Do these conditions always imply convergence?

Question

Let's say I have a continuous function $f:[1,\infty)\to[1,\infty)$. Does this imply that either $\int_{1}^{\infty}{f(x)}dx$ or $\int_{1}^{\infty}{1/f(x)}dx$ converges?

I'm quite certain this is true but struggle to prove it rigorously. Any insight is appreciated!

Answer 1

$f(x)=x$ restricted to the given domain (and thus range) satisfy the hypothesis but fail the convergence of both integrals.

As stated, it is false by counter-example.

Answer 2

Since $f$ is continuous on $[1,\infty)$ and $f(x)\geq 1$ for all $x\in[1,\infty)$, we have:

$$\int_{1}^{\infty}f(x)dx \geq \int_{1}^{\infty}dx$$

which is a divergent integral. Therefore, if $\int_{1}^{\infty}1/f(x)dx$ converges, then $\int_{1}^{\infty}f(x)dx$ must also converge.

On the other hand, if $\int_{1}^{\infty}1/f(x)dx$ diverges, then since $1/f(x)\leq 1$ for all $x\in[1,\infty)$, we have:

$$\int_{1}^{\infty}f(x)dx \leq \int_{1}^{\infty}dx$$

which is a convergent integral. Therefore, if $\int_{1}^{\infty}f(x)dx$ converges, then $\int_{1}^{\infty}1/f(x)dx$ must also converge.

Therefore, we have shown that if $f:[1,\infty)\to[1,\infty)$ is a continuous function, then either $\int_{1}^{\infty}f(x)dx$ or $\int_{1}^{\infty}1/f(x)dx$ converges, but not necessarily both of them.

An Example (Double divergence)

Consider the function $f(x) = \frac{1}{x}$ for $x \in [1,\infty)$. We have

$$\int_1^\infty f(x)dx = \int_1^\infty \frac{1}{x}dx = \lim_{t \to \infty} \ln(t) = \infty,$$

which diverges. However, we also have

$$\int_1^\infty \frac{1}{f(x)}dx = \int_1^\infty xdx = \lim_{t \to \infty} \frac{t^2}{2} - \frac{1}{2} = \infty,$$

which also diverges. Therefore, neither $\int_1^\infty f(x)dx$ nor $\int_1^\infty \frac{1}{f(x)}dx$ converges.

Another Example (One convergence and One divergence)

Consider the function $f(x) = e^{-x}$ defined on the interval $[1, \infty)$.

We can see that $f$ is continuous on $[1,\infty)$ and also bounded below by $e^{-1}$ since $f(x) \geq e^{-x}$ for all $x \in [1, \infty)$. Thus, $\int_{1}^{\infty} f(x) dx$ may converge.

However, we have

$$\int_{1}^{\infty} \frac{1}{f(x)} dx = \int_{1}^{\infty} \frac{1}{e^{-x}} dx = \int_{1}^{\infty} e^{x} dx = \infty,$$

which diverges. Therefore, not both $\int_{1}^{\infty} f(x) dx$ and $\int_{1}^{\infty} \frac{1}{f(x)} dx$ can converge.

Answer 3

It is not true that either one or the other integrals must converge. As commented, both integrals diverge in the case $f: x \mapsto 1$.

Perhaps a more interesting question is can both integrals converge — even when the range of $f$ is in $(0,\infty)$. The answer is no.

Since $a + 1/a \geqslant 2$ for any positive real number $a$, we have

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

and, thus,

$$\lim_{c \to \infty}\left[ \int_1^c f(x) \, dx + \int_1^c \frac{dx}{f(x)} \right] = +\infty$$

This implies that if $\int_1^\infty f(x) \, dx$ is finite then $\int_1^\infty \frac{dx}{f(x)}$ is infinite and vice versa.