2 should read



Λ2(p21+p22+zΛ2+(1z)m2)2dp1dp2



which after a transformation to polar coordinates (p1,p2)(ρ,ϕ) becomes



00Λ2(p21+p22+zΛ2+(1z)m2)2dp1dp2=2π00Λ2ρ(ρ2+zΛ2+(1z)m2)2ρdρdϕ=2π0Λ2ρ(ρ2+zΛ2+(1z)m2)2ρdρ =2π(121ρ2+zΛ2+(1z)m2)|ρρ=0=2π(Λ2121zΛ2+(1z)m2)



Finally, dividing by 4π2 and integrating over z from 0 to 1 yields the coveted result.


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