= Solution
In four dimensions set $b=3/(16\pi^2)$ and retain the one-loop <beta function> $dg/d\log\mu=bg^2$. On an interval where $g\ne0$, differentiate its inverse:
$$
\frac{d(1/g)}{d\log\mu}=-b.
$$
Integrating between $\mu_s$ and $\mu$ yields
$$
\frac1{g(\mu)}=\frac1{g_s}-b\log\frac\mu{\mu_s},\qquad
\boxed{g(\mu)=\frac{g_s}{1-\dfrac{3g_s}{16\pi^2}\log(\mu/\mu_s)}.}
$$
The $g_s=0$ solution is identically zero and is also the continuous limit of this expression. Expanding the denominator gives $g(\mu)=g_s+bg_s^2\log(\mu/\mu_s)+\cdots$, directly checking the perturbative sign.
For positive $g_s$, the <running coupling> decreases towards the infrared and increases towards the ultraviolet. The one-loop expression has a <Landau pole> at
$$
\mu_L=\mu_s\exp\left(\frac{16\pi^2}{3g_s}\right).
$$
This is the scale where the truncated solution diverges, not a controlled perturbative prediction of a physical singularity at arbitrarily strong coupling. The formula is quantitatively reliable while the running coupling remains small enough that omitted higher-loop terms are negligible. It resums the leading logarithms of that approximate <renormalization group> equation; it is not the exact all-orders solution of the theory.
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