# Dimensional regularization - Expansion of powers of $\epsilon$ turns into logarithms

Looking into Schwartz's book on QFT at the appendices, it seems that when doing a dimensional regularization, one expands around $$\epsilon=0$$ and usually obtains $$x^\epsilon=\log x+O(\epsilon),$$ or in an example $$\bigg(\frac{4\pi\mu^2}{\Delta}\bigg)^\epsilon=\log4\pi+\log\mu^2-log\Delta+O(\epsilon)=\log\bigg(\frac{4\pi\mu^2}{\Delta}\bigg)+O(\epsilon)$$ How is that?

I was looking around for the answer in the book and it seems to not be given anywhere as to why. Note: I'm not exactly sure if there are other terms of order $$\epsilon$$ as there are on the $$\Gamma(\epsilon)$$ and the $$log$$'s terms may have been mixed with the $$\Gamma(\epsilon)$$ so I just added them here just in case.

## 2 Answers

The usual way to do it is to write the quantity like $$x^\epsilon$$. Then use the relation that

$$x^\epsilon = e^{\epsilon \ln(x)}$$

and then use the usual exponential expansion so that

$$e^{\epsilon \ln(x)} = 1 + \epsilon\ln(x) + \mathcal{O}(\epsilon^2)$$

• Where did the factor of $\epsilon$ go though? – Gradient137 Dec 28 '18 at 6:45
• What do you mean – InertialObserver Dec 28 '18 at 6:51
• like $x^\epsilon=1+\epsilon\log x$ to first order, there is a factor of $\epsilon$ in the log which makes sense, but then in Schwarz's (or any other books featuring regularization) there isn't a factor of $\epsilon$ in the log. – Gradient137 Dec 28 '18 at 7:16
• That's because they usually expand, not just what you wrote, but the product of other factors multiplying what you wrote (all of which have $\epsilon$s). The expansion I've written is correct. Write out the full expression and expand each of them in $\epsilon$. As mentioned in @CAF's answer the original expression you wrote down isn't correct. There will be powers of $\frac{1}{\epsilon}$ in your expansion as well and so you'll see the cancellations you're looking for. – InertialObserver Dec 28 '18 at 7:20

Hint: $$x^\epsilon = e^{\log x^\epsilon} = e^{\epsilon \log x},$$ then proceed in expansion of the exponential for small $$\epsilon$$.

Note the expressions in OP are not quite correct.