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If $\int_0^1 1 / \sqrt{x}\,\mathrm dx$ Riemann integrable then using second fundamental theorem of calculus I can easily say that $\sqrt{x}$ is uniformly continuous.

Basically it has one point i.e. $0$ where it diverges. Otherwise I can use the theorem that a function with finite number of discontinuous points is Riemann integrable. Can I take zero as point of discontinuity?

José Carlos Santos
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That function is unbounded (whatever way you choose it to extend it to $0$), and therefore it is not Riemann-integrable (by definition, every Riemann-integrable function is bounded). However, the improper integral $\int_0^1\frac1{\sqrt x}\,\mathrm dx$ converges.

The theorem that you mentioned holds for bounded functions, not in general.

José Carlos Santos
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  • This means when we talk about Riemann integrable functions we mean that it has to be bounded ? in other words we consider only bounded functions in study of Riemann integration ? –  Nov 27 '18 at 13:16
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    Yes, you are absolutely right. – José Carlos Santos Nov 27 '18 at 13:18
  • Also one more doubt - suppose a function is unbounded at a point ( like the function in this question is unbounded at $0$ ) then can we say the function is discontinuous at that point ? (This doubt is irrelevant to this question) –  Nov 27 '18 at 13:27
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    Yes, that is correct. To be more precise: if $a$ belongs to the domain $D_f$ of $f$ and if, for each $r>0$, the restriction of $f$ to $(a-r,a+r)\cap D_f$ is unbounded, then $f$ is discontinuous at $a$. – José Carlos Santos Nov 27 '18 at 13:30
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    @Upstart Yes, I mean what I say: the improper integral $\int_0^1\frac1{\sqrt x},\mathrm dx$ converges, since\begin{align}\int_0^1\frac1{\sqrt x},\mathrm dx&=\lim_{\varepsilon\to0}\int_\varepsilon^1\frac1{\sqrt x},\mathrm dx\&=\lim_{\varepsilon\to0}\left(2-2\sqrt\varepsilon\right)\&=2.\end{align} – José Carlos Santos Nov 10 '20 at 19:26
  • No No, what I was trying to ask here is given the OP's question if we add the part that $f(x)=0$ for $x=0$, then also it is unbounded since given any bound $M$ we can always find an $x=\frac{1}{(M+1)^2}$ which violates the condition of boundedness. Which is what you wanted to say I guess. To make it short even if we define like this even the the function is bounded. – Upstart Nov 10 '20 at 19:38
  • I wrote in my answer that “That function is unbounded (whatever way you choose it to extend it to $0$)”. – José Carlos Santos Nov 10 '20 at 19:42
  • Yes you are right, after my comment, I thought over your answer, and found my reasoning flawed. – Upstart Nov 10 '20 at 19:43
  • +1 for the answer and more importantly for the nice discussion in comments. – Paramanand Singh Nov 11 '20 at 06:01
  • If that function is not Riemann integrable then what kind of integrability method is that? – William Aug 27 '22 at 02:22
  • @William It's an improper integral. Did you read the comments? – José Carlos Santos Aug 27 '22 at 06:22
  • Just did and I looked it up. From what I understand, it's not a method of integration. It's more of a type of integral that is extended from a method of integration (like Riemann method or Lebesgue method etc), am I on the right track? – William Aug 27 '22 at 07:59
  • Yes, that is correct. The function is unbounded, but, for any $\varepsilon\in(0,1)$, its restriction to $[\varepsilon,1]$ is bounded. Actually, it is Riemann-integrable. And so we define the improper inter $\int_0^1\frac1{\sqrt x},\mathrm dx$ by$$\int_0^1\frac1{\sqrt x},\mathrm dx=\lim_{\varepsilon\to0}\int_\varepsilon^1\frac1{\sqrt x},\mathrm dx.$$ – José Carlos Santos Aug 27 '22 at 08:05
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$\frac{1}{\sqrt{x}}$ is not bounded on $(0,1)$, and therefore, it cannot be Riemann integrable.

However, $\sqrt{x}$ is continuous on $[0,1]$ and $[0,1]$ is a compact set, therefore, $\sqrt x$ is uniformly continuous on $[0,1]$. Also, the derivative of $\sqrt{x}$ is bounded on $[1, \infty)$, so $\sqrt{x}$ is uniformly continuous on $[1, \infty)$.

5xum
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It's unbounded so no it's not Riemann integrable. It is improperly Riemann integrable though.

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The problem here is that the integrand is unbounded in the domain of integration so it is not Riemann integrable.

The improper integral does exist if it is perrceived as the limit $$\displaystyle\lim_{a\to0^+}\int_a^1\dfrac{1}{\sqrt{x}}=\lim_{a\to0^+}(2-2\sqrt{a})=2$$

Yadati Kiran
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  • Yadati the function isn't even defined at all points in the domain $[0,1]$ so should any further investigation be done, since it fails the definition of being a function since you have no image of $0$ under $f$. – Upstart Nov 10 '20 at 18:58