Integrability of the derivative of $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise.












-1














why is the derivative of the function $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise, not integrable?






real-analysis calculus integration riemann-integration







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  • The least you can do is compute the derivative first. Then we can think about why it is not integrable.
    – астон вілла олоф мэллбэрг
    Dec 6 at 11:22












  • Have you plotted this derivative and seen what it looks like?
    – Arthur
    Dec 6 at 11:22










  • I have computed it in my notebook @астонвіллаолофмэллбэрг
    – hopefully
    Dec 6 at 11:27






  • 1




    @hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
    – астон вілла олоф мэллбэрг
    Dec 6 at 11:29






  • 1




    I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
    – Daphna Keidar
    Dec 6 at 11:35
















-1














why is the derivative of the function $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise, not integrable?






real-analysis calculus integration riemann-integration







share|cite|improve this question
























  • The least you can do is compute the derivative first. Then we can think about why it is not integrable.
    – астон вілла олоф мэллбэрг
    Dec 6 at 11:22












  • Have you plotted this derivative and seen what it looks like?
    – Arthur
    Dec 6 at 11:22










  • I have computed it in my notebook @астонвіллаолофмэллбэрг
    – hopefully
    Dec 6 at 11:27






  • 1




    @hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
    – астон вілла олоф мэллбэрг
    Dec 6 at 11:29






  • 1




    I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
    – Daphna Keidar
    Dec 6 at 11:35














-1












-1








-1







why is the derivative of the function $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise, not integrable?






real-analysis calculus integration riemann-integration







share|cite|improve this question















why is the derivative of the function $f(x) = x^2 sin (1/x^2)$ if $x ne 0$ and 0 otherwise, not integrable?






real-analysis calculus integration riemann-integration




real-analysis calculus integration riemann-integration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 at 11:21









Larry

1,3862722




1,3862722










asked Dec 6 at 11:11









hopefully

129112




129112












  • The least you can do is compute the derivative first. Then we can think about why it is not integrable.
    – астон вілла олоф мэллбэрг
    Dec 6 at 11:22












  • Have you plotted this derivative and seen what it looks like?
    – Arthur
    Dec 6 at 11:22










  • I have computed it in my notebook @астонвіллаолофмэллбэрг
    – hopefully
    Dec 6 at 11:27






  • 1




    @hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
    – астон вілла олоф мэллбэрг
    Dec 6 at 11:29






  • 1




    I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
    – Daphna Keidar
    Dec 6 at 11:35


















  • The least you can do is compute the derivative first. Then we can think about why it is not integrable.
    – астон вілла олоф мэллбэрг
    Dec 6 at 11:22












  • Have you plotted this derivative and seen what it looks like?
    – Arthur
    Dec 6 at 11:22










  • I have computed it in my notebook @астонвіллаолофмэллбэрг
    – hopefully
    Dec 6 at 11:27






  • 1




    @hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
    – астон вілла олоф мэллбэрг
    Dec 6 at 11:29






  • 1




    I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
    – Daphna Keidar
    Dec 6 at 11:35
















The least you can do is compute the derivative first. Then we can think about why it is not integrable.
– астон вілла олоф мэллбэрг
Dec 6 at 11:22






The least you can do is compute the derivative first. Then we can think about why it is not integrable.
– астон вілла олоф мэллбэрг
Dec 6 at 11:22














Have you plotted this derivative and seen what it looks like?
– Arthur
Dec 6 at 11:22




Have you plotted this derivative and seen what it looks like?
– Arthur
Dec 6 at 11:22












I have computed it in my notebook @астонвіллаолофмэллбэрг
– hopefully
Dec 6 at 11:27




I have computed it in my notebook @астонвіллаолофмэллбэрг
– hopefully
Dec 6 at 11:27




1




1




@hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
– астон вілла олоф мэллбэрг
Dec 6 at 11:29




@hopefully Whatever computation you have done should have come in the question itself. Remember, the more effort you show in solving your question, the more help you get here. Besides, the answer below is fine.
– астон вілла олоф мэллбэрг
Dec 6 at 11:29




