If $\Phi\geq 0$ is non-decreasing, does $\int_1^\infty \frac{\Phi(x)}{x^2}\,dx=\infty$ imply $\int_0^\infty e^{-\Phi(x)}\,dx<\infty$?

We may assume $\Phi(1)=0$. Let $\Phi(x)=\int_1^x f(t) dt$ for some nonnegative function $f$. By assumption, we have $$ \int_1^{\infty} \int_1^x \frac{f(t)}{x^2}dtdx =\int_1^{\infty}\int_t^{\infty} \frac{f(t)}{x^2} dxdt = \int_1^{\infty} \frac{f(t)}tdt=\infty. $$ Let $$ a_{\ell} = \int_{2^{\ell-1}}^{2^{\ell}} \frac{f(t)}t dt. $$ We have $\sum_{\ell=1}^{\infty} a_{\ell} = \infty$ and $$ \int_{2^{\ell-1}}^{2^{\ell}} f(t) dt \leq 2^{\ell}\int_{2^{\ell-1}}^{2^{\ell}} \frac{f(t)}t dt=2^{\ell}a_{\ell}. $$ On the other hand, by Cauchy condensation test, the convergence of $\int_1^{\infty} e^{-\Phi(x)} dx$ is equivalent to the convergence of LHS of $$ \sum_{k=1}^{\infty} 2^k e^{-\Phi(2^k)} \geq \sum_{k=1}^{\infty} 2^k e^{-\sum_{\ell\leq k} 2^{\ell}a_{\ell} }. $$ Let $a_{\ell}$ be a characteristic function of a very rapidly increasing sequence $\{b_n\}$ of natural numbers.

If $b_m\leq k < b_{m+1}$, we have $$ 2^k e^{-\sum_{\ell\leq k} 2^{\ell}a_{\ell} }\geq 2^{b_m} e^{-2^{1+b_m}}. $$ Then the sum over $k$ in this interval is $$ \geq (b_{m+1}-b_m)2^{b_m} e^{-2^{1+b_m}}. $$ For each $m$, we take $b_{m+1}$ large enough that $$ (b_{m+1}-b_m)2^{b_m} e^{-2^{1+b_m}}\geq 1. $$ Then we have the divergence of the series $$ \sum_{k=1}^{\infty} 2^k e^{-\sum_{\ell\leq k} 2^{\ell}a_{\ell} }. $$ Hence, we have the divergence of $$ \int_1^{\infty} e^{-\Phi(x)} dx. $$