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I was asked to prove or disprove the following:

If $\int_{1}^{\infty}f(x)dx$ converges then $\lim_{x\to\infty}f(x) = 0$.

I said that this is false and gave this example:

$f(x) = \left\{ \begin{array}{ll} 1 & \mbox{if } x \in \mathbb{Q} \\ -1 & \mbox{if } x \notin \mathbb{Q} \end{array} \right.$

$\int_{1}^{\infty}f(x)dx = 0$ and $\lim_{x\to\infty}f(x)$ does not exist.

Was my example correct? And could anyone please elaborate more on this topic and write a little about how I should think in case I encounter similar T/F questions in the future?

Amir
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    Dirichlet function is not Riemann integrable – MathematicsStudent1122 Jun 18 '16 at 16:42
  • Can you explain more? I was not told $f$ is continuous – Amir Jun 18 '16 at 16:45
  • $f$ does not have to be continuous. But it must be integrable, obviously, otherwise the question makes no sense. – MathematicsStudent1122 Jun 18 '16 at 16:46
  • @MathematicsStudent1122 I edited my post... Can you please explain why the function I provided is not Reimann integrable? – Amir Jun 18 '16 at 16:58
  • Continuous example(s): https://en.wikipedia.org/wiki/Fresnel_integral – Winther Jun 18 '16 at 16:58
  • For the OP function, the integral does not converge. – GEdgar Jun 18 '16 at 17:21
  • Why? Isn't $\delta x = \frac{b-a}{n} = 0$ always? – Amir Jun 18 '16 at 17:24
  • A function is Riemann integrable if and only if its set of discontinuous points has measure $0$. Your function is discontinuous at every point. In case you have not learned about measure, the real line does not have measure $0$. – J126 Jun 18 '16 at 17:54
  • See the counterexample mentioned in this answer http://math.stackexchange.com/a/1823476/72031 Here the counter examples takes non-negative values only. Thus infinite integrals (integrals on unbounded intervals) are somewhat similar to infinite series but there are some differences (the $n$-th term of convergent series goes to zero, but a similar counterpart does not exist in the theory of convergent integrals). – Paramanand Singh Jun 19 '16 at 05:50

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This is an old friend. If $f$ may be discontinuous, as a counterexample you can propose $$ f(x) = \chi_{\{n \mid n \in \mathbb{N}\}}(x) = \begin{cases} 1 &\text{if $x \in \mathbb{N}$}\\ 0 &\text{otherwise}. \end{cases} $$ If you want a continuous counterxample, you must play with "bump" functions, for instance a function that is mostly zero but that has small bumps of smaller and smaller area.

Siminore
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  • Thanks! What is this function you wrote? I'm not familiar with it. And, can you provide a continuous counterexample please? – Amir Jun 18 '16 at 16:49
  • @JaneR You should try to construct a counterexample by yourself. Start from the discontinuous example above, and draw small triangle with one side on the $x$-axis. Now you should make these triangles smaller and smaller, so that the infinite sum of their areas is convergent. See also http://math.stackexchange.com/questions/85975/integrable-function-f-on-mathbb-r-does-not-imply-that-limit-fx-is-zero – Siminore Jun 18 '16 at 17:09
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The statement is false in general. See Fresnel-Integrals $$\int_{0}^{\infty}\cos(t^2)=\int_{0}^{\infty}\sin(t^2)=\frac{\sqrt{2\pi}}{4} $$

Soumik Mukherjee
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MrYouMath
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