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A smoothed form of the summatory totient function satisfiesUse the Dirichlet series $\zeta (s-1)/\zeta (s)$ and the kernel $x^s/[s(s+1)]$ to write the left side as a vertical contour integral. Move the line from real part greater than two to $3/2$. The only crossed pole is at two, with residue $x^3/(6\zeta (2))=x^3/{\pi }^2$; the remaining vertical integral is $O(x^{5/2})$ by the standard zeta bounds in the half-strip. If $A(x)={\sum }_{n\le x}\phi (n)$ and $S(x)={\int }_0^{x}A(t)dt$, positivity of the summands makes $A$ nondecreasing. The backward and forward difference quotients of $S$ bound $A(x)$. Insert $S(x)\sim x^3/{\pi }^2$, then let the relative difference-quotient step tend to zero, proving the displayed asymptotic. This monotone integral Tauberian lemma justifies the passage from a smoothed estimate to an unsmoothed one.

The improper integral is assumed to have a finite limit. Fix $\lambda >1$ and put $I(U,V)={\int }_U^V(\beta (x)-x)x^{-2}dx$. Convergence implies $I(X,\lambda X)\to 0$ and $I(X/\lambda ,X)\to 0$ as $X\to \infty$. By monotonicity,andThusBoth outer bounds tend to one as $\lambda \downarrow 1$, giving $\boxed{\beta (X)/X\to 1}$. This monotone integral Tauberian lemma uses monotonicity to control the value at a point from integral averages on adjacent multiplicative intervals; convergence of the integral alone would not control arbitrary local spikes.