Is $1 / sqrt{x}$ Riemann integrable on $[0,1]$?











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If $int_0^1 1 / sqrt{x}$ 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 ?






real-analysis riemann-integration







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    Is $dfrac{1}{sqrt{x}}$ bounded at $x=0$?
    – Yadati Kiran
    Nov 27 at 13:11








  • 3




    $sqrt x$ is continuous on the compact interval $[0, 1]$ – that already implies uniform continuity on the interval. You don't need an integral for that argument.
    – Martin R
    Nov 27 at 13:13






  • 1




    Note that $sqrt{x}$ is uniformly continuous on all of $mathbb{R}^{+}$. math.stackexchange.com/questions/569928/…
    – Euler....IS_ALIVE
    Nov 27 at 13:14






  • 2




    Looks like an XY problem ...
    – Martin R
    Nov 27 at 13:16










  • @MartinR No .X was my main problem i just stated Y (Y can be solved by this way also )
    – Amit
    Nov 27 at 13:18















up vote
2
down vote

favorite
1












If $int_0^1 1 / sqrt{x}$ 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 ?






real-analysis riemann-integration







share|cite|improve this question




















  • 2




    Is $dfrac{1}{sqrt{x}}$ bounded at $x=0$?
    – Yadati Kiran
    Nov 27 at 13:11








  • 3




    $sqrt x$ is continuous on the compact interval $[0, 1]$ – that already implies uniform continuity on the interval. You don't need an integral for that argument.
    – Martin R
    Nov 27 at 13:13






  • 1




    Note that $sqrt{x}$ is uniformly continuous on all of $mathbb{R}^{+}$. math.stackexchange.com/questions/569928/…
    – Euler....IS_ALIVE
    Nov 27 at 13:14






  • 2




    Looks like an XY problem ...
    – Martin R
    Nov 27 at 13:16










  • @MartinR No .X was my main problem i just stated Y (Y can be solved by this way also )
    – Amit
    Nov 27 at 13:18













up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





If $int_0^1 1 / sqrt{x}$ 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 ?






real-analysis riemann-integration







share|cite|improve this question















If $int_0^1 1 / sqrt{x}$ 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 ?






real-analysis riemann-integration




real-analysis riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 at 13:13

























asked Nov 27 at 13:09









Amit

1398




1398








  • 2




    Is $dfrac{1}{sqrt{x}}$ bounded at $x=0$?
    – Yadati Kiran
    Nov 27 at 13:11








  • 3




    $sqrt x$ is continuous on the compact interval $[0, 1]$ – that already implies uniform continuity on the interval. You don't need an integral for that argument.
    – Martin R
    Nov 27 at 13:13






  • 1




    Note that $sqrt{x}$ is uniformly continuous on all of $mathbb{R}^{+}$. math.stackexchange.com/questions/569928/…
    – Euler....IS_ALIVE
    Nov 27 at 13:14






  • 2




    Looks like an XY problem ...
    – Martin R
    Nov 27 at 13:16










  • @MartinR No .X was my main problem i just stated Y (Y can be solved by this way also )
    – Amit
    Nov 27 at 13:18














  • 2




    Is $dfrac{1}{sqrt{x}}$ bounded at $x=0$?
    – Yadati Kiran
    Nov 27 at 13:11








  • 3




    $sqrt x$ is continuous on the compact interval $[0, 1]$ – that already implies uniform continuity on the interval. You don't need an integral for that argument.
    – Martin R
    Nov 27 at 13:13






  • 1




    Note that $sqrt{x}$ is uniformly continuous on all of $mathbb{R}^{+}$. math.stackexchange.com/questions/569928/…
    – Euler....IS_ALIVE
    Nov 27 at 13:14






  • 2




    Looks like an XY problem ...
    – Martin R
    Nov 27 at 13:16










  • @MartinR No .X was my main problem i just stated Y (Y can be solved by this way also )
    – Amit
    Nov 27 at 13:18








2




2




Is $dfrac{1}{sqrt{x}}$ bounded at $x=0$?
– Yadati Kiran
Nov 27 at 13:11






Is $dfrac{1}{sqrt{x}}$ bounded at $x=0$?
– Yadati Kiran
Nov 27 at 13:11






3




3




$sqrt x$ is continuous on the compact interval $[0, 1]$ – that already implies uniform continuity on the interval. You don't need an integral for that argument.
– Martin R
Nov 27 at 13:13




$sqrt x$ is continuous on the compact interval $[0, 1]$ – that already implies uniform continuity on the interval. You don't need an integral for that argument.
– Martin R
Nov 27 at 13:13




1




1




Note that $sqrt{x}$ is uniformly continuous on all of $mathbb{R}^{+}$. math.stackexchange.com/questions/569928/…
– Euler....IS_ALIVE
Nov 27 at 13:14




Note that $sqrt{x}$ is uniformly continuous on all of $mathbb{R}^{+}$. math.stackexchange.com/questions/569928/…
– Euler....IS_ALIVE
Nov 27 at 13:14




2




2




Looks like an XY problem ...
– Martin R
Nov 27 at 13:16




Looks like an XY problem ...
– Martin R
Nov 27 at 13:16












@MartinR No .X was my main problem i just stated Y (Y can be solved by this way also )
– Amit
Nov 27 at 13:18




@MartinR No .X was my main problem i just stated Y (Y can be solved by this way also )
– Amit
Nov 27 at 13:18










4 Answers
4






active

oldest

votes

















up vote
5
down vote



accepted










That function is unbounded (whatever way you chhose 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 $int_0^1frac1{sqrt x},mathrm dx$ converges.



