Is $f(x)=sin(1/x),;xne0$, $f(0)=0$, Riemann integrable on $[0,1]$?












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Is the function $sin frac{1}{x}$ Riemann integrable on an interval containing $0$?






calculus real-analysis integration







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  • $begingroup$
    Consider a change of variables.
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    – Pedro Tamaroff♦
    Mar 27 '14 at 4:17






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    $begingroup$
    A function is Riemann integrable iff it is bounded and continuous almost everywhere.
    $endgroup$
    – Paul Siegel
    Mar 27 '14 at 4:41
















1












$begingroup$


Is the function $sin frac{1}{x}$ Riemann integrable on an interval containing $0$?






calculus real-analysis integration







share|cite|improve this question











$endgroup$












  • $begingroup$
    Consider a change of variables.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 27 '14 at 4:17






  • 3




    $begingroup$
    A function is Riemann integrable iff it is bounded and continuous almost everywhere.
    $endgroup$
    – Paul Siegel
    Mar 27 '14 at 4:41














1












1








1


1



$begingroup$


Is the function $sin frac{1}{x}$ Riemann integrable on an interval containing $0$?






calculus real-analysis integration







share|cite|improve this question











$endgroup$




Is the function $sin frac{1}{x}$ Riemann integrable on an interval containing $0$?






calculus real-analysis integration




calculus real-analysis integration






share|cite|improve this question















share|cite|improve this question













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edited Jan 9 at 11:14









Larry

2,0912826




2,0912826










asked Mar 27 '14 at 4:12









user134070user134070

387521




387521












  • $begingroup$
    Consider a change of variables.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 27 '14 at 4:17






  • 3




    $begingroup$
    A function is Riemann integrable iff it is bounded and continuous almost everywhere.
    $endgroup$
    – Paul Siegel
    Mar 27 '14 at 4:41


















  • $begingroup$
    Consider a change of variables.
    $endgroup$
    – Pedro Tamaroff♦
    Mar 27 '14 at 4:17






  • 3




    $begingroup$
    A function is Riemann integrable iff it is bounded and continuous almost everywhere.
    $endgroup$
    – Paul Siegel
    Mar 27 '14 at 4:41
















$begingroup$
Consider a change of variables.
$endgroup$
– Pedro Tamaroff♦
Mar 27 '14 at 4:17




$begingroup$
Consider a change of variables.
$endgroup$
– Pedro Tamaroff♦
Mar 27 '14 at 4:17




3




3




$begingroup$
A function is Riemann integrable iff it is bounded and continuous almost everywhere.
$endgroup$
– Paul Siegel
Mar 27 '14 at 4:41




$begingroup$
A function is Riemann integrable iff it is bounded and continuous almost everywhere.
$endgroup$
– Paul Siegel
Mar 27 '14 at 4:41










2 Answers
2






active

oldest

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7












$begingroup$

There are 2 proofs: a short one and a longer one.



First Proof: f is bounded on [0,1] since |sin 1/x| is less than or equal to 1. f is continuous except at x=0 (since limit as x-> 0 doesn’t exist), so the set of discontinuities is the single point x=0 and this set has measure zero. There is a well known Thm. that if a function is bounded and continuous except on a set of measure zero it is (Riemann) integrable. If you’re not familiar with measure, this answer probably won’t help you much, however.



The second proof is based on the upper U and lower L sums. f is integrable if given any e (epsilon) greater than zero there is a partition such that U – L is less than e. So given e, consider the intervals [0,e] and [e,1]. Since f is bounded and continuous on [e,1] it is integrable, so there is a partition P1 such that U-L






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    This is late but will be useful for future readers.



