# Dimensional regularization

1. Aug 1, 2007

### ice109

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.

2. Aug 2, 2007

### ice109

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

3. Aug 2, 2007

### Hurkyl

Staff Emeritus
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)

Last edited: Aug 2, 2007
4. Aug 2, 2007

### ice109

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(