How to find lower Riemann integral in given function?











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$f(x)$ defined on $[0,1]$ as following -
$$
begin{align}
f(x) = begin{cases}
0 & text{if $x=0$}\
frac{1}{n} & text{if $1/(n+1)<xle 1/n$}
end{cases}
end{align}
$$



How to find lower Riemann integral of $f(x)$ from $0$ to $1$.
My question is different from How to find the Riemann integral of following function?
Since we know $f(x)$ has countable number of discontinuities hence it is Riemann integrable and we can find it's upper integral for getting the answer .But how to find lower Riemann integral of this function ?



EDIT - I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.










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  • Since it is Riemann integrable the lower integral is same as upper integral.
    – Paramanand Singh
    1 hour ago










  • @ParamanandSingh I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.
    – Amit
    1 hour ago















up vote
3
down vote

favorite












$f(x)$ defined on $[0,1]$ as following -
$$
begin{align}
f(x) = begin{cases}
0 & text{if $x=0$}\
frac{1}{n} & text{if $1/(n+1)<xle 1/n$}
end{cases}
end{align}
$$



How to find lower Riemann integral of $f(x)$ from $0$ to $1$.
My question is different from How to find the Riemann integral of following function?
Since we know $f(x)$ has countable number of discontinuities hence it is Riemann integrable and we can find it's upper integral for getting the answer .But how to find lower Riemann integral of this function ?



EDIT - I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.










share|cite|improve this question
























  • Since it is Riemann integrable the lower integral is same as upper integral.
    – Paramanand Singh
    1 hour ago










  • @ParamanandSingh I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.
    – Amit
    1 hour ago













up vote
3
down vote

favorite









up vote
3
down vote

favorite











$f(x)$ defined on $[0,1]$ as following -
$$
begin{align}
f(x) = begin{cases}
0 & text{if $x=0$}\
frac{1}{n} & text{if $1/(n+1)<xle 1/n$}
end{cases}
end{align}
$$



How to find lower Riemann integral of $f(x)$ from $0$ to $1$.
My question is different from How to find the Riemann integral of following function?
Since we know $f(x)$ has countable number of discontinuities hence it is Riemann integrable and we can find it's upper integral for getting the answer .But how to find lower Riemann integral of this function ?



EDIT - I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.










share|cite|improve this question















$f(x)$ defined on $[0,1]$ as following -
$$
begin{align}
f(x) = begin{cases}
0 & text{if $x=0$}\
frac{1}{n} & text{if $1/(n+1)<xle 1/n$}
end{cases}
end{align}
$$



How to find lower Riemann integral of $f(x)$ from $0$ to $1$.
My question is different from How to find the Riemann integral of following function?
Since we know $f(x)$ has countable number of discontinuities hence it is Riemann integrable and we can find it's upper integral for getting the answer .But how to find lower Riemann integral of this function ?



EDIT - I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.







real-analysis measure-theory riemann-integration






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edited 1 hour ago

























asked 1 hour ago









Amit

828




828












  • Since it is Riemann integrable the lower integral is same as upper integral.
    – Paramanand Singh
    1 hour ago










  • @ParamanandSingh I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.
    – Amit
    1 hour ago


















  • Since it is Riemann integrable the lower integral is same as upper integral.
    – Paramanand Singh
    1 hour ago










  • @ParamanandSingh I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.
    – Amit
    1 hour ago
















Since it is Riemann integrable the lower integral is same as upper integral.
– Paramanand Singh
1 hour ago




Since it is Riemann integrable the lower integral is same as upper integral.
– Paramanand Singh
1 hour ago












@ParamanandSingh I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.
– Amit
1 hour ago




@ParamanandSingh I know since $f(x)$ is Riemann integrable hence it's lower Riemann integral is same as upper Riemann integral.But how to find it by partitioning the domain or in other words by using definition of lower Riemann integral.
– Amit
1 hour ago










3 Answers
3






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Consider the partitions $P_N={0}cup {frac{1}{n}: 1leq n leq N}$. Then the lower sum is $$L(f;P_N)=sum limits _{i=1}^N m_ileft(frac{1}{i}-frac{1}{i+1}right)+m_0(frac{1}{N}-0)$$



Then given any partition $P$ you can find an $N$ such that $L(f;P)leq L(f,P_N)$ (why?)



