= Solution
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 $I(X,\lambda X)\to0$ and $I(X/\lambda,X)\to0$ as $X\to\infty$. By monotonicity,
$$
I(X,\lambda X)\ge\beta(X)\int_X^{\lambda X}x^{-2}dx-\log\lambda
=(1-\lambda^{-1})\frac{\beta(X)}X-\log\lambda,
$$
and
$$
I(X/\lambda,X)\le(\lambda-1)\frac{\beta(X)}X-\log\lambda.
$$
Thus
$$
\frac{\log\lambda}{\lambda-1}\le\liminf_{X\to\infty}\frac{\beta(X)}X
\le\limsup_{X\to\infty}\frac{\beta(X)}X\le\frac{\log\lambda}{1-\lambda^{-1}}.
$$
Both outer bounds tend to one as $\lambda\downarrow1$, giving $\boxed{\beta(X)/X\to1}$. 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.
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