    PROOF



    Let $epsilon>0$ be given but arbitrarily small. Choose $x_1in(0,1]$ such that $1-x_1<epsilon/2.$ Then, the set $P={0,x_1,1} $ is a patition. The upper and lower Darboux sums are given by



    begin{align}U(f,P)=sum^{2}_{k=1}M_k (x_k-x_{k+1}),;;text{where};;M_{k}= sup{f(x):;x_{k-1}leq x<x_{k}},end{align}
    begin{align}L(f,P)=sum^{2}_{k=1}m_k (x_k-x_{k+1}),;;text{where};;m_{k}= inf{f(x):;x_{k-1}leq x<x_{k}}.end{align}
    By computation
    begin{align}U(f,P)&=sum^{2}_{k=1}M_k (x_k-x_{k+1})\&=M_1 (x_1-x_{0})+M_2 (x_2-x_{1})\&= 1-x_{1},;;text{where};;M_{1}=0,,M_{2}=1,end{align}
    begin{align}L(f,P)&=sum^{2}_{k=1}m_k (x_k-x_{k+1})\&=m_1 (x_1-x_{0})+m_2 (x_2-x_{1})\&=-(1-x_{1}),;;text{where};;m_{1}=0,,m_{2}=-1.end{align}
    Thus,
    begin{align}U(f,P)-L(f,P)&=2(1-x_{1})<epsilon.end{align}
    So, given any $epsilon$, choose $P={0,x_1,1} $ such that $1-x_1<epsilon/2.$ Then,
    begin{align}U(f,P)-L(f,P)<epsilon.end{align}
    Hence, $f$ is integrable.






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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      7












      $begingroup$

      There are 2 proofs: a short one and a longer one.



      First Proof: f is bounded on [0,1] since |sin 1/x| is less than or equal to 1. f is continuous except at x=0 (since limit as x-> 0 doesn’t exist), so the set of discontinuities is the single point x=0 and this set has measure zero. There is a well known Thm. that if a function is bounded and continuous except on a set of measure zero it is (Riemann) integrable. If you’re not familiar with measure, this answer probably won’t help you much, however.



      The second proof is based on the upper U and lower L sums. f is integrable if given any e (epsilon) greater than zero there is a partition such that U – L is less than e. So given e, consider the intervals [0,e] and [e,1]. Since f is bounded and continuous on [e,1] it is integrable, so there is a partition P1 such that U-L






      share|cite|improve this answer









      $endgroup$


















        7












        $begingroup$

        There are 2 proofs: a short one and a longer one.



        First Proof: f is bounded on [0,1] since |sin 1/x| is less than or equal to 1. f is continuous except at x=0 (since limit as x-> 0 doesn’t exist), so the set of discontinuities is the single point x=0 and this set has measure zero. There is a well known Thm. that if a function is bounded and continuous except on a set of measure zero it is (Riemann) integrable. If you’re not familiar with measure, this answer probably won’t help you much, however.



        The second proof is based on the upper U and lower L sums. f is integrable if given any e (epsilon) greater than zero there is a partition such that U – L is less than e. So given e, consider the intervals [0,e] and [e,1]. Since f is bounded and continuous on [e,1] it is integrable, so there is a partition P1 such that U-L






        share|cite|improve this answer









        $endgroup$
















          7












          7








          7





          $begingroup$

          There are 2 proofs: a short one and a longer one.



          First Proof: f is bounded on [0,1] since |sin 1/x| is less than or equal to 1. f is continuous except at x=0 (since limit as x-> 0 doesn’t exist), so the set of discontinuities is the single point x=0 and this set has measure zero. There is a well known Thm. that if a function is bounded and continuous except on a set of measure zero it is (Riemann) integrable. If you’re not familiar with measure, this answer probably won’t help you much, however.



          The second proof is based on the upper U and lower L sums. f is integrable if given any e (epsilon) greater than zero there is a partition such that U – L is less than e. So given e, consider the intervals [0,e] and [e,1]. Since f is bounded and continuous on [e,1] it is integrable, so there is a partition P1 such that U-L






          share|cite|improve this answer









          $endgroup$



          There are 2 proofs: a short one and a longer one.