1




1




I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
– Daphna Keidar
Dec 6 at 11:35




I actually think it can be possible to answer this without calculating the derivative, using the fundamental theorem of calculus and the fact that f(x) (as the antiderivative of its derivative ) does not follow the properties of an integral..?
– Daphna Keidar
Dec 6 at 11:35










2 Answers
2






active

oldest

votes


















3














For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:



$f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,



thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.






share|cite|improve this answer





























    0














    What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.






    share|cite|improve this answer























    • I want to show that the derivative is not integrable on [-1,1].
      – hopefully
      Dec 6 at 12:01










    • Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
      – user539887
      Dec 6 at 12:54










    • In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
      – MANI SHANKAR PANDEY
      Dec 6 at 16:35











    Your Answer





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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    3














    For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:



    $f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,



    thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



    Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



    Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.






    share|cite|improve this answer


























      3














      For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:



      $f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,



      thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



      Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



      Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.






      share|cite|improve this answer
























        3












        3








        3






        For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:



        $f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,



        thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



        Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



        Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.






        share|cite|improve this answer












        For $x ne 0$ we have $f'(x)=2x sin(1/x^2)-frac{1}{x}cos(1/x^2)$. Then:



        $f'(frac{1}{sqrt{n pi}})=-sqrt{n pi}cos(n pi)=(-1)^{n+1}sqrt{n pi}$,



        thus $|f'(frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



        Similar: $|f'(-frac{1}{sqrt{n pi}})|=sqrt{n pi} to infty$ as $n to infty.$



        Conclusion: if $[a,b]$ is an intervall with $0 in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 at 11:23









        Fred

        44.1k1645




        44.1k1645























            0














            What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.






            share|cite|improve this answer























            • I want to show that the derivative is not integrable on [-1,1].
              – hopefully
              Dec 6 at 12:01










            • Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
              – user539887
              Dec 6 at 12:54










            • In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
              – MANI SHANKAR PANDEY
              Dec 6 at 16:35
















            0














            What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.






            share|cite|improve this answer























            • I want to show that the derivative is not integrable on [-1,1].
              – hopefully
              Dec 6 at 12:01










            • Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
              – user539887
              Dec 6 at 12:54










            • In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
              – MANI SHANKAR PANDEY
              Dec 6 at 16:35














            0












            0








            0






            What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.






            share|cite|improve this answer














            What is your domain? If it is $mathbf{R}$ then $f$ is also not integrable. It is by the definition of Riemann integrability. Now if your domain is $[0,1]$ then $f$ is differentiable but its derivative $f'(x)=2xsin(frac{1}{x^2})-frac{cos(frac{1}{x^2})}{x}$ is not bounded so not differentiable.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 6 at 12:15









            Tianlalu

            3,01021038




            3,01021038










            answered Dec 6 at 11:30









            MANI SHANKAR PANDEY

            276




            276












            • I want to show that the derivative is not integrable on [-1,1].
              – hopefully
              Dec 6 at 12:01










            • Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
              – user539887
              Dec 6 at 12:54










            • In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
              – MANI SHANKAR PANDEY
              Dec 6 at 16:35


















            • I want to show that the derivative is not integrable on [-1,1].
              – hopefully
              Dec 6 at 12:01










            • Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
              – user539887
              Dec 6 at 12:54










            • In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
              – MANI SHANKAR PANDEY
              Dec 6 at 16:35
















            I want to show that the derivative is not integrable on [-1,1].
            – hopefully
            Dec 6 at 12:01




            I want to show that the derivative is not integrable on [-1,1].
            – hopefully
            Dec 6 at 12:01












            Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
            – user539887
            Dec 6 at 12:54




            Who upvoted the above answer? It is partly missing the point, and partly plainly wrong.
            – user539887
            Dec 6 at 12:54












            In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
            – MANI SHANKAR PANDEY
            Dec 6 at 16:35




            In this domain [-1,1] also, the same problem will occur unboundedness of the function $f'$.
            – MANI SHANKAR PANDEY
            Dec 6 at 16:35


















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