How do logarithms show up in the one loop calculation of the vacuum polarization in QED?

Question

I am following Peskin with the computation of the vacuum polarization in QED and there is one thing I do not see.

Equation (7.90) reads

$$\frac{-8e^2}{(4\pi)^{d/2}}\int_0^1dx\,x(1-x)\frac{\Gamma(2-\frac{d}{2})}{\Delta^{2-d/2}}$$

now we make $d\to4-\epsilon$ where $\epsilon$ is infinitesimal. This leads $\Gamma(2-\frac{d}{2})\to\Gamma(\epsilon/2)$ and we can use the following expansion given by equation (7.83).

$$\Gamma(\epsilon/2)=\frac{2}{\epsilon}-\gamma+O(\epsilon)$$

where $\gamma$ is the Euler-Mascheroni constant. Doing the computation with these formulae leads me to

$$\frac{-8e^2}{(4\pi)^{2}}\int_0^1dx\,x(1-x)(\frac{4\pi}{\Delta})^{\epsilon/2}(\frac{2}{\epsilon}-\gamma+O(\epsilon))$$

nonetheless the last part of equation equation (7.90) claims

$$\frac{-2\alpha}{\pi}\int_0^1dx\,x(1-x)(\frac{2}{\epsilon}-\log\Delta-\gamma+\log(4\pi))$$

My question is simple. How do the logarithms in the end show up?

Answer 1

You might end up slapping yourself in the forehead because it's probably simpler than what you were thinking.

$$\left(\frac{4\pi}{\Delta}\right)^{\epsilon/2} = \exp\left[\frac{\epsilon}{2}\log \left(\frac{4\pi}{\Delta}\right)\right] = 1 + \frac{\epsilon}{2}\log \left(\frac{4\pi}{\Delta}\right) +\mathcal{O}(\epsilon^2)$$ whence the result follows.