 Solution

At one loop, writing $t=\log \mu$ makes the renormalization-group beta function $d{g}_i/dt=b_i{g}_i^3$. For $b_i>0$ a positive running coupling grows towards high energies and decreases towards the infrared. Extrapolation of the one-loop expression gives a finite ultraviolet Landau pole; perturbation theory fails before it reaches that pole. For $b_i<0$, the coupling decreases towards high energies, giving asymptotic freedom, and grows towards the infrared. The formal infrared pole identifies a strong-coupling scale, where the weak-coupling approximation no longer applies. These conclusions concern the small-coupling branch; a pole in the perturbative solution does not establish a pole in the exact theory. If $b_i=0$, higher-order terms decide the running.

From ${\alpha }_i=g_i^2/(4\pi )$,Integrating the one-loop beta function givesThese formulas retain the same particle content and neglect threshold corrections throughout the interval. For $b_i>0$, the formal pole is $\mu =m_Z\exp [1/(8\pi b_i{\alpha }_i(m_Z))]$; for $b_i<0$ the analogous scale lies below $m_Z$.

For one-loop normalized hypercharge unification, the normalized hypercharge coupling is ${\alpha }_Y^{\mathrm{GUT}}=(5/3){\alpha }_1$, so its inverse and one-loop slope are $(3/5){\alpha }_1^{-1}$ and $(3/5)b_1$. DefineEquality of the three normalized inverses at the unification scale givesEliminating $L_G$ provesThis expression assumes $(3/5)b_1\ne b_2$. If these slopes coincide, unification first requires $U_1=U_2$, and the displayed division is unavailable. A unification scale above $m_Z$ additionally requires the inferred $L_G$ to be positive. The relation is a consistency condition under the stated one-loop assumptions, not proof that the measured couplings unify without threshold effects.

For the two-loop running coupling, the claimed logarithmic asymptotic concerns the asymptotically free branch with ${\beta }_0>0$ and $\mu /\Lambda \to \infty$. SetThe differential equation becomes $dy/dL={\beta }_0+{\beta }_1/y={\beta }_0(1+c/y)$. Separating variables yieldsA change of the strong-coupling scale $\Lambda$ absorbs any additive constant $K$. Choose that scale so that $K=-c\log {\beta }_0$. On the large positive-$y$ branch the exact implicit relation is thenIt first gives $y\sim {\beta }_{0}L$. Substituting this back into the logarithm gives $y={\beta }_{0}L+c\log L+o(1)$. To determine the error rather than assume it, write $y={\beta }_{0}L+c\log L+r(L)$. Expansion of the exact implicit relation yieldsConsequently the mathematically correct two-loop asymptotic isMore precisely, the next term is $\boxed{{\beta }_1^2(\log L+1)/({\beta }_0^{3}L)}$. For ${\beta }_1\ne 0$ this is not $O(1/L)$: multiplying the remainder by $L$ makes it grow as $({\beta }_1^2/{\beta }_0^3)\log L$. No fixed change of $\Lambda$ can remove this term. Such a change adds a constant to $L$ and changes only constant or $1/L$ contributions, rather than the coefficient of $\log L/L$. Thus the two leading terms requested are correct, but the literal remainder printed in the PDF is too small. This is the two-loop inverse-coupling logarithmic remainder.

If ${\beta }_1=0$, the one-loop result $y={\beta }_{0}L$ is exact after choosing $\Lambda$. If ${\beta }_0=0$, the displayed expansion is undefined and the differential equation instead gives $y^2=2{\beta }_1\log \mu +\text{constant}$. For ${\beta }_0<0$, a positive weak coupling does not approach zero at arbitrarily large $\mu$ along the branch used above. The asymptotic assumptions therefore matter as well as the remainder.