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






share|cite|improve this answer





















  • 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 ?
    – Amit
    Nov 27 at 13:16






  • 1




    Yes, you are absolutely right.
    – José Carlos Santos
    Nov 27 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)
    – Amit
    Nov 27 at 13:27








  • 1




    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 at 13:30


















up vote
3
down vote













$frac{1}{sqrt{x}}$ 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 $[1, infty)$.






share|cite|improve this answer




























    up vote
    2
    down vote













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






    share|cite|improve this answer




























      up vote
      2
      down vote













      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 $$displaystylelim_{ato0^+}int_a^1dfrac{1}{sqrt{x}}=lim_{ato0^+}(2-2sqrt{a})=2$$






      share|cite|improve this answer





















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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






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        oldest

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        active

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        up vote
        5
        down vote



        accepted










        That function is unbounded (whatever way you chhose 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 $int_0^1frac1{sqrt x},mathrm dx$ converges.



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






        share|cite|improve this answer





















        • 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 ?
          – Amit
          Nov 27 at 13:16






        • 1




          Yes, you are absolutely right.
          – José Carlos Santos
          Nov 27 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)
          – Amit
          Nov 27 at 13:27








        • 1




          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 at 13:30















        up vote
        5
        down vote



        accepted










        That function is unbounded (whatever way you chhose 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 $int_0^1frac1{sqrt x},mathrm dx$ converges.



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






        share|cite|improve this answer





















        • 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 ?
          – Amit
          Nov 27 at 13:16






        • 1




          Yes, you are absolutely right.
          – José Carlos Santos
          Nov 27 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)
          – Amit
          Nov 27 at 13:27








        • 1




          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 at 13:30













        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        That function is unbounded (whatever way you chhose 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 $int_0^1frac1{sqrt x},mathrm dx$ converges.



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






        share|cite|improve this answer












        That function is unbounded (whatever way you chhose 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 $int_0^1frac1{sqrt x},mathrm dx$ converges.



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







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 27 at 13:11









        José Carlos Santos

        146k22116215




        146k22116215












        • 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 ?
          – Amit
          Nov 27 at 13:16






        • 1




          Yes, you are absolutely right.
          – José Carlos Santos
          Nov 27 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)
          – Amit
          Nov 27 at 13:27








        • 1




          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 at 13:30


















        • 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 ?
          – Amit
          Nov 27 at 13:16






        • 1




          Yes, you are absolutely right.
          – José Carlos Santos
          Nov 27 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)
          – Amit
          Nov 27 at 13:27








        • 1




          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 at 13:30
















        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 ?
        – Amit
        Nov 27 at 13:16




        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 ?
        – Amit
        Nov 27 at 13:16




        1




        1




        Yes, you are absolutely right.
        – José Carlos Santos
        Nov 27 at 13:18




        Yes, you are absolutely right.
        – José Carlos Santos
        Nov 27 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)
        – Amit
        Nov 27 at 13:27






        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)
        – Amit
        Nov 27 at 13:27






        1




        1




        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 at 13:30




        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 at 13:30










        up vote
        3
        down vote













        $frac{1}{sqrt{x}}$ 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 $[1, infty)$.






        share|cite|improve this answer

























          up vote
          3
          down vote













          $frac{1}{sqrt{x}}$ 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 $[1, infty)$.






          share|cite|improve this answer























            up vote
            3
            down vote










            up vote
            3
            down vote









            $frac{1}{sqrt{x}}$ 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 $[1, infty)$.






            share|cite|improve this answer












            $frac{1}{sqrt{x}}$ 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 $[1, infty)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 27 at 13:13









            5xum

            89.4k393161




            89.4k393161






















                up vote
                2
                down vote













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






                share|cite|improve this answer

























                  up vote
                  2
                  down vote













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






                  share|cite|improve this answer























                    up vote
                    2
                    down vote










                    up vote
                    2
                    down vote









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






                    share|cite|improve this answer












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







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 27 at 13:11









                    Zachary Selk

                    536311




                    536311






















                        up vote
                        2
                        down vote













                        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 $$displaystylelim_{ato0^+}int_a^1dfrac{1}{sqrt{x}}=lim_{ato0^+}(2-2sqrt{a})=2$$






                        share|cite|improve this answer

























                          up vote
                          2
                          down vote













                          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 $$displaystylelim_{ato0^+}int_a^1dfrac{1}{sqrt{x}}=lim_{ato0^+}(2-2sqrt{a})=2$$






                          share|cite|improve this answer























                            up vote
                            2
                            down vote










                            up vote
                            2
                            down vote









                            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 $$displaystylelim_{ato0^+}int_a^1dfrac{1}{sqrt{x}}=lim_{ato0^+}(2-2sqrt{a})=2$$






                            share|cite|improve this answer












                            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 $$displaystylelim_{ato0^+}int_a^1dfrac{1}{sqrt{x}}=lim_{ato0^+}(2-2sqrt{a})=2$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 27 at 13:17









                            Yadati Kiran

                            1,327418




                            1,327418






























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                                Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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                                3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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