We know that $1 + \frac{x f'}{f} < 0$.

I'm unsure how to proceed but guessing the proof involves knowing that:

If $\lim\_{x \to b} \frac{f(x)}{g(x)}$ exists and is greater than 0, then $\int\_a^b g(x)$ converges if and only if $\int\_a^b f(x)$ converges.

If $\lim\_{x \to b^-} (f(x))^\frac{1}{x}$ exists in the extended reals, then $\int\_a^b f(x)$ converges if the limit is less than 1 and diverges otherwise.