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












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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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  • $begingroup$
    The least you can do is compute the derivative first. Then we can think about why it is not integrable.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 11:22












  • $begingroup$
    Have you plotted this derivative and seen what it looks like?
    $endgroup$
    – Arthur
    Dec 6 '18 at 11:22










  • $begingroup$
    I have computed it in my notebook @астонвіллаолофмэллбэрг
    $endgroup$
    – hopefully
    Dec 6 '18 at 11:27






  • 1




    $begingroup$
    @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.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 11:29






  • 1




    $begingroup$
    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..?
    $endgroup$
    – Daphna Keidar
    Dec 6 '18 at 11:35
















-1












$begingroup$


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











$endgroup$












  • $begingroup$
    The least you can do is compute the derivative first. Then we can think about why it is not integrable.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 11:22












  • $begingroup$
    Have you plotted this derivative and seen what it looks like?
    $endgroup$
    – Arthur
    Dec 6 '18 at 11:22










  • $begingroup$
    I have computed it in my notebook @астонвіллаолофмэллбэрг
    $endgroup$
    – hopefully
    Dec 6 '18 at 11:27






  • 1




    $begingroup$
    @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.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 11:29






  • 1




    $begingroup$
    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..?
    $endgroup$
    – Daphna Keidar
    Dec 6 '18 at 11:35














-1












-1








-1





$begingroup$


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











$endgroup$




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 '18 at 11:21









Larry

2,31431028




2,31431028










asked Dec 6 '18 at 11:11









hopefullyhopefully

273113




273113












  • $begingroup$
    The least you can do is compute the derivative first. Then we can think about why it is not integrable.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 11:22












  • $begingroup$
    Have you plotted this derivative and seen what it looks like?
    $endgroup$
    – Arthur
    Dec 6 '18 at 11:22










  • $begingroup$
    I have computed it in my notebook @астонвіллаолофмэллбэрг
    $endgroup$
    – hopefully
    Dec 6 '18 at 11:27






  • 1




    $begingroup$
    @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.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 11:29






  • 1




    $begingroup$
    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..?
    $endgroup$
    – Daphna Keidar
    Dec 6 '18 at 11:35


















  • $begingroup$
    The least you can do is compute the derivative first. Then we can think about why it is not integrable.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 11:22












  • $begingroup$
    Have you plotted this derivative and seen what it looks like?
    $endgroup$
    – Arthur
    Dec 6 '18 at 11:22










  • $begingroup$
    I have computed it in my notebook @астонвіллаолофмэллбэрг
    $endgroup$
    – hopefully
    Dec 6 '18 at 11:27






  • 1




    $begingroup$
    @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.
    $endgroup$
    – астон вілла олоф мэллбэрг
    Dec 6 '18 at 11:29






  • 1




    $begingroup$
    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..?
    $endgroup$
    – Daphna Keidar
    Dec 6 '18 at 11:35
















$begingroup$
The least you can do is compute the derivative first. Then we can think about why it is not integrable.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 6 '18 at 11:22






$begingroup$
The least you can do is compute the derivative first. Then we can think about why it is not integrable.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 6 '18 at 11:22














$begingroup$
Have you plotted this derivative and seen what it looks like?
$endgroup$
– Arthur
Dec 6 '18 at 11:22




$begingroup$
Have you plotted this derivative and seen what it looks like?
$endgroup$
– Arthur
Dec 6 '18 at 11:22












$begingroup$
I have computed it in my notebook @астонвіллаолофмэллбэрг
$endgroup$
– hopefully
Dec 6 '18 at 11:27




$begingroup$
I have computed it in my notebook @астонвіллаолофмэллбэрг
$endgroup$
– hopefully
Dec 6 '18 at 11:27




1




1




$begingroup$
@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.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 6 '18 at 11:29




$begingroup$
@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.
$endgroup$
– астон вілла олоф мэллбэрг
Dec 6 '18 at 11:29




1




1




$begingroup$
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..?
$endgroup$
– Daphna Keidar
Dec 6 '18 at 11:35




$begingroup$
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..?
$endgroup$
– Daphna Keidar
Dec 6 '18 at 11:35










1 Answer
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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 $[a,b]$.






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    $begingroup$

    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









    $endgroup$


















      3












      $begingroup$

      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









      $endgroup$
















        3












        3








        3





        $begingroup$

        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









        $endgroup$



        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 '18 at 11:23









        FredFred

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