Convergence of an integral involving the radical of an integer versus the convergence of $\int_2^\infty\frac{dx}{x(\log x)^2}$

As pointed out by other users, we may study the summation version instead. Let $\mathbb{P} = \{p_1,p_2,\cdots\}$ be the set of primes and $\mathbb{N} = \{1,2,\cdots\}$ be the set of positive integers. Then for $\alpha, \beta > 0$ let us consider the sum

$$ S_{\alpha,\beta} := \sum_{n=2}^{\infty} \frac{1}{(\operatorname{rad}n)^{\alpha} (\log n)^{\beta}} = \frac{1}{\Gamma(\beta)} \sum_{n=2}^{\infty} \frac{1}{(\operatorname{rad}n)^{\alpha}} \int_{0}^{\infty} t^{\beta-1} n^{-t} \, dt. $$

Here, the equality follows from the gamma integral $ \int_{0}^{\infty} t^{\beta-1} e^{-st} \, dt = \Gamma(\beta)s^{-\beta}$ for $\beta, s > 0$. Now we partition $n$'s according to the value of radicals. Then resulting classes can be naturally indexed by non-empty finite subsets $I$ of $\mathbb{N}$ in the following way.

$$ \bigg\{ n \in \mathbb{N} : \operatorname{rad}(n) = \prod_{i\in I} p_i \bigg\} = \bigg\{ \prod_{i\in I} p_i^{k_i} : (k_i)_{i\in I} \in \mathbb{N}^I \bigg\} $$

Using this partition, we find that

\begin{align*} S_{\alpha,\beta} &= \frac{1}{\Gamma(\beta)} \sum_{\substack{I \subset \mathbb{N} \\ 0<|I|<\infty}} \sum_{(k_i)_{i\in I} \in \mathbb{N}^{I}} \int_{0}^{\infty} t^{\beta-1} \left( \prod_{i\in I} p_i^{-\alpha-k_i t} \right) \, dt \\ &= \frac{1}{\Gamma(\beta)} \sum_{\substack{I \subset \mathbb{N} \\ 0<|I|<\infty}} \int_{0}^{\infty} t^{\beta-1} \left( \prod_{i\in I} \frac{1}{p_i^{\alpha}(p_i^t - 1)} \right) \, dt \tag{1} \\ &= \frac{1}{\Gamma(\beta)} \int_{0}^{\infty} t^{\beta-1} \left[ \prod_{i=1}^{\infty} \left( 1 + \frac{1}{p_i^{\alpha}(p_i^t - 1)} \right) - 1 \right] \, dt. \tag{2} \end{align*}

Here, interchanging the order of summations and integrals is justified by the Tonelli's theorem. Also we utilized geometric series formula to obtain $\text{(1)}$.

Now using the inequality $e^x - 1 \leq x e^x$, we find that each integral in $\text{(1)}$ satisfies

$$ \int_{0}^{\infty} t^{\beta-1} \left( \prod_{i\in I} \frac{1}{p_i^{\alpha}(p_i^t - 1)} \right) \, dt \geq \int_{0}^{\infty} t^{\beta-|I|-1} \left( \prod_{i\in I} \frac{1}{p_i^{\alpha +t} \log p_i} \right) \, dt, $$

which diverses if $|I| \geq \beta$. So the sum diverges for all $\alpha, \beta > 0$ and hence the same is true for the integral:

$$ \int_{2}^{\infty} \frac{dx}{(\operatorname{rad}\lfloor x\rfloor)^{\alpha}(\log x)^{\beta}} = \infty. $$