If $\Phi\geq 0$ is non-decreasing, does $\int_1^\infty \frac{\Phi(x)}{x^2}\,dx=\infty$ imply $\int_0^\infty e^{-\Phi(x)}\,dx<\infty$? 
Q. Suppose $\Phi : [0, \infty) \to [0, \infty)$ is a non-decreasing function. If
$$ \int_{1}^{\infty} \frac{\Phi(x)}{x^2} \, \mathrm{d}x = \infty, $$
then does the inequality
$$ \int_{0}^{\infty} e^{-\Phi(x)} \, \mathrm{d}x < \infty $$
hold?

This question is motivated by this posting, but it became a statement of independent interest to me.
I suspect that the statement is true, based on a naive observation that a higher growth rate of $\Phi$ is in favor of both conditions and a slower growth rate is against them.
One obvious observation is that $\Phi(x)$ must tend to $\infty$ as $x \to \infty$. But, to be honest, I have no clue as to where I should start. So I would like to invite you to this strange yet interesting problem!

Edit. The user @Sungjin Kim showed that the answer is false by providing a family of counter-examples. Here is a simplification of his answer:
For each strictly increasing sequence $(b_k)_{k=1}^{\infty}$ in $[1, \infty)$, let
$$ \Phi(t) = \sum_{k \mathop{:} b_k \leq t} b_k = \sum_{k=1}^{\infty} b_k \mathbf{1}_{[b_k, \infty)}(t). $$
Then it follows that
$$ \int_{1}^{\infty} \frac{\Phi(t)}{t^2} \, \mathrm{d}t
= \sum_{k=1}^{\infty} b_k \int_{b_k}^{\infty} \frac{\mathrm{d}t}{t^2}
= \sum_{k=1}^{\infty} 1
= \infty. $$
On the other hand, since $\Phi(t) = \sum_{l=1}^{k} b_l$ for each $t \in [b_k, b_{k+1})$,
\begin{align*}
\int_{0}^{\infty} e^{-\Phi(t)} \, \mathrm{d}t
\geq \sum_{k=1}^{\infty} \int_{b_k}^{b_{k+1}} e^{-\Phi(t)} \, \mathrm{d}t
= \sum_{k=1}^{\infty} (b_{k+1} - b_k) e^{-\sum_{l=1}^{k} b_l}.
\end{align*}
So by choosing $(b_{k+1})$ so that it satisfies $(b_{k+1} - b_k) e^{-\sum_{l=1}^{k} b_l} \geq 1$ for each $k$, we have
$$ \int_{0}^{\infty} e^{-\Phi(t)} \, \mathrm{d}t = \infty. $$
 A: 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
$$
\int_1^{\infty} e^{-\Phi(x)} dx.
$$
