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An Issue about Riemann Integral



The Next CEO of Stack OverflowProving a function is boundedComputing a limes via integrationChecking Riemann integrabilityExpectation defined as Riemann integralExpressing Limit as Riemann IntegralRiemann Integral : $lim_{nrightarrow infty}frac{1}{n}sum_{k=1}^{n}f'(frac{k}{3n})$Confusion about the Riemann IntegralInfinite Riemann integral implies infinite lower Riemann sum?Limit of Riemann Sum as Definite IntegralRiemann Sum - correct formula for an integral?Limit, Riemann Sum, Integration, Natural logarithmEvaluating an Integral as a Riemann sum












2












$begingroup$


By the denition of Riemann integral, we have
$$lim_{nrightarrow infty}sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} = int_{a}^{b}f(x)dx$$



In a particular problem, for $k<n$, $f$ is defined, and for $k=n$, $f$ is not defined, so my question is that




In this situation, is the following relationship valid?
$$lim_{nrightarrow infty}sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n} = int_{a}^{b}f(x)dx$$







calculus integration riemann-integration riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} =sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n}+f(b)frac{b-a}{n} $.
    $endgroup$
    – Riemann
    May 13 '18 at 7:08










  • $begingroup$
    @Riemann $f(b)$ is not defined. Indeed, in my problem, $b=1$ and $f(x)=frac{1}{1-x}$, thus $f(b) = frac{1}{1-b} = frac{1}{1-1} = frac{1}{0}$.
    $endgroup$
    – Hasan Heydari
    May 13 '18 at 7:23


















2












$begingroup$


By the denition of Riemann integral, we have
$$lim_{nrightarrow infty}sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} = int_{a}^{b}f(x)dx$$



In a particular problem, for $k<n$, $f$ is defined, and for $k=n$, $f$ is not defined, so my question is that




In this situation, is the following relationship valid?
$$lim_{nrightarrow infty}sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n} = int_{a}^{b}f(x)dx$$







calculus integration riemann-integration riemann-sum







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} =sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n}+f(b)frac{b-a}{n} $.
    $endgroup$
    – Riemann
    May 13 '18 at 7:08










  • $begingroup$
    @Riemann $f(b)$ is not defined. Indeed, in my problem, $b=1$ and $f(x)=frac{1}{1-x}$, thus $f(b) = frac{1}{1-b} = frac{1}{1-1} = frac{1}{0}$.
    $endgroup$
    – Hasan Heydari
    May 13 '18 at 7:23
















2












2








2





$begingroup$


By the denition of Riemann integral, we have
$$lim_{nrightarrow infty}sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} = int_{a}^{b}f(x)dx$$



In a particular problem, for $k<n$, $f$ is defined, and for $k=n$, $f$ is not defined, so my question is that




In this situation, is the following relationship valid?
$$lim_{nrightarrow infty}sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n} = int_{a}^{b}f(x)dx$$







calculus integration riemann-integration riemann-sum







share|cite|improve this question











$endgroup$




By the denition of Riemann integral, we have
$$lim_{nrightarrow infty}sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} = int_{a}^{b}f(x)dx$$



In a particular problem, for $k<n$, $f$ is defined, and for $k=n$, $f$ is not defined, so my question is that




In this situation, is the following relationship valid?
$$lim_{nrightarrow infty}sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n} = int_{a}^{b}f(x)dx$$







calculus integration riemann-integration riemann-sum




calculus integration riemann-integration riemann-sum






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 16 at 5:55









RRL

53.1k52574




53.1k52574










asked May 13 '18 at 6:08









Hasan HeydariHasan Heydari

907521




907521












  • $begingroup$
    $sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} =sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n}+f(b)frac{b-a}{n} $.
    $endgroup$
    – Riemann
    May 13 '18 at 7:08










  • $begingroup$
    @Riemann $f(b)$ is not defined. Indeed, in my problem, $b=1$ and $f(x)=frac{1}{1-x}$, thus $f(b) = frac{1}{1-b} = frac{1}{1-1} = frac{1}{0}$.
    $endgroup$
    – Hasan Heydari
    May 13 '18 at 7:23




















