Integral of Fractional Part $int_{0}^{1} { frac{1}{x} }dx$












1












$begingroup$


Does the integral exist? $displaystyleint_{0}^{1}{frac{1}{x}}dx,quad$ where {x} is the fractional part.



I have broken it into $$displaystyleint_{0}^{1}frac{1}{x}-lfloor frac{1}{x} rfloor dx = lim_{xto 0}-ln{x} -lim_{ntoinfty} H_n+1$$



This looks very close to a negative Euler-Mascheroni const + 1, although $gamma$ is defined as $$lim_{ntoinfty}quad H_n - ln{n}$$



Are $$lim_{xtoinfty}quad ln{x} text{ and } lim_{xto 0}quad -ln{x}$$ interchangeable in these cases? If so, it would seem then the integral would evaluate to $-gamma+1$, as is shown from an area approximation from .001 to 1 on my calculator.



Any help is appreciated, thanks!






definite-integrals improper-integrals fractional-part







share|cite|improve this question









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  • 3




    $begingroup$
    I notice there is no tag for Euler-Mascheroni constant. Perhaps someone would add it?
    $endgroup$
    – Jasper Braun
    Jun 19 '16 at 2:30






  • 3




    $begingroup$
    $$int_0^1( frac{1}{x}- lfloor frac{1}{x}rfloor )dx = int_1^infty (x- lfloor x rfloor) frac{dx}{x^2} = lim_{n to infty} int_1^n ( x- lfloor x rfloor) frac{dx}{x^2} =lim_{n to infty} ln n - sum_{k=1}^{n-1} int_k^{k+1} frac{k}{x^2} dx = ln n - sum_{k=1}^{n-1} k (frac{1}{k}-frac{1}{k+1}) =lim_{n to infty} ln n - sum_{k=1}^{n-1} frac{1}{k+1} = