# Question about DGLAP evolution equation

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:
$$(2)~\mu \frac{d}{d\mu}f_i(x,\mu) = \frac{\alpha_s}{\pi} \int_x^1 \frac{d\xi}{\xi} f_i(\xi, \mu) P_{qq}(\frac{x}{\xi})$$.
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?
• I think the answer is very simple: the PDF under the integral in your first equation should be evaluated at $\mu^2$ and not $\mu^2_1$. – ptaels Feb 1 at 20:18
• 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. – lomby Feb 4 at 11:39