And how $f$ is constant in each interval of the form $[1/(n+1),1/n]$ the infimum's $m_i$ are just $f(1/i)=1/i$ and $m_0=0$... We have
$$L(f;P)= sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)$$



Therefore,
$$sup _P L(f;P)leq sup _{P_N} L(f;P_N)= lim _{Nto infty} sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right) = sum limits _{i=1}^{infty} frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)=frac{pi^2}{6} -1$$



and you also can readly prove $geq$ to conclude equality.






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    up vote
    1
    down vote













    For every partition $P= {x_j}_0^n$ where $x_0 = 0, x_n = 1$, there is some $N$ s.t. $x_1 in (1/(N+1), 1/N]$. Then on $[0, x_1]$, the infimum is $0$. On $[1/(N+1), 1]$, $f$ is a step function, i.e. piecewise constant function, then the lower Riemann sum is easy to find:
    $$
    {underline int}_{x_1}^1 f leqslant underline int_{1/(N+1)}^1 f = sum_1^N int_{1/(n+1)}^{1/n} f = sum_1^N left(frac 1n - frac 1{n+1}right) frac 1n = -1 + frac 1{N+1} +sum_1^N frac 1{n^2}.
    $$

    Thus
    $$
    -1 +frac 1{N+1} + sum_1^N frac 1{n^2}underline int_{1/(N+1)}^1 f leqslant underline int_0^1 f leqslant underline int_0^{x_1} f +underline int_{x_1}^1 f leqslant x_1 + underline int_{1/(N+1)}^1 f = x_1 - 1+frac 1{N+1} + sum_1^N frac 1{n^2}.
    $$



    Now let the mesh $delta$ of $P$ goes to $0$. Since $x_1 leqslant delta$, $x_1 to 0$ as well, hence $1/(N+1) to 0$, then $Nto infty$. Take the limit $delta to 0^+$ w.r.t. the inequalities, we have
    $$
    underline int_0^1 f = -1+sum_1^infty frac 1{n^2} =- 1 + frac {pi^2}6.
    $$






    share|cite|improve this answer






























      up vote
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      The function $f$ is riemann integrable , so that the function $F:[0,1]rightarrow Bbb R$ defined by $F(x)={underline int}_x^1 f(t)dt $ is continuous. In particular, the limit $lim_{xrightarrow 0} F(x)$ exists and equals to $F(0)$. That is $${underline int}_0^1f(x)dx=F(0)=lim_{nrightarrow infty} {underline int}_{frac{1}{n+1}}^1f(t)dt.$$ Now notice that in the interval $[frac{1}{n+1},1]$ number of discontinuities of $f$ is finite, so that the riemann integral ${underline int}_{frac{1}{n+1}}^1f(t)dt=sum_{k=1}^nfrac{1}{k}(frac{1}{k}-frac{1}{k+1})=(sum_{k=1}^nfrac{1}{k^2})-(1-frac{1}{n+1})$. Therefore , $${underline int}_0^1f(t)dt=frac{pi^2}{6}-1.$$






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        3 Answers
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        3 Answers
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        up vote
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        down vote













        Consider the partitions $P_N={0}cup {frac{1}{n}: 1leq n leq N}$. Then the lower sum is $$L(f;P_N)=sum limits _{i=1}^N m_ileft(frac{1}{i}-frac{1}{i+1}right)+m_0(frac{1}{N}-0)$$



        Then given any partition $P$ you can find an $N$ such that $L(f;P)leq L(f,P_N)$ (why?)



        And how $f$ is constant in each interval of the form $[1/(n+1),1/n]$ the infimum's $m_i$ are just $f(1/i)=1/i$ and $m_0=0$... We have
        $$L(f;P)= sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)$$



        Therefore,
        $$sup _P L(f;P)leq sup _{P_N} L(f;P_N)= lim _{Nto infty} sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right) = sum limits _{i=1}^{infty} frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)=frac{pi^2}{6} -1$$



        and you also can readly prove $geq$ to conclude equality.






        share|cite|improve this answer

























          up vote
          1
          down vote













          Consider the partitions $P_N={0}cup {frac{1}{n}: 1leq n leq N}$. Then the lower sum is $$L(f;P_N)=sum limits _{i=1}^N m_ileft(frac{1}{i}-frac{1}{i+1}right)+m_0(frac{1}{N}-0)$$



          Then given any partition $P$ you can find an $N$ such that $L(f;P)leq L(f,P_N)$ (why?)



