Renormalization of phi^4 Theory d dimensional Integral, Derivation in Peskin Schröder

1. Mar 13, 2009

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

2. Mar 13, 2009

Avodyne

Re: Renormalization of phi^4 Theory d dimensional Integral, Derivation in Peskin Schr

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.

Last edited: Mar 14, 2009
3. Mar 14, 2009

Phileas.Fogg

Re: Renormalization of phi^4 Theory d dimensional Integral, Derivation in Peskin Schr

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

4. Mar 14, 2009

Avodyne

Re: Renormalization of phi^4 Theory d dimensional Integral, Derivation in Peskin Schr

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