Which of the following functions is Riemann- integrable in $[0,1]$ ?

  1. $f(x)= \sin(\ln x ) ) , x\neq 0 $ ,$ f(0)=0 $ .

  2. $ f(x) = \frac{1}{x} \sin(\frac{1}{x} ) , x\neq 0 $ ,$ f(0)=0 $ .

  3. $f(x)= \frac{\sin(x)}{x} , x\neq 0 $ , $f(0)=0$.

all the functions here are bounded, but none is continuous. How can I determine whether or not they are integrable (I should not check Riemann sums...) ?

Maybe I can calculate these integrals somehow?

Thanks in advance !

  • $\begingroup$ They all look integrable to me. The set of discontinuities of each function is a set of measure zero, thus they are integrable. $\endgroup$ – Wintermute Jan 12 '14 at 21:51
  • $\begingroup$ I don't think the function in 2. is bounded. $\endgroup$ – David Mitra Jan 12 '14 at 21:52
  • $\begingroup$ @DavidMitra $\displaystyle{\large 2.}$ becomes $\displaystyle{\large \int_{1}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x}$ with the change $\displaystyle{\large x \to {1 \over x}}$. $\endgroup$ – Felix Marin Jan 12 '14 at 22:19
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    $\begingroup$ @FelixMarin Yes. But the function in 2. is not Riemann integrable over $[0,1]$, since it's unbounded there. $\endgroup$ – David Mitra Jan 12 '14 at 22:29
  • $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Jan 13 '14 at 5:36

A bounded function is Riemann integrable if and only if its set of discontinuities is very small (namely, Lebesgue measure zero; but countable or finite does it). Each of your functions is a composition and/or multiplication of continuous functions away from $0$, so each function is certainly continuous on $(0, 1]$, so the set of discontinuities for each is contained in $\{0\}$, which certainly doesn't change integrability.

Function $(1)$ is certainly bounded, since $\sin$ is on the outside. Function $(3)$ is bounded because

$$\lim_{x \to 0} \frac{\sin x }{x} = 1$$

However, function $(2)$ is not bounded; you may find it useful to consider the behaviour of

$$ t \sin t$$

as $t \to \infty$.

  • $\begingroup$ Can you tell me how is your statement for Lebesgue criteria differs from the my downvoted and deleted answer? You are using "iff" and "finite or countable does it". By the way, you are one of those who deleted the answer. $\endgroup$ – Mhenni Benghorbal Jan 13 '14 at 5:25
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    $\begingroup$ @MhenniBenghorbal Your answer stated that a function is Riemann integrable on $[a, b]$ iff i) it's bounded and ii) It has a countable number of discontinuities. This is false, leading to my downvotes: there are functions with uncountably many points of discontinuity that are still Riemann integrable (e.g. the indicator function on the Cantor set). My answer gives the statement that whenever the set of discontinuities is finite or countable, then the function is integrable, and does not assert anything more. $\endgroup$ – user61527 Jan 13 '14 at 5:51
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    $\begingroup$ @MhenniBenghorbal In short, the 'if' is true and the 'iff' is false. $\endgroup$ – user61527 Jan 13 '14 at 5:52
  • $\begingroup$ Dear all, I now understand 1 and 3, but not 2. How does the statement that a Riemann integrable function must be bounded agree with that the function $\frac{1}{\sqrt{x} }$ is integrable in [0,1] but not bounded near 0? Thanks ! $\endgroup$ – Integral Professor Jan 13 '14 at 17:39
  • $\begingroup$ @IntegralProfessor It's not Riemann integrable in the classical sense, but rather in the usual sense of an improper integral. $\endgroup$ – user61527 Jan 13 '14 at 18:40

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