Proving the Renormalization of Phi^4 Theory: A Challenge for Mr. Fogg

[Phileas.Fogg]

Hello,
to understand the renormalization of phi^4 theory, I read Peskin Schröder and Ryder. In both books important steps are left out. I found the following identity in Peskin Schöder "An Introduction to Quantum Field Theory" on page page 808, equation A.52 (Appendix)

\frac{\Gamma(2 - \frac{d}{2})}{(4\pi)^{\frac{d}{2}}} \left( \frac{1}{\Delta} \right)^{2-\frac{d}{2}} = \frac{1}{(4\pi)^2} \left( \frac{2}{\epsilon} - log(\Delta) - \gamma + log(4\pi) + O(\epsilon)\right)

Now I want to prove that explicitly, but I don't know how to start and how the logarithm on the right hand side appears.

Could anyone help me?

Regards,
Mr. Fogg

 

[Avodyne]

to understand the renormalization of phi^4 theory, I read Peskin Schröder and Ryder. In both books important steps are left out.

That's why you should read a book like Srednicki that doesn't leave steps out.

For any nonzero A and small x,

A^x=\exp(x\ln A)=1+x\ln A + O(x^2)

\Gamma(x) = {1\over x}-\gamma+O(x)

These are equations 14.33 and 14.26 in Srednicki.

 

[Phileas.Fogg]

Thank You,

so I get this expression:

\frac{1}{(4\pi)^2} \left( \frac{2}{\epsilon} - \gamma \right) \left(1 + \frac{\epsilon}{2} ln(\frac{4 \pi}{\Delta})\right)

But that's not the equation from Peskin & Schröder, isn't it?

How do I go on to get it finally?

Regards,
Mr. Fogg

 

[Avodyne]

It's the same. Just multiply it out, and use ln(a/b)=ln(a)-ln(b).