- #1

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## 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

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

which is divergent

but in arbitrary dimensions

you get

[tex]\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}}[/tex]

and setting [tex] \epsilon = 4-D[/tex]

and then letting D approach 4 we get

[tex]\frac{1}{(4\pi)^{2}}(\frac{2}{\epsilon}-log\Delta - \gamma +log(4\pi) + O(\epsilon))[/tex]

where

[tex]\gamma \approx .5772[/tex]

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.

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

which is divergent

but in arbitrary dimensions

you get

[tex]\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}}[/tex]

and setting [tex] \epsilon = 4-D[/tex]

and then letting D approach 4 we get

[tex]\frac{1}{(4\pi)^{2}}(\frac{2}{\epsilon}-log\Delta - \gamma +log(4\pi) + O(\epsilon))[/tex]

where

[tex]\gamma \approx .5772[/tex]

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.