Two-loop asymptotic running coupling ($\frac{4\pi }{{\alpha }_{\lambda }}={\beta }_{0}L+\frac{{\beta }_1}{{\beta }_0}\log L+O(\log L/L)$)

For $\beta (\lambda )=-{\beta }_0{\lambda }^3/(4\pi {)}^2-{\beta }_1{\lambda }^5/(4\pi {)}^4+O({\lambda }^7)$, define $y=(4\pi {)}^2/{\lambda }^2$ and $L=\log ({\mu }^2/{\Lambda }^2)$. Then $dy/dL={\beta }_0+{\beta }_1/y+O(y^{-2})$. In the asymptotically free regime ${\beta }_0>0$, successive substitution gives the displayed expansion. If the beta function is exactly truncated to two loops, then $y-\frac{{\beta }_1}{{\beta }_0}\log [(y+{\beta }_1/{\beta }_0)/{\beta }_0]={\beta }_{0}L$, after choosing the integration constant. Its next term is ${\beta }_1^2(\log L+1)/({\beta }_0^{3}L)$. Thus the generic remainder after the first two terms is not $O(1/L)$ when ${\beta }_1\ne 0$. Higher-loop terms change the constant coefficient of $1/L$, but not the forced $\log L/L$ term.