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Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The chain rule turns the renormalization-group beta function intoIntegrating gives the one-loop inverse gauge coupling evolutionThis expression applies over a range with unchanged active field content and in the perturbative regime. It uses $\log \mu$, not $\log {\mu }^2$.

For gauge coupling unification, define the normalized inverse couplings $A_1=3/(5{\alpha }_1)$, $A_2=1/{\alpha }_2$, $A_3=1/{\alpha }_3$, and slopes $b_1=(3/5){\beta }_1$, $b_2={\beta }_2$, $b_3={\beta }_3$. At the common scale each inverse coupling is $A_U$. With $L=\log (M_{\mathrm{GUT}}/M_Z)$,Subtract the equations for factors 1 and 2 to solve $L=2\pi (A_1-A_2)/(b_1-b_2)$, then insert this into the difference of factors 3 and 2:This one-loop normalized hypercharge unification relation requires $b_1\ne b_2$. A scale above $M_Z$ also requires the inferred $L$ to be positive. If those two slopes coincide, unification instead requires $A_1=A_2$ and their difference cannot fix the scale.

In the convention $Q=T_3+Y$, one matter generation and one Higgs doublet have the following Standard Model representations. Spin is in units of $\hslash$; L/R describe chirality, not the spin projection. The $U(1{)}_Y$ entry is its hypercharge eigenvalue, defining the one-dimensional representation $e^{iY\omega }$.

There is no right-handed neutrino in the minimal field content being counted. Right-handed matter can alternatively be written as left-handed charge-conjugate Weyl fermions, in conjugate colour representations with opposite hypercharges. The quadratic indices and squared charges in the beta coefficients are unchanged.

For $SU(3)$, one generation supplies two colour fundamentals from $Q_L$ and one each from $u_R,d_R$. With three generations there are $n_f=3(2+1+1)=12$ Weyl fermions counted as fundamental copies, and no coloured scalar. ThusFor $SU(2)$, the three colours of $Q_L$ give three weak-doublet copies and $L_L$ gives one. There are $n_f=3(3+1)=12$ doublets across three generations and $n_s=1$ complex Higgs doublet, givingThese are spectator multiplicities in gauge beta functions: the colour index multiplies the weak contribution, while the weak index multiplies the colour contribution. Each two-component fermion is counted once; it must not be counted again as a separate antiparticle.

For $U(1{)}_Y$, sum squared hypercharges over all colour and weak components. One generation givesThe two complex Higgs components give ${\sum }_s{Y}_s^2=2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction contribution. ThereforeThe Higgs is counted only once for the whole model, whereas the fermionic generation is repeated three times. These are the Standard Model one-loop gauge coefficients in the unnormalized hypercharge convention. The grand-unified normalization instead has $b_1=41/10$. The ratio of slopes in the unification relation is $({\beta }_3-{\beta }_2)/(b_1-{\beta }_2)=-115/218$.

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.