Improper Integral of a Monotonic Function

[Bashyboy]

Homework Statement

Let $f: (1, \infty) \to [0,\infty)$ be a function such that the improper integral $\int_{1}^{\infty} f(x)dx$ converges. If $f$ is monotonically decreasing, then $\lim_{x \to \infty} f(x)$ exists.

Homework Equations

The Attempt at a Solution

This problem doesn't come from a textbook, so there are no restrictions on what theorems can be used. However, I would prefer an elementary solution. My strategy is to show that if $f$ is not bounded below, then $\int_{1}^{\infty} f(x)dx$ couldn't possibly converge. This would mean that $f$ would have to bounded below in addition to being monotonically decreasing, which I believe would imply that $\lim_{x \to \infty} f(x)$ exists. I could use some hints.

 

[Dick]

Homework Statement

Let $f: (1, \infty) \to [0,\infty)$ be a function such that the improper integral $\int_{1}^{\infty} f(x)dx$ converges. If $f$ is monotonically decreasing, then $\lim_{x \to \infty} f(x)$ exists.

Homework Equations

The Attempt at a Solution

This problem doesn't come from a textbook, so there are no restrictions on what theorems can be used. However, I would prefer an elementary solution. My strategy is to show that if $f$ is not bounded below, then $\int_{1}^{\infty} f(x)dx$ couldn't possibly converge. This would mean that $f$ would have to bounded below in addition to being monotonically decreasing, which I believe would imply that $\lim_{x \to \infty} f(x)$ exists. I could use some hints.

If $f$ is not bounded below then there exists a $c$ such that $f(x)<-1$ for all $x>c$, right?

 

[Ray Vickson]

Homework Statement

Let $f: (1, \infty) \to [0,\infty)$ be a function such that the improper integral $\int_{1}^{\infty} f(x)dx$ converges. If $f$ is monotonically decreasing, then $\lim_{x \to \infty} f(x)$ exists.

Homework Equations

The Attempt at a Solution

This problem doesn't come from a textbook, so there are no restrictions on what theorems can be used. However, I would prefer an elementary solution. My strategy is to show that if $f$ is not bounded below, then $\int_{1}^{\infty} f(x)dx$ couldn't possibly converge. This would mean that $f$ would have to bounded below in addition to being monotonically decreasing, which I believe would imply that $\lim_{x \to \infty} f(x)$ exists. I could use some hints.

There is nothing to prove: you said that $f(x) \in [0,\infty)$, so $f$ is automatically bounded below by 0.

 

[Dick]

There is nothing to prove: you said that $f(x) \in [0,\infty)$, so $f$ is automatically bounded below by 0.

Good point.

 

[Bashyboy]

There is nothing to prove: you said that $f(x) \in [0,\infty)$, so $f$ is automatically bounded below by 0.

Ah! Very keen eye! So the fact that the interval converges has no bearing on whether $\lim_{x \to \infty} f(x)$ exists? What if we were to remove the condition and just stipulate that $f$ is a function from $(1,\infty)$ to $\Bbb{R}$? Would I follow Dick's suggestion in that case?

 

[Dick]

Ah! Very keen eye! So the fact that the interval converges has no bearing on whether $\lim_{x \to \infty} f(x)$ exists? What if we were to remove the condition and just stipulate that $f$ is a function from $(1,\infty)$ to $\Bbb{R}$? Would I follow Dick's suggestion in that case?

I would.

 

[Ray Vickson]

Ah! Very keen eye! So the fact that the interval converges has no bearing on whether $\lim_{x \to \infty} f(x)$ exists? What if we were to remove the condition and just stipulate that $f$ is a function from $(1,\infty)$ to $\Bbb{R}$? Would I follow Dick's suggestion in that case?

I don't know what you mean when you say "the interval converges". Intervals do not converge; they just sit there, silently. However, integrals may converge.

Of course $\lim_{x \to \infty} f(x)$ exists, because $f$ is a bounded monotonic function. (There are theorems about that; you should look them up.)

The only remaining question is what the limit actually turns out to be.