- #1

RicardoMP

- 49

- 2

- Homework Statement
- I need to determine Feynman one-loop integrals to work out some Feynman diagrams, in particular ##I_{2,1}##.

- Relevant Equations
- $$I_{n,m}=\frac{1}{(4\pi)^2}\frac{\Gamma(m+2-\frac{\epsilon}{2})}{\Gamma(2-\frac{\epsilon}{2})\Gamma(n)}\frac{1}{\Delta^{n-m-2}}(\frac{4\pi M^2}{\Delta})^{\frac{\epsilon}{2}}\Gamma(n-m-2+\frac{\epsilon}{2})$$

Starting from the general formula:

$$I_{n,m}=\frac{1}{(4\pi)^2}\frac{\Gamma(m+2-\frac{\epsilon}{2})}{\Gamma(2-\frac{\epsilon}{2})\Gamma(n)}\frac{1}{\Delta^{n-m-2}}(\frac{4\pi M^2}{\Delta})^{\frac{\epsilon}{2}}\Gamma(n-m-2+\frac{\epsilon}{2})$$

I arrived to the following:

$$I_{2,1}=\frac{\Delta}{(4\pi)^2}\frac{(2-\frac{\epsilon}{2})}{(\epsilon-1)}[\frac{2}{\epsilon}-\gamma+ln(\frac{4\pi M^2}{\Delta})-\gamma\frac{\epsilon}{2}ln(\frac{4\pi M^2}{\Delta})+O(\epsilon)]$$

The term ##\frac{1}{\epsilon-1}## is giving me some trouble so I expanded it and, after removing terms proportional to ##\epsilon##, finally got:

$$I_{2,1}=-2\Delta I_{20}-\frac{\Delta}{(4\pi)^2}$$

Can someone confirm if this is the correct result?

$$I_{n,m}=\frac{1}{(4\pi)^2}\frac{\Gamma(m+2-\frac{\epsilon}{2})}{\Gamma(2-\frac{\epsilon}{2})\Gamma(n)}\frac{1}{\Delta^{n-m-2}}(\frac{4\pi M^2}{\Delta})^{\frac{\epsilon}{2}}\Gamma(n-m-2+\frac{\epsilon}{2})$$

I arrived to the following:

$$I_{2,1}=\frac{\Delta}{(4\pi)^2}\frac{(2-\frac{\epsilon}{2})}{(\epsilon-1)}[\frac{2}{\epsilon}-\gamma+ln(\frac{4\pi M^2}{\Delta})-\gamma\frac{\epsilon}{2}ln(\frac{4\pi M^2}{\Delta})+O(\epsilon)]$$

The term ##\frac{1}{\epsilon-1}## is giving me some trouble so I expanded it and, after removing terms proportional to ##\epsilon##, finally got:

$$I_{2,1}=-2\Delta I_{20}-\frac{\Delta}{(4\pi)^2}$$

Can someone confirm if this is the correct result?