- #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?