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

$f:\mathbb{R}\to \mathbb{R}$ is Riemann integrable on any bounded interval and $\lim_{x\to\infty} f(x)=0$.

Define $g(x)=\int_{x}^{x+1}f(t)dt$, we need to show $\lim_{x\to\infty} g(x)=0$.

Please give me hint, I want to try myself. Thank you, I was trying to apply fundamental theorem of calculus, like taking the derivative of $g$ which is $f(x+1)-f(x)$ but then don't know what to do.

share|cite|improve this question
The proof I just gave holds if $f$ is continious – Amr Jan 13 '13 at 10:06
@Amr you should have edited it instead of deleting the whole answer.anyway thank you for response. – Un Chien Andalou Jan 13 '13 at 10:08
I will edit it now – Amr Jan 13 '13 at 10:08
fixed it ${}{}{}{}{}{}{}{}$ – Amr Jan 13 '13 at 10:12
@Amr I don't see what's wrong, i.e. where you require $f$ to be continuous ... – Hagen von Eitzen Jan 13 '13 at 10:13
up vote 2 down vote accepted


Choose a large number $M$ such that $\forall x>M [|f(x)|<\epsilon]$. Now it follows thatfor all $x>M$:

$$\left|\int_x^{x+1} f(x) \right|\leq \int_x^{x+1} |f(x)|dx \leq \int_x^{x+1} \epsilon \,dx=\epsilon $$

share|cite|improve this answer

I guess you could use as well the following (part of the integral MVT):

$$|g(x)|\leq (x+1-x)\cdot \max_{t\in [x,x+1]}|f(t)|\xrightarrow[x\to\infty\Longrightarrow t\to\infty]{}0$$

since $\,t\to\infty\,$ as $\,x\to\infty\,$ and $\,f(x)\xrightarrow[x\to\infty]{}0\,$

share|cite|improve this answer
This was my first solution. Integral MVT holds for continuous functions. It is not mentioned in the question that $f$ is continuous . – Amr Jan 21 '13 at 19:38
I shall delete the MVT reference to avoid misunderstandings, but the above's true for functions as given in the OP. Put $$M_x:=\max_{t\in[x,x+1]}|f(t)|\Longrightarrow M_x\xrightarrow[x\to\infty]{}0$$ and from here $$|g(x)|\leq\int\limits_x^{x+1}|f(t)|\,dt\leq M_x\xrightarrow [x\to\infty]{} 0$$ – DonAntonio Jan 21 '13 at 20:18

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.