# Surface terms in loop integrals (2D)

1. May 23, 2012

### nidnus

Hi everyone,

Say I have a 2D one loop integral of the form

$$I^s_n(\Delta)=\int d^2 l \frac{(l^2)^s}{(l^2-\Delta)^n}$$

Using that $1 = \frac{1}{2} \partial_\mu l^\mu$, I can relate say

$$I^1_2(\Delta)=I^0_1(\Delta)$$
+ total derivative term.

In dimensional regularization one usually throw away surface terms meaning that

$$I^1_2(\Delta)=I^0_1(\Delta)$$

However, looking the appendix of, say, Peskin & Schröder they state the actual value of the different $I^s_n(\Delta)$ integrals in terms of $\Gamma$ functions for general D. Expanding $I^1_2$ and $I^0_1$ show that they agree up to a linear $i \pi$ term.

Thus this term has to correspond to a surface term and one should be able to throw it away on physical grounds. However, all this seem very basic and seem rather an important concept every time you want to make an actual calculation. How come I have never seen this? I don't think it's even mentioned in P&S.

What is more, what happens in a hard cutoff $\Lambda$? If one can use the same arguments as in dim reg, then it follows that $\Lambda^2$ terms from 1-loop tadpoles correspond to surface terms (i.e they don't exist)! (this can be seen by relating $2I^1_1=I^2_2$+ total derivative).

Can someone who knows these kind of things in some detail clarify this for me?

Thanks alot,

Nid

Last edited: May 23, 2012