  • $begingroup$
    $sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} =sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n}+f(b)frac{b-a}{n} $.
    $endgroup$
    – Riemann
    May 13 '18 at 7:08










  • $begingroup$
    @Riemann $f(b)$ is not defined. Indeed, in my problem, $b=1$ and $f(x)=frac{1}{1-x}$, thus $f(b) = frac{1}{1-b} = frac{1}{1-1} = frac{1}{0}$.
    $endgroup$
    – Hasan Heydari
    May 13 '18 at 7:23


















$begingroup$
$sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} =sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n}+f(b)frac{b-a}{n} $.
$endgroup$
– Riemann
May 13 '18 at 7:08




$begingroup$
$sum_{k=1}^{n}f(a + frac{b-a}{n}k)frac{b-a}{n} =sum_{k=1}^{n-1}f(a + frac{b-a}{n}k)frac{b-a}{n}+f(b)frac{b-a}{n} $.
$endgroup$
– Riemann
May 13 '18 at 7:08












$begingroup$
@Riemann $f(b)$ is not defined. Indeed, in my problem, $b=1$ and $f(x)=frac{1}{1-x}$, thus $f(b) = frac{1}{1-b} = frac{1}{1-1} = frac{1}{0}$.
$endgroup$
– Hasan Heydari
May 13 '18 at 7:23






$begingroup$
@Riemann $f(b)$ is not defined. Indeed, in my problem, $b=1$ and $f(x)=frac{1}{1-x}$, thus $f(b) = frac{1}{1-b} = frac{1}{1-1} = frac{1}{0}$.
$endgroup$
– Hasan Heydari
May 13 '18 at 7:23












1 Answer
1






active

oldest

votes


















6












$begingroup$

If $f$ is unbounded on $[0,1]$ then it is not Riemann integrable, as shown here, and we can't be sure that Riemann sums converge to anything.



If the improper integral is convergent,



$$lim_{c to 1}int_0^c f(x) , dx = underbrace{int_0^1 f(x) , dx }_{text{improper integral}},$$



then it is possible for the Riemann sums to converge if the intermediate points (tags) are carefully selected. This would be the case for $f(x) = 1/sqrt{1-x},$ where



$$lim_{n to infty} frac{1}{n} sum_{k=1}^n frac{1}{sqrt{1 - (k-1)/n}} = lim_{c to 1}int_0^c frac{dx}{sqrt{1-x}} = lim_{c to 1}(2 - 2sqrt{1-c}) = 2$$



Note that it is not entirely trivial to prove that the limit of the sum on the LHS is $2$. It is proved here for the equivalent improper integral $displaystyleint_0^1 frac{dx}{sqrt{x}}$.



However, in your example $f(x) = 1/(1-x)$, not even the improper integral is convergent, since



$$lim_{c to 1}int_0^c frac{dx}{1-x} = lim_{c to 1}[log(1-0)-log(1-c)] = +infty$$






share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






    active

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    active

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    active

    oldest

    votes









    6












    $begingroup$

    If $f$ is unbounded on $[0,1]$ then it is not Riemann integrable, as shown here, and we can't be sure that Riemann sums converge to anything.



    If the improper integral is convergent,



    $$lim_{c to 1}int_0^c f(x) , dx = underbrace{int_0^1 f(x) , dx }_{text{improper integral}},$$



    then it is possible for the Riemann sums to converge if the intermediate points (tags) are carefully selected. This would be the case for $f(x) = 1/sqrt{1-x},$ where



    $$lim_{n to infty} frac{1}{n} sum_{k=1}^n frac{1}{sqrt{1 - (k-1)/n}} = lim_{c to 1}int_0^c frac{dx}{sqrt{1-x}} = lim_{c to 1}(2 - 2sqrt{1-c}) = 2$$



    Note that it is not entirely trivial to prove that the limit of the sum on the LHS is $2$. It is proved here for the equivalent improper integral $displaystyleint_0^1 frac{dx}{sqrt{x}}$.



