Trying to integrate Thomae's Function with Riemann sums definition











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Let $f:[1,2] to mathbb{R}, f(x) = 0$ at irrationals and $f(x)=frac1q$ at rationals $frac{p}{q},q>0$ in lowest terms. (Thomae function in $[1,2])$



I need to prove that $int_{1}^{2}dx = 0$. I tried to do this using the riemann sums definition of integration, even though it seems easier using the lower and upper integral definition.



But things got weird. My attempt:
Consider the partition $[1,2] = [1,1+frac1n]cup[1+frac1n,1+frac2n]cup...cup[1+frac{n-1}{n},2]$, where I choose the left endpoints $x_k^* = 1+frac{k}{n}$ and $Delta x = (1 + frac1n) - 1 = frac1n$



$f(x_k^*)=f(frac{n+k}{n}) = frac1s$



Here is my biggest problem, I think. $gcd(n+k,n)=gcd(k,n)$, but how can I write this $s$ such that $f(frac{n+k}{n})=frac1s$?



If $f(x_k^*)=frac1n$, then
$int_{1}^{2}f(x)dx = lim sum_{k=1}^n f(x_k^*)Delta x=lim sum_{k=1}^{n} frac{1}{n^2}=infty$, so there is a mistake. I think it is at the $gcd$ part.



Any help in solving this problem by this method would be appreciated.



Thanks.






real-analysis integration riemann-integration







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    down vote

    favorite












    Let $f:[1,2] to mathbb{R}, f(x) = 0$ at irrationals and $f(x)=frac1q$ at rationals $frac{p}{q},q>0$ in lowest terms. (Thomae function in $[1,2])$



    I need to prove that $int_{1}^{2}dx = 0$. I tried to do this using the riemann sums definition of integration, even though it seems easier using the lower and upper integral definition.



    But things got weird. My attempt:
    Consider the partition $[1,2] = [1,1+frac1n]cup[1+frac1n,1+frac2n]cup...cup[1+frac{n-1}{n},2]$, where I choose the left endpoints $x_k^* = 1+frac{k}{n}$ and $Delta x = (1 + frac1n) - 1 = frac1n$



    $f(x_k^*)=f(frac{n+k}{n}) = frac1s$



    Here is my biggest problem, I think. $gcd(n+k,n)=gcd(k,n)$, but how can I write this $s$ such that $f(frac{n+k}{n})=frac1s$?



    If $f(x_k^*)=frac1n$, then
    $int_{1}^{2}f(x)dx = lim sum_{k=1}^n f(x_k^*)Delta x=lim sum_{k=1}^{n} frac{1}{n^2}=infty$, so there is a mistake. I think it is at the $gcd$ part.



    Any help in solving this problem by this method would be appreciated.



    Thanks.






    real-analysis integration riemann-integration







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Let $f:[1,2] to mathbb{R}, f(x) = 0$ at irrationals and $f(x)=frac1q$ at rationals $frac{p}{q},q>0$ in lowest terms. (Thomae function in $[1,2])$



      I need to prove that $int_{1}^{2}dx = 0$. I tried to do this using the riemann sums definition of integration, even though it seems easier using the lower and upper integral definition.



      But things got weird. My attempt:
      Consider the partition $[1,2] = [1,1+frac1n]cup[1+frac1n,1+frac2n]cup...cup[1+frac{n-1}{n},2]$, where I choose the left endpoints $x_k^* = 1+frac{k}{n}$ and $Delta x = (1 + frac1n) - 1 = frac1n$



      $f(x_k^*)=f(frac{n+k}{n}) = frac1s$



      Here is my biggest problem, I think. $gcd(n+k,n)=gcd(k,n)$, but how can I write this $s$ such that $f(frac{n+k}{n})=frac1s$?



      If $f(x_k^*)=frac1n$, then
      $int_{1}^{2}f(x)dx = lim sum_{k=1}^n f(x_k^*)Delta x=lim sum_{k=1}^{n} frac{1}{n^2}=infty$, so there is a mistake. I think it is at the $gcd$ part.



      Any help in solving this problem by this method would be appreciated.



      Thanks.






      real-analysis integration riemann-integration







      share|cite|improve this question













      Let $f:[1,2] to mathbb{R}, f(x) = 0$ at irrationals and $f(x)=frac1q$ at rationals $frac{p}{q},q>0$ in lowest terms. (Thomae function in $[1,2])$



      I need to prove that $int_{1}^{2}dx = 0$. I tried to do this using the riemann sums definition of integration, even though it seems easier using the lower and upper integral definition.



      But things got weird. My attempt:
      Consider the partition $[1,2] = [1,1+frac1n]cup[1+frac1n,1+frac2n]cup...cup[1+frac{n-1}{n},2]$, where I choose the left endpoints $x_k^* = 1+frac{k}{n}$ and $Delta x = (1 + frac1n) - 1 = frac1n$



      $f(x_k^*)=f(frac{n+k}{n}) = frac1s$



      Here is my biggest problem, I think. $gcd(n+k,n)=gcd(k,n)$, but how can I write this $s$ such that $f(frac{n+k}{n})=frac1s$?



      If $f(x_k^*)=frac1n$, then
      $int_{1}^{2}f(x)dx = lim sum_{k=1}^n f(x_k^*)Delta x=lim sum_{k=1}^{n} frac{1}{n^2}=infty$, so there is a mistake. I think it is at the $gcd$ part.



      Any help in solving this problem by this method would be appreciated.



      Thanks.






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      real-analysis integration riemann-integration






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      asked Nov 22 at 19:59









      dude3221

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