# Improper Integral of a Monotonic Function

## 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.

## 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 Helper

## 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.

## 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 Helper
Dearly Missed

## 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.

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

Good point.

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
Homework Helper
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.

Bashyboy
Ray Vickson
Homework Helper
Dearly Missed
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.