# Dimensional regularization

## Main Question or Discussion Point

i'm doing an integral for my advisor that is way beyond me but i have pages from a textbook that tell me how to do it so here goes

$$\int\frac{d^4\ell}{(2\pi)^4}\frac{1}{(\ell^2+A^2)^2} = \frac{1}{2}B(0,2)$$

which is divergent

but in arbitrary dimensions

you get

$$\int\frac{d^4\ell}{(2\pi)^4}\frac{1}{(\ell^2+A^2)^2} = \frac{1}{(4\pi)^\frac{D}{2}}\frac{\Gamma (2-\frac{D}{2})}{\Gamma (2)}(\frac{1}{A})^{2-\frac{D}{2}}$$

and setting $$\epsilon = 4-D$$

and then letting D approach 4 we get

$$\frac{1}{(4\pi)^{2}}(\frac{2}{\epsilon}-log\Delta - \gamma +log(4\pi) + O(\epsilon))$$

where

$$\gamma \approx .5772$$

the thing is i only understand why this works on a very very superficial level. something along the lines of taking a limit and studying behavior as the limit approaches a "pole." but i really have no intuitive clue as to what an integral in arbitrary space even means. can anyone give me some sense of what's going on.

anybody? anybody? maybe this should be moved into atomic physics

Hurkyl
Staff Emeritus
Gold Member
4 dimensional calculus is the same as 3 dimensional calculus, except for the extra dimension.

You appear to be leaving out some details. I imagine that l is supposed to be a vector quantity, and that you have some particular regions of integration in mind that you haven't shared with us.

I don't know what "arbitrary dimensions" means, or from where D and $\Delta$ came. The meaning of B is not obvious to me from the context either.

LaTeX tip: use \left( and \right) to make large parentheses. (You can replace parentheses with other symbols too)

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4 dimensional calculus is the same as 3 dimensional calculus, except for the extra dimension.

You appear to be leaving out some details. I imagine that l is supposed to be a vector quantity, and that you have some particular regions of integration in mind that you haven't shared with us.

I don't know what "arbitrary dimensions" means, or from where D and $\Delta$ came. The meaning of B is not obvious to me from the context either.

LaTeX tip: use \left( and \right) to make large parentheses. (You can replace parentheses with other symbols too)
ahh i've made lots of mistakes in the write up

$$\int\frac{d^D\ell}{(2\pi)^D}\frac{1}{(\ell^2+\Delta)^ 2} = \frac{1}{(4\pi)^\frac{D}{2}}\frac{\Gamma (2-\frac{D}{2})}{\Gamma (2)}\left(\frac{1}{\Delta}\right)^{2-\frac{D}{2}}$$

region of integration is all of space. yes $\ell$ is a vector,arbitrary dimensions means just that, instead of 3 or 4 , D and it comes from how to solve that integral.
$\Delta$ is just a constant and B is the beta function.

don't mind this stuff, something i started typing out before i realized my initial write up was completely retarded. if you need to know how to work out the integral i'll finish that up.

[tex]\int\frac{d^D\ell}{(2\pi)^D}\frac{1}{(\ell^2+\Delta)^ 2} = \int\frac{d\Omega_{D}}{(2\pi^D)}}\int^{0}_{\infty}d\ell\frac{\ell^{D-1}}{(\ell^2+\Delta)^2}= \frac{2(\sqrt{\pi})^D}{\Gamma(\frac{D}{2})}\left(