          First Proof: f is bounded on [0,1] since |sin 1/x| is less than or equal to 1. f is continuous except at x=0 (since limit as x-> 0 doesn’t exist), so the set of discontinuities is the single point x=0 and this set has measure zero. There is a well known Thm. that if a function is bounded and continuous except on a set of measure zero it is (Riemann) integrable. If you’re not familiar with measure, this answer probably won’t help you much, however.



          The second proof is based on the upper U and lower L sums. f is integrable if given any e (epsilon) greater than zero there is a partition such that U – L is less than e. So given e, consider the intervals [0,e] and [e,1]. Since f is bounded and continuous on [e,1] it is integrable, so there is a partition P1 such that U-L







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 21 '14 at 7:29









          honghong

          7112




          7112























              3












              $begingroup$

              This is late but will be useful for future readers.



              PROOF



              Let $epsilon>0$ be given but arbitrarily small. Choose $x_1in(0,1]$ such that $1-x_1<epsilon/2.$ Then, the set $P={0,x_1,1} $ is a patition. The upper and lower Darboux sums are given by



              begin{align}U(f,P)=sum^{2}_{k=1}M_k (x_k-x_{k+1}),;;text{where};;M_{k}= sup{f(x):;x_{k-1}leq x<x_{k}},end{align}
              begin{align}L(f,P)=sum^{2}_{k=1}m_k (x_k-x_{k+1}),;;text{where};;m_{k}= inf{f(x):;x_{k-1}leq x<x_{k}}.end{align}
              By computation
              begin{align}U(f,P)&=sum^{2}_{k=1}M_k (x_k-x_{k+1})\&=M_1 (x_1-x_{0})+M_2 (x_2-x_{1})\&= 1-x_{1},;;text{where};;M_{1}=0,,M_{2}=1,end{align}
              begin{align}L(f,P)&=sum^{2}_{k=1}m_k (x_k-x_{k+1})\&=m_1 (x_1-x_{0})+m_2 (x_2-x_{1})\&=-(1-x_{1}),;;text{where};;m_{1}=0,,m_{2}=-1.end{align}
              Thus,
              begin{align}U(f,P)-L(f,P)&=2(1-x_{1})<epsilon.end{align}
              So, given any $epsilon$, choose $P={0,x_1,1} $ such that $1-x_1<epsilon/2.$ Then,
              begin{align}U(f,P)-L(f,P)<epsilon.end{align}
              Hence, $f$ is integrable.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                This is late but will be useful for future readers.



                PROOF



                Let $epsilon>0$ be given but arbitrarily small. Choose $x_1in(0,1]$ such that $1-x_1<epsilon/2.$ Then, the set $P={0,x_1,1} $ is a patition. The upper and lower Darboux sums are given by



                begin{align}U(f,P)=sum^{2}_{k=1}M_k (x_k-x_{k+1}),;;text{where};;M_{k}= sup{f(x):;x_{k-1}leq x<x_{k}},end{align}
                begin{align}L(f,P)=sum^{2}_{k=1}m_k (x_k-x_{k+1}),;;text{where};;m_{k}= inf{f(x):;x_{k-1}leq x<x_{k}}.end{align}
                By computation
                begin{align}U(f,P)&=sum^{2}_{k=1}M_k (x_k-x_{k+1})\&=M_1 (x_1-x_{0})+M_2 (x_2-x_{1})\&= 1-x_{1},;;text{where};;M_{1}=0,,M_{2}=1,end{align}
                begin{align}L(f,P)&=sum^{2}_{k=1}m_k (x_k-x_{k+1})\&=m_1 (x_1-x_{0})+m_2 (x_2-x_{1})\&=-(1-x_{1}),;;text{where};;m_{1}=0,,m_{2}=-1.end{align}
                Thus,
                begin{align}U(f,P)-L(f,P)&=2(1-x_{1})<epsilon.end{align}
                So, given any $epsilon$, choose $P={0,x_1,1} $ such that $1-x_1<epsilon/2.$ Then,
                begin{align}U(f,P)-L(f,P)<epsilon.end{align}
                Hence, $f$ is integrable.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  This is late but will be useful for future readers.



