An integral in Srednicki's book

[nklohit]

Hi all, I have just started to study QFT myself with Srednicki's book but there are some points that aren't clear to me.
First, I search for the proof of the integral in eq. 14.27
$$\int \frac{d^{d}\bar{q}}{(2\pi)^d} \frac{(\bar{q}^2)^a}{(\bar{q}^2+D)^b} = \frac{\Gamma(b-a-\frac{d}{2})\Gamma(a+\frac{d}{2})}{(4\pi)^{d/2}\Gamma(b)\Gamma(\frac{d}{2})}$$
but find nothing about it. Can anyone give a hint how to prove it?

Second, I'm very confusing that instead of putting the cut-off into the integration of feynman propagator, he use the factor $$(\frac{\Lambda^2}{k^2+\Lambda^2 -i\epsilon})$$ . Are there any reasons to do that?

Thank you for every answer :)

 

[vanhees71]

I've tried to put together all these techniques in my qft manuscript (but I don't use cutoff renormalization in there very much since I don't like it too much :-)):

http://theorie.physik.uni-giessen.de/~hees/publ/lect.pdf

About the Gamma function and its use in dimensional regularization for the evaluation of Feynman diagrams (loop integrals) you find there from page 143 on.

Concerning your 2nd question: Instead of introducing a sharp momentum cutoff by multiplying the whole integrand with such a factor, is a clever method since a sharp cutoff destroys nearly all nice symmetries as Lorentz invariance etc. This makes it more difficult to deal with the infinities afterwards and then letting the cutoff going to infinity for the renormlized quantities. If you introduce Lorentz-invariant form factors instead (this can be even physical if you deal with extended objects like atomic nuclei or hadrons instead of elementary "pointlike" particles), you avoid a lot of problems, which you would introduce with a sharp cutoff.

 

[nklohit]

Thank you very much :) Your manuscript will be very useful for me.
Do I understand right that by introducing that cutoff factor is to maintain the Lorentz invarince manipulation? And it appears in this form because the asymptotic properties that it become 1 when the $$\Lambda$$ goes to infinity?

 

[vanhees71]

Yes, that's what's behind a "form factor".

 

[Avodyne]

The angular part is done elsewhere in the book. The radial part is a standard integral that you can look up.