    However, in your example $f(x) = 1/(1-x)$, not even the improper integral is convergent, since



    $$lim_{c to 1}int_0^c frac{dx}{1-x} = lim_{c to 1}[log(1-0)-log(1-c)] = +infty$$






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      If $f$ is unbounded on $[0,1]$ then it is not Riemann integrable, as shown here, and we can't be sure that Riemann sums converge to anything.



      If the improper integral is convergent,



      $$lim_{c to 1}int_0^c f(x) , dx = underbrace{int_0^1 f(x) , dx }_{text{improper integral}},$$



      then it is possible for the Riemann sums to converge if the intermediate points (tags) are carefully selected. This would be the case for $f(x) = 1/sqrt{1-x},$ where



      $$lim_{n to infty} frac{1}{n} sum_{k=1}^n frac{1}{sqrt{1 - (k-1)/n}} = lim_{c to 1}int_0^c frac{dx}{sqrt{1-x}} = lim_{c to 1}(2 - 2sqrt{1-c}) = 2$$



      Note that it is not entirely trivial to prove that the limit of the sum on the LHS is $2$. It is proved here for the equivalent improper integral $displaystyleint_0^1 frac{dx}{sqrt{x}}$.



      However, in your example $f(x) = 1/(1-x)$, not even the improper integral is convergent, since



      $$lim_{c to 1}int_0^c frac{dx}{1-x} = lim_{c to 1}[log(1-0)-log(1-c)] = +infty$$






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        If $f$ is unbounded on $[0,1]$ then it is not Riemann integrable, as shown here, and we can't be sure that Riemann sums converge to anything.



        If the improper integral is convergent,



        $$lim_{c to 1}int_0^c f(x) , dx = underbrace{int_0^1 f(x) , dx }_{text{improper integral}},$$



        then it is possible for the Riemann sums to converge if the intermediate points (tags) are carefully selected. This would be the case for $f(x) = 1/sqrt{1-x},$ where



        $$lim_{n to infty} frac{1}{n} sum_{k=1}^n frac{1}{sqrt{1 - (k-1)/n}} = lim_{c to 1}int_0^c frac{dx}{sqrt{1-x}} = lim_{c to 1}(2 - 2sqrt{1-c}) = 2$$



        Note that it is not entirely trivial to prove that the limit of the sum on the LHS is $2$. It is proved here for the equivalent improper integral $displaystyleint_0^1 frac{dx}{sqrt{x}}$.



        However, in your example $f(x) = 1/(1-x)$, not even the improper integral is convergent, since



        $$lim_{c to 1}int_0^c frac{dx}{1-x} = lim_{c to 1}[log(1-0)-log(1-c)] = +infty$$






        share|cite|improve this answer











        $endgroup$



        If $f$ is unbounded on $[0,1]$ then it is not Riemann integrable, as shown here, and we can't be sure that Riemann sums converge to anything.



        If the improper integral is convergent,



        $$lim_{c to 1}int_0^c f(x) , dx = underbrace{int_0^1 f(x) , dx }_{text{improper integral}},$$



        then it is possible for the Riemann sums to converge if the intermediate points (tags) are carefully selected. This would be the case for $f(x) = 1/sqrt{1-x},$ where



        $$lim_{n to infty} frac{1}{n} sum_{k=1}^n frac{1}{sqrt{1 - (k-1)/n}} = lim_{c to 1}int_0^c frac{dx}{sqrt{1-x}} = lim_{c to 1}(2 - 2sqrt{1-c}) = 2$$



        Note that it is not entirely trivial to prove that the limit of the sum on the LHS is $2$. It is proved here for the equivalent improper integral $displaystyleint_0^1 frac{dx}{sqrt{x}}$.



        However, in your example $f(x) = 1/(1-x)$, not even the improper integral is convergent, since



        $$lim_{c to 1}int_0^c frac{dx}{1-x} = lim_{c to 1}[log(1-0)-log(1-c)] = +infty$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 13 '18 at 8:15

























        answered May 13 '18 at 7:35









        RRLRRL

        53.1k52574




        53.1k52574






























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