 One-loop running of the strong coupling

With ${\beta }_0>0$, differentiating ${\alpha }_s=g^2/(4\pi )$ under $\mu dg/d\mu =-{\beta }_0{g}^3/(16{\pi }^2)$ gives $d(1/{\alpha }_s)/d\ln \mu ={\beta }_0/(2\pi )$. Integration defines the strong-coupling scale $\Lambda$ and gives the displayed positive branch for $\mu >\Lambda$. The pole signals breakdown of perturbative quantum field theory. The equivalence between specifying ${\alpha }_s$ and specifying $\Lambda$ is dimensional transmutation.