          And how $f$ is constant in each interval of the form $[1/(n+1),1/n]$ the infimum's $m_i$ are just $f(1/i)=1/i$ and $m_0=0$... We have
          $$L(f;P)= sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)$$



          Therefore,
          $$sup _P L(f;P)leq sup _{P_N} L(f;P_N)= lim _{Nto infty} sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right) = sum limits _{i=1}^{infty} frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)=frac{pi^2}{6} -1$$



          and you also can readly prove $geq$ to conclude equality.






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            Consider the partitions $P_N={0}cup {frac{1}{n}: 1leq n leq N}$. Then the lower sum is $$L(f;P_N)=sum limits _{i=1}^N m_ileft(frac{1}{i}-frac{1}{i+1}right)+m_0(frac{1}{N}-0)$$



            Then given any partition $P$ you can find an $N$ such that $L(f;P)leq L(f,P_N)$ (why?)



            And how $f$ is constant in each interval of the form $[1/(n+1),1/n]$ the infimum's $m_i$ are just $f(1/i)=1/i$ and $m_0=0$... We have
            $$L(f;P)= sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)$$



            Therefore,
            $$sup _P L(f;P)leq sup _{P_N} L(f;P_N)= lim _{Nto infty} sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right) = sum limits _{i=1}^{infty} frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)=frac{pi^2}{6} -1$$



            and you also can readly prove $geq$ to conclude equality.






            share|cite|improve this answer












            Consider the partitions $P_N={0}cup {frac{1}{n}: 1leq n leq N}$. Then the lower sum is $$L(f;P_N)=sum limits _{i=1}^N m_ileft(frac{1}{i}-frac{1}{i+1}right)+m_0(frac{1}{N}-0)$$



            Then given any partition $P$ you can find an $N$ such that $L(f;P)leq L(f,P_N)$ (why?)



            And how $f$ is constant in each interval of the form $[1/(n+1),1/n]$ the infimum's $m_i$ are just $f(1/i)=1/i$ and $m_0=0$... We have
            $$L(f;P)= sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)$$



            Therefore,
            $$sup _P L(f;P)leq sup _{P_N} L(f;P_N)= lim _{Nto infty} sum limits _{i=1}^N frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right) = sum limits _{i=1}^{infty} frac{1}{i}left(frac{1}{i}-frac{1}{i+1}right)=frac{pi^2}{6} -1$$



            and you also can readly prove $geq$ to conclude equality.







            share|cite|improve this answer












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            answered 58 mins ago









            Robson

            751221




            751221






















                up vote
                1
                down vote













                For every partition $P= {x_j}_0^n$ where $x_0 = 0, x_n = 1$, there is some $N$ s.t. $x_1 in (1/(N+1), 1/N]$. Then on $[0, x_1]$, the infimum is $0$. On $[1/(N+1), 1]$, $f$ is a step function, i.e. piecewise constant function, then the lower Riemann sum is easy to find:
                $$
                {underline int}_{x_1}^1 f leqslant underline int_{1/(N+1)}^1 f = sum_1^N int_{1/(n+1)}^{1/n} f = sum_1^N left(frac 1n - frac 1{n+1}right) frac 1n = -1 + frac 1{N+1} +sum_1^N frac 1{n^2}.
                $$

                Thus
                $$
                -1 +frac 1{N+1} + sum_1^N frac 1{n^2}underline int_{1/(N+1)}^1 f leqslant underline int_0^1 f leqslant underline int_0^{x_1} f +underline int_{x_1}^1 f leqslant x_1 + underline int_{1/(N+1)}^1 f = x_1 - 1+frac 1{N+1} + sum_1^N frac 1{n^2}.
                $$



