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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.

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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

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

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  • $\begingroup$ Where did the factor of $\epsilon$ go though? $\endgroup$ – Gradient137 Dec 28 '18 at 6:45
  • $\begingroup$ What do you mean $\endgroup$ – InertialObserver Dec 28 '18 at 6:51
  • $\begingroup$ 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. $\endgroup$ – Gradient137 Dec 28 '18 at 7:16
  • $\begingroup$ 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. $\endgroup$ – InertialObserver Dec 28 '18 at 7:20
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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.

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