                  PROOF



                  Let $epsilon>0$ be given but arbitrarily small. Choose $x_1in(0,1]$ such that $1-x_1<epsilon/2.$ Then, the set $P={0,x_1,1} $ is a patition. The upper and lower Darboux sums are given by



                  begin{align}U(f,P)=sum^{2}_{k=1}M_k (x_k-x_{k+1}),;;text{where};;M_{k}= sup{f(x):;x_{k-1}leq x<x_{k}},end{align}
                  begin{align}L(f,P)=sum^{2}_{k=1}m_k (x_k-x_{k+1}),;;text{where};;m_{k}= inf{f(x):;x_{k-1}leq x<x_{k}}.end{align}
                  By computation
                  begin{align}U(f,P)&=sum^{2}_{k=1}M_k (x_k-x_{k+1})\&=M_1 (x_1-x_{0})+M_2 (x_2-x_{1})\&= 1-x_{1},;;text{where};;M_{1}=0,,M_{2}=1,end{align}
                  begin{align}L(f,P)&=sum^{2}_{k=1}m_k (x_k-x_{k+1})\&=m_1 (x_1-x_{0})+m_2 (x_2-x_{1})\&=-(1-x_{1}),;;text{where};;m_{1}=0,,m_{2}=-1.end{align}
                  Thus,
                  begin{align}U(f,P)-L(f,P)&=2(1-x_{1})<epsilon.end{align}
                  So, given any $epsilon$, choose $P={0,x_1,1} $ such that $1-x_1<epsilon/2.$ Then,
                  begin{align}U(f,P)-L(f,P)<epsilon.end{align}
                  Hence, $f$ is integrable.






                  share|cite|improve this answer











                  $endgroup$



                  This is late but will be useful for future readers.



                  PROOF



                  Let $epsilon>0$ be given but arbitrarily small. Choose $x_1in(0,1]$ such that $1-x_1<epsilon/2.$ Then, the set $P={0,x_1,1} $ is a patition. The upper and lower Darboux sums are given by



                  begin{align}U(f,P)=sum^{2}_{k=1}M_k (x_k-x_{k+1}),;;text{where};;M_{k}= sup{f(x):;x_{k-1}leq x<x_{k}},end{align}
                  begin{align}L(f,P)=sum^{2}_{k=1}m_k (x_k-x_{k+1}),;;text{where};;m_{k}= inf{f(x):;x_{k-1}leq x<x_{k}}.end{align}
                  By computation
                  begin{align}U(f,P)&=sum^{2}_{k=1}M_k (x_k-x_{k+1})\&=M_1 (x_1-x_{0})+M_2 (x_2-x_{1})\&= 1-x_{1},;;text{where};;M_{1}=0,,M_{2}=1,end{align}
                  begin{align}L(f,P)&=sum^{2}_{k=1}m_k (x_k-x_{k+1})\&=m_1 (x_1-x_{0})+m_2 (x_2-x_{1})\&=-(1-x_{1}),;;text{where};;m_{1}=0,,m_{2}=-1.end{align}
                  Thus,
                  begin{align}U(f,P)-L(f,P)&=2(1-x_{1})<epsilon.end{align}
                  So, given any $epsilon$, choose $P={0,x_1,1} $ such that $1-x_1<epsilon/2.$ Then,
                  begin{align}U(f,P)-L(f,P)<epsilon.end{align}
                  Hence, $f$ is integrable.







                  share|cite|improve this answer














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                  share|cite|improve this answer








                  edited Jan 9 at 12:31

























                  answered Jan 9 at 11:30









                  Omojola MichealOmojola Micheal

                  1,814324




                  1,814324






























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