Convergence of improper integral?

Question

Consider an improper integral such that: $$I = \int_0^{+\infty} \frac{f(x)}{x}dx.$$

If $\int_0^{+\infty}f(x)dx < + \infty$, Can we conclude that the integral I converges? Thanks for any answer or suggestion.

Answer 1

No. Consider $f=1_{[0,1]}$, that is, $f(x)=1$ when $x\in[0,1]$, and $0$ otherwise.

Answer 2

Other example. Take $f(x)=e^{-x}$.

Answer 3

The answer is a partial YES and a partial NO.

If you split the Riemann improper integral into two pieces:

$$\int_0^\infty \frac{f(x)}{x} dx \stackrel{def}{=} \lim_{\Lambda \to \infty, \lambda \to 0} \int_\lambda^\Lambda \frac{f(x)}{x} dx = \lim_{\Lambda\to\infty} \int_1^\Lambda \frac{f(x)}{x} dx + \lim_{\lambda\to 0} \int_\lambda^1 \frac{f(x)}{x}dx $$ The first piece for large $x$ exists. This is because the condition $$\int_0^\infty f(x)dx \stackrel{def}{=} \lim_{\Lambda\to\infty,\lambda \to 0} \int_\lambda^\Lambda f(x)dx < \infty$$ implies as a function of $\Lambda$, the integral $\displaystyle\;\int_1^\Lambda f(x) dx\;$ converges and hence bounded as $\Lambda \to \infty$.

This in turn implies the collection of integrals $\displaystyle\;\int_a^b f(x) dx\;$ for $(a,b) \subset (1,\infty)$ are uniformly bounded. Since $\frac{1}{x}$ is monotonic decreasing to $0$ as $x \to \infty$, Dirichlet's test for improper integral tell us following limit exists: $$\lim_{\Lambda\to\infty} \int_1^\Lambda \frac{f(x)}{x} dx$$

However, this doesn't mean $\displaystyle\;\int_0^\infty \frac{f(x)}{x} dx\;$ exists. This is because the second piece for small $x$, $\displaystyle\;\lim_{\lambda\to 0}\int_\lambda^1 \frac{f(x)}{x} dx\;$ can blow up. A simple example is take $f(x)$ to be any function which equal to $1$ on $[0,1]$. Independent how well behaved is $f(x)$ for $x > 1$, we have

$$\int_\lambda^1 \frac{f(x)}{x}dx = \int_\lambda^1 \frac{1}{x} dx = -\log\lambda \to +\infty \quad\text{ as }\quad\lambda \to 0$$