Suppose a positive function $f:[1,\infty) \to (0,\infty)$ has $\int_1^\infty \frac{f(x)}{x} dx < +\infty$. Is it also possible that $\int_1^\infty \frac{1}{xf(x)} dx$ converges?

I can make progress when $f(x)$ is monotone decreasing. Since $f(x)/x$ is also decreasing and the improper integral converges it must hold that $\lim_{x \to \infty} xf(x)/x = \lim_{x \to \infty} f(x) = 0$. Then there exists $C > 0$ such that $f(x) < 1$ and $1/(xf(x)) > 1/x$ for $x > C$. By the comparison test the integral of $1/(xf(x))$ diverges since $\int_C^\infty dx/x = + \infty$.


Note that as $c \to \infty$,

$$\int_1^c \frac{f(x)}{x} \, dx + \int_1^c \frac{dx}{x f(x)} = \int_1^c\left(f(x) + \frac{1}{f(x)} \right) \frac{dx}{x} \geqslant 2 \int_1^c\frac{dx}{x} \to +\infty$$

Both integrals on the LHS cannot converge.


if the integral $\int_1^\infty \frac{f(x)}{x} dx$ would converge, we would need

1.$f(x)$ to be differentiable, specially, $f'(x)$ to be bounded, $|f(M)-f(N)|<C $.Where M,N are boundaries of the integral.

  1. 1/x to drop, monotonicly. Combining 1 and 2 we can conclude the integral converges by the dirihles condition. Using the fact $ f(x) $ must be bounded we can conclude. With this, the integral $\int_1^\infty \frac{1}{xf(x)} dx$ can be bounded from above with $C \cdot\int_1^\infty \frac{1}{x} dx$ and this integral obviously diverges, as it goes to log(+infinity).
  • $\begingroup$ Completely illogical answer. $\endgroup$ – Kavi Rama Murthy Oct 17 '19 at 23:42
  • $\begingroup$ To me , it is completely logical. Such integrals like $\int_1^\infty \frac{f(x)}{x} dx$ usually converge only if $f(x)$ is bounded, such as sinx,cosx...Taking into the consideration this, integral $\int_1^\infty \frac{1}{xf(x)} dx$ cannot converge. $\endgroup$ – Vuk Stojiljkovic Oct 18 '19 at 7:33

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