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\begin{equation} \begin{split} \frac{d(g_b\mu^{\epsilon})}{d\log\mu^2}&=\frac{\mu}{2}\frac{d(g_b\mu^{\epsilon})}{d\mu}\\ &=\frac{\mu}{2}\left[\mu^{\epsilon}\frac{dg_b}{d\mu}+g_b\frac{d\mu^{\epsilon}}{d\mu}\right] \end{split} \end{equation} By definition, the bare coupling does not depend on the renormalization scale $\mu$. Hence \begin{equation} \...



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