= Solution
{c}
At the stated order the <renormalization-group beta function> is separable:
$$
\int_{\alpha_i}^{\alpha_f}\frac{d\alpha}{\alpha^2}
=b_0\int_{\mu_i}^{\mu_f}\frac{d\mu}{\mu}.
$$
Consequently
$$
\boxed{\frac1{\alpha_f}=\frac1{\alpha_i}-b_0\ln\frac{\mu_f}{\mu_i},\qquad
\alpha_f=\frac{\alpha_i}{1-b_0\alpha_i\ln(\mu_f/\mu_i)}.}
$$
For a positive initial <running coupling>, this expression is used on the connected interval containing $\mu_i$ where its denominator is positive and <perturbation theory> remains valid.
If $b_0<0$, the <running coupling> decreases at high <renormalization scale> and tends to zero: this is <asymptotic freedom>. Decreasing the <renormalization scale> increases the <running coupling>, with a formal infrared pole at $\Lambda=\mu_i\exp[1/(b_0\alpha_i)]$. Near $\Lambda$, <perturbation theory> fails; the pole alone is not a proof of <confinement>. Both <QCD> and the <SU(2)> weak <gauge coupling> have this sign for the <Standard Model> particle content above the relevant thresholds.
If $b_0>0$, the <running coupling> decreases towards low <renormalization scale> and grows towards high <renormalization scale>. The formal ultraviolet <Landau pole> is $\mu_L=\mu_i\exp[1/(b_0\alpha_i)]$. The pole signals failure of this perturbative evolution, rather than a controlled physical infinity. The <hypercharge> <gauge coupling> and ordinary <QED> with charged matter have this sign.
If $b_0=0$, the <running coupling> is constant \b[at this order]; higher-order terms may still make it run. No <Standard Model> gauge factor has a vanishing one-<loop order> coefficient for the usual particle content.
For precise normalization, take $\alpha_i=g_i^2/(4\pi)$ and $Q=T_3+Y$. The <Standard Model one-loop gauge coefficients> then give
$$
(b_0^Y,b_0^2,b_0^3)=\frac1{2\pi}\left(\frac{41}{6},-\frac{19}{6},-7\right).
$$
These coefficients use $d/d\ln\mu$, not $d/d\ln\mu^2$. Threshold matching changes the active particle content, so the single constant-$b_0$ expression is applied piecewise across thresholds.
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