                Now let the mesh $delta$ of $P$ goes to $0$. Since $x_1 leqslant delta$, $x_1 to 0$ as well, hence $1/(N+1) to 0$, then $Nto infty$. Take the limit $delta to 0^+$ w.r.t. the inequalities, we have
                $$
                underline int_0^1 f = -1+sum_1^infty frac 1{n^2} =- 1 + frac {pi^2}6.
                $$






                share|cite|improve this answer



























                  up vote
                  1
                  down vote













                  For every partition $P= {x_j}_0^n$ where $x_0 = 0, x_n = 1$, there is some $N$ s.t. $x_1 in (1/(N+1), 1/N]$. Then on $[0, x_1]$, the infimum is $0$. On $[1/(N+1), 1]$, $f$ is a step function, i.e. piecewise constant function, then the lower Riemann sum is easy to find:
                  $$
                  {underline int}_{x_1}^1 f leqslant underline int_{1/(N+1)}^1 f = sum_1^N int_{1/(n+1)}^{1/n} f = sum_1^N left(frac 1n - frac 1{n+1}right) frac 1n = -1 + frac 1{N+1} +sum_1^N frac 1{n^2}.
                  $$

                  Thus
                  $$
                  -1 +frac 1{N+1} + sum_1^N frac 1{n^2}underline int_{1/(N+1)}^1 f leqslant underline int_0^1 f leqslant underline int_0^{x_1} f +underline int_{x_1}^1 f leqslant x_1 + underline int_{1/(N+1)}^1 f = x_1 - 1+frac 1{N+1} + sum_1^N frac 1{n^2}.
                  $$



                  Now let the mesh $delta$ of $P$ goes to $0$. Since $x_1 leqslant delta$, $x_1 to 0$ as well, hence $1/(N+1) to 0$, then $Nto infty$. Take the limit $delta to 0^+$ w.r.t. the inequalities, we have
                  $$
                  underline int_0^1 f = -1+sum_1^infty frac 1{n^2} =- 1 + frac {pi^2}6.
                  $$






                  share|cite|improve this answer

























                    up vote
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                    down vote










                    up vote
                    1
                    down vote









                    For every partition $P= {x_j}_0^n$ where $x_0 = 0, x_n = 1$, there is some $N$ s.t. $x_1 in (1/(N+1), 1/N]$. Then on $[0, x_1]$, the infimum is $0$. On $[1/(N+1), 1]$, $f$ is a step function, i.e. piecewise constant function, then the lower Riemann sum is easy to find:
                    $$
                    {underline int}_{x_1}^1 f leqslant underline int_{1/(N+1)}^1 f = sum_1^N int_{1/(n+1)}^{1/n} f = sum_1^N left(frac 1n - frac 1{n+1}right) frac 1n = -1 + frac 1{N+1} +sum_1^N frac 1{n^2}.
                    $$

                    Thus
                    $$
                    -1 +frac 1{N+1} + sum_1^N frac 1{n^2}underline int_{1/(N+1)}^1 f leqslant underline int_0^1 f leqslant underline int_0^{x_1} f +underline int_{x_1}^1 f leqslant x_1 + underline int_{1/(N+1)}^1 f = x_1 - 1+frac 1{N+1} + sum_1^N frac 1{n^2}.
                    $$



                    Now let the mesh $delta$ of $P$ goes to $0$. Since $x_1 leqslant delta$, $x_1 to 0$ as well, hence $1/(N+1) to 0$, then $Nto infty$. Take the limit $delta to 0^+$ w.r.t. the inequalities, we have
                    $$
                    underline int_0^1 f = -1+sum_1^infty frac 1{n^2} =- 1 + frac {pi^2}6.
                    $$






                    share|cite|improve this answer














                    For every partition $P= {x_j}_0^n$ where $x_0 = 0, x_n = 1$, there is some $N$ s.t. $x_1 in (1/(N+1), 1/N]$. Then on $[0, x_1]$, the infimum is $0$. On $[1/(N+1), 1]$, $f$ is a step function, i.e. piecewise constant function, then the lower Riemann sum is easy to find:
                    $$
                    {underline int}_{x_1}^1 f leqslant underline int_{1/(N+1)}^1 f = sum_1^N int_{1/(n+1)}^{1/n} f = sum_1^N left(frac 1n - frac 1{n+1}right) frac 1n = -1 + frac 1{N+1} +sum_1^N frac 1{n^2}.
                    $$

                    Thus
                    $$
                    -1 +frac 1{N+1} + sum_1^N frac 1{n^2}underline int_{1/(N+1)}^1 f leqslant underline int_0^1 f leqslant underline int_0^{x_1} f +underline int_{x_1}^1 f leqslant x_1 + underline int_{1/(N+1)}^1 f = x_1 - 1+frac 1{N+1} + sum_1^N frac 1{n^2}.
                    $$



