Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is not homework.

I'd love your help proving that if $f$ is an unbounded monotonic increasing function, then $$\lim_{x \to \infty} \frac{1}{x}\int_{0}^{x} \ f(t)\, dt=\infty.$$ I want to use it in a couple of proofs, but I can't prove it by myself.

Thanks a lot.

share|cite|improve this question
If you say "$f$ is an unbounded monotonic increasing function on $[0,\infty)$", then you rule out the special cases that make your statement false. In particular, this guarantees that $f$ is bounded below, and therefore unbounded above. – TonyK Dec 7 '11 at 19:14
up vote 4 down vote accepted

I think the l'Hôpital's rule may be helpful here if your $f$ satisfies the requirements of the rule. And say the antiderviative of $f$ is $F(x)$, then

$$ \lim_{x\to\infty}\frac{1}{x}\int_{0}^{x}f(t)dt = \lim_{x\to\infty}\frac{F(x) - F(0)}{x} = \lim_{x\to\infty}f(x) .$$

Since your tag is calculus, not something like real analysis, I assume your function is a relatively normal function involving Riemann integral.

share|cite|improve this answer

You need to assume that the integral $\int_0^x f(t) dt$ is always finite ; it can be possible that it is $-\infty$, because $f$ might go to $-\infty$ when $x \to 0^+$ and this might cause problems. Assume that the integral from $0$ to some point $a$ is finite, so that we don't have any problems. Thus without loss of generality we can assume that $f$ is positive, because at some point $c$, $f(x)$ must be positive for all $x > c$ and we can separate the integral in two parts : the one before $c$ and the one after. Since the integral from $0$ to $c$ will be finite, we have that

Then $$ \frac 1x \int_c^x f(t) dt \ge \frac 1x \int_{x/2}^x f(t) \, dt \ge \frac {(x/2) f(x/2)}{x} = \frac {f(x/2)}2 \to \infty. $$

Hope that helps,

share|cite|improve this answer

Hint If $f$ is unbounded and increasing function, prove that $\lim_{x \to \infty} f(x) =\infty$.

Then L'Hôpital solves the problem.

share|cite|improve this answer
You named it first. – dfeuer Dec 5 '11 at 7:41

$\newcommand{\d}{\;\mathrm{d}}$My try.

First of all, note that you have to assume that: $$\forall x>0,\quad \int_0^x f(t)\d t>-\infty\; ,$$ for otherwise the limit is $-\infty$ and the claim is not true.

Moreover, w.l.o.g. you can assume that $f\,$ has finite integral over each interval $[0,x]$ (because otherwise your limit is trivial).

Since $f\,$ is unbounded from above and increasing, when you choose $n\in \mathbb{N}$ you can always find $x_n\geq 0$ such that for all $x\geq x_n$, $$f(x)\geq n\qquad \text{and}\qquad \int_0^x f(t)\d t\geq 0\; ;$$ then for $x\geq x_n$ you get: $$\frac{1}{x}\int_{x_n}^x f(t)\d t \geq \frac{n(x-x_n)}{x}\; .$$ Thus: $$\frac{1}{x}\int_0^x f(t)\d t \geq \frac{1}{x}\int_{x_n}^x f(t)\d t\geq \frac{n(x-x_n)}{x}$$ and: $$\tag{1} \liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t \geq \liminf_{x\to \infty} \frac{n(x-x_n)}{x} =n\; ;$$ inequality (1) proves that $\liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\ \text{d} t$ exceeds $n$ for all $n\in \mathbb{N}$, hence you necessarily have: $$\liminf_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t =+\infty$$ therefore: $$\lim_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t =+\infty$$ as you claimed.

Just a small remark. Under your assumptions on $f$, the function $F(x):=\int_0^x f(t)\ \text{d} t$ is convex. Thus relation: $$\lim_{x\to \infty} \frac{1}{x}\int_0^x f(t)\d t=\infty$$ expresses the fact that $F$ is also coercive.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.