# Convergence of related improper integrals

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)|.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).
• Completely illogical answer. – Kavi Rama Murthy Oct 17 '19 at 23:42
• 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. – Vuk Stojiljkovic Oct 18 '19 at 7:33