                    Now let the mesh $delta$ of $P$ goes to $0$. Since $x_1 leqslant delta$, $x_1 to 0$ as well, hence $1/(N+1) to 0$, then $Nto infty$. Take the limit $delta to 0^+$ w.r.t. the inequalities, we have
                    $$
                    underline int_0^1 f = -1+sum_1^infty frac 1{n^2} =- 1 + frac {pi^2}6.
                    $$







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                    edited 31 mins ago

























                    answered 47 mins ago









                    xbh

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                        1
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                        The function $f$ is riemann integrable , so that the function $F:[0,1]rightarrow Bbb R$ defined by $F(x)={underline int}_x^1 f(t)dt $ is continuous. In particular, the limit $lim_{xrightarrow 0} F(x)$ exists and equals to $F(0)$. That is $${underline int}_0^1f(x)dx=F(0)=lim_{nrightarrow infty} {underline int}_{frac{1}{n+1}}^1f(t)dt.$$ Now notice that in the interval $[frac{1}{n+1},1]$ number of discontinuities of $f$ is finite, so that the riemann integral ${underline int}_{frac{1}{n+1}}^1f(t)dt=sum_{k=1}^nfrac{1}{k}(frac{1}{k}-frac{1}{k+1})=(sum_{k=1}^nfrac{1}{k^2})-(1-frac{1}{n+1})$. Therefore , $${underline int}_0^1f(t)dt=frac{pi^2}{6}-1.$$






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                          down vote













                          The function $f$ is riemann integrable , so that the function $F:[0,1]rightarrow Bbb R$ defined by $F(x)={underline int}_x^1 f(t)dt $ is continuous. In particular, the limit $lim_{xrightarrow 0} F(x)$ exists and equals to $F(0)$. That is $${underline int}_0^1f(x)dx=F(0)=lim_{nrightarrow infty} {underline int}_{frac{1}{n+1}}^1f(t)dt.$$ Now notice that in the interval $[frac{1}{n+1},1]$ number of discontinuities of $f$ is finite, so that the riemann integral ${underline int}_{frac{1}{n+1}}^1f(t)dt=sum_{k=1}^nfrac{1}{k}(frac{1}{k}-frac{1}{k+1})=(sum_{k=1}^nfrac{1}{k^2})-(1-frac{1}{n+1})$. Therefore , $${underline int}_0^1f(t)dt=frac{pi^2}{6}-1.$$






                          share|cite|improve this answer

























                            up vote
                            1
                            down vote










                            up vote
                            1
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                            The function $f$ is riemann integrable , so that the function $F:[0,1]rightarrow Bbb R$ defined by $F(x)={underline int}_x^1 f(t)dt $ is continuous. In particular, the limit $lim_{xrightarrow 0} F(x)$ exists and equals to $F(0)$. That is $${underline int}_0^1f(x)dx=F(0)=lim_{nrightarrow infty} {underline int}_{frac{1}{n+1}}^1f(t)dt.$$ Now notice that in the interval $[frac{1}{n+1},1]$ number of discontinuities of $f$ is finite, so that the riemann integral ${underline int}_{frac{1}{n+1}}^1f(t)dt=sum_{k=1}^nfrac{1}{k}(frac{1}{k}-frac{1}{k+1})=(sum_{k=1}^nfrac{1}{k^2})-(1-frac{1}{n+1})$. Therefore , $${underline int}_0^1f(t)dt=frac{pi^2}{6}-1.$$






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                            The function $f$ is riemann integrable , so that the function $F:[0,1]rightarrow Bbb R$ defined by $F(x)={underline int}_x^1 f(t)dt $ is continuous. In particular, the limit $lim_{xrightarrow 0} F(x)$ exists and equals to $F(0)$. That is $${underline int}_0^1f(x)dx=F(0)=lim_{nrightarrow infty} {underline int}_{frac{1}{n+1}}^1f(t)dt.$$ Now notice that in the interval $[frac{1}{n+1},1]$ number of discontinuities of $f$ is finite, so that the riemann integral ${underline int}_{frac{1}{n+1}}^1f(t)dt=sum_{k=1}^nfrac{1}{k}(frac{1}{k}-frac{1}{k+1})=(sum_{k=1}^nfrac{1}{k^2})-(1-frac{1}{n+1})$. Therefore , $${underline int}_0^1f(t)dt=frac{pi^2}{6}-1.$$







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                            answered 1 hour ago









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