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I am reading chapter 32.2 of Schwartz's QFT book, where he defines the renormalized PDFs $f_i(x, \mu)$. This leads to an equation (32.48), which relates PDFs at different scales $\mu, \mu_1$:
$f_i(x,\mu_1) = f_i(x,\mu) + \frac{\alpha_s}{2\pi} \int_x^1 \frac{d\xi}{\xi} f_i(\xi, \mu_1) P_{qq}(\frac{x}{\xi}) ln(\frac{\mu_1^2}{\mu^2})$
When I apply $\mu \frac{d}{d\mu}$ to this equation I get:
$(1)~\mu \frac{d}{d\mu}f_i(x,\mu) = \frac{\alpha_s}{\pi} \int_x^1 \frac{d\xi}{\xi} f_i(\xi, \mu_1) P_{qq}(\frac{x}{\xi})$.
However according to the book the correct equation is:
$(2)~\mu \frac{d}{d\mu}f_i(x,\mu) = \frac{\alpha_s}{\pi} \int_x^1 \frac{d\xi}{\xi} f_i(\xi, \mu) P_{qq}(\frac{x}{\xi})$.
I am very confused about this. My questions is:

Why do we have $\mu$ in the argument of $f_i$ in the RHS of equation $(2)$?

Doesn't it follow that $\frac{\alpha_s}{\pi} \int_x^1 \frac{d\xi}{\xi} f_i(\xi, \mu) P_{qq}(\frac{x}{\xi})$ is independent of $ \mu$, since equation $(1)$ should be valid for any $\mu_1$? This doesn't make a lot of sense to me.
My other thought was that in order for the perturbation expansion to be good we have to choose $\mu_1 \sim \mu$ so that the logarithm is small. So we can get from $(1)$ to $(2)$ by saying that they are approximately equal?

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  • $\begingroup$ I think the answer is very simple: the PDF under the integral in your first equation should be evaluated at $\mu^2$ and not $\mu^2_1$. $\endgroup$ – ptaels Feb 1 at 20:18
  • $\begingroup$ Thank you, for your answer! So this would mean that there is an error in the book? If it is as you say, as far as I can see, there is still a term that comes from product rule, i.e. $ \frac{\alpha_s}{2\pi} \int_x^1 \frac{d\xi}{\xi} (\mu \frac{d\mu}{\mu} f_i(x, \mu)) P_{qq}(\frac{x}{\xi}) ln(\frac{\mu_1^2}{\mu^2})$, which should vanish to produce Eq (2). However I can't see how this term vanishes. $\endgroup$ – lomby Feb 4 at 11:39

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