Suppose $ f(x) >0$ and $\int_{a}^{\infty} \frac{1}{f(x)} = \infty$. Is it necessarily true that $\int_{a}^{\infty} \frac{1}{\ln(x)f(x)} = \infty$? I assume the answer is false but I can't think of a counterexample.
More generally is there a function $g(x)>0$ with the property that $\lim_{x \to \infty} g(x) = \infty$ and $\int_{a}^{\infty} \frac{1}{f(x)} = \infty$ implies that $\int_{a}^{\infty} \frac{1}{g(x)f(x)} = \infty$
Here is a basic counter example:
Let $u=\ln(x)$ be a substitution.
$$\int_e^\infty\frac1{x\ln(x)}\ dx=\int_1^\infty\frac1u\ du\to\infty$$
But, on the contrary,
$$\int_e^\infty\frac1{x\ln(x)\ln(x)}\ dx=\int_1^\infty\frac1{u^2}\ du=1<\infty$$
As per the second problem, just have $f(x)g(x)=x^ph(x)$, thus, $f(x)=\frac{x^ph(x)}{g(x)}$. For $g(x)\to\infty$, appropriately choose $p,h(x)$ such that $f(x)\to\infty$ and $\int_a^\infty\frac1{f(x)}\ dx$ diverges while $\int_a^\infty\frac1{f(x)g(x)}\ dx$ converges.