How to solve Stieltjes integral $int_0^n f(x) d lfloor x rfloor$?












2












$begingroup$


I'm trying to solve the following



$$
int_0^n f(x) d lfloor x rfloor
$$



where $lfloor x rfloor$ is the floor function, $f : [0,n] rightarrow mathbb{R}$ is continuous, and $n in mathbb{N}$.



I know $frac{d}{dx} lfloor x rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.



Can anyone give some hints?










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  • 1




    $begingroup$
    The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
    $endgroup$
    – Kavi Rama Murthy
    Aug 31 '18 at 8:06








  • 3




    $begingroup$
    @KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
    $endgroup$
    – Batominovski
    Aug 31 '18 at 8:08


















2












$begingroup$


I'm trying to solve the following



$$
int_0^n f(x) d lfloor x rfloor
$$



where $lfloor x rfloor$ is the floor function, $f : [0,n] rightarrow mathbb{R}$ is continuous, and $n in mathbb{N}$.



I know $frac{d}{dx} lfloor x rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.



Can anyone give some hints?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
    $endgroup$
    – Kavi Rama Murthy
    Aug 31 '18 at 8:06








  • 3




    $begingroup$
    @KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
    $endgroup$
    – Batominovski
    Aug 31 '18 at 8:08
















2












2








2





$begingroup$


I'm trying to solve the following



$$
int_0^n f(x) d lfloor x rfloor
$$



where $lfloor x rfloor$ is the floor function, $f : [0,n] rightarrow mathbb{R}$ is continuous, and $n in mathbb{N}$.



I know $frac{d}{dx} lfloor x rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.



Can anyone give some hints?










share|cite|improve this question









$endgroup$




I'm trying to solve the following



$$
int_0^n f(x) d lfloor x rfloor
$$



where $lfloor x rfloor$ is the floor function, $f : [0,n] rightarrow mathbb{R}$ is continuous, and $n in mathbb{N}$.



I know $frac{d}{dx} lfloor x rfloor$ itself cannot be secured (i.e. not differentiable). I cannot proceed further.



Can anyone give some hints?







calculus integration stieltjes-integral






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asked Aug 31 '18 at 8:01









MoreblueMoreblue

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8931217








  • 1




    $begingroup$
    The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
    $endgroup$
    – Kavi Rama Murthy
    Aug 31 '18 at 8:06








  • 3




    $begingroup$
    @KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
    $endgroup$
    – Batominovski
    Aug 31 '18 at 8:08
















  • 1




    $begingroup$
    The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
    $endgroup$
    – Kavi Rama Murthy
    Aug 31 '18 at 8:06








  • 3




    $begingroup$
    @KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
    $endgroup$
    – Batominovski
    Aug 31 '18 at 8:08










1




1




$begingroup$
The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
$endgroup$
– Kavi Rama Murthy
Aug 31 '18 at 8:06






$begingroup$
The answer is $f(1)+...+f(n-1)$. Just write down Riemann - Steiltjes sums and take the limit.
$endgroup$
– Kavi Rama Murthy
Aug 31 '18 at 8:06






3




3




$begingroup$
@KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
$endgroup$
– Batominovski
Aug 31 '18 at 8:08






$begingroup$
@KaviRamaMurthy Shouldn't it be $$int_0^n,f(x),text{d}lfloor xrfloor = f(1)+f(2)+ldots+f(n),?$$ I think we have $$int_0^n,f(x),text{d}lceil xrceil=f(0)+f(1)+ldots+f(n-1),.$$
$endgroup$
– Batominovski
Aug 31 '18 at 8:08












1 Answer
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$begingroup$

Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:



$$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$



$$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$



$$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$



$$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$



since the associated measure for the floor-function is



$$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$






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    3












    $begingroup$

    Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:



    $$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$



    $$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$



    $$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$



    $$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$



    since the associated measure for the floor-function is



    $$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:



      $$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$



      $$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$



      $$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$



      $$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$



      since the associated measure for the floor-function is



      $$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:



        $$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$



        $$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$



        $$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$



        $$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$



        since the associated measure for the floor-function is



        $$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$






        share|cite|improve this answer











        $endgroup$



        Note that there is a difference between integrating $[0,n]$, $[0,n)$, $(0,n]$ and $(0,n)$:



        $$int_{[0,n]}f(x)dlfloor xrfloor =sum_{i=0}^nf(i)$$



        $$int_{[0,n)}f(x)dlfloor xrfloor =sum_{i=0}^{n-1}f(i)$$



        $$int_{(0,n]}f(x)dlfloor xrfloor =sum_{i=1}^nf(i)$$



        $$int_{(0,n)}f(x)dlfloor xrfloor =sum_{i=1}^{n-1}f(i)$$



        since the associated measure for the floor-function is



        $$mu_{lfloorcdotrfloor}=sum_{k=-infty}^infty delta_k$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 16 at 22:01

























        answered Jan 12 at 23:20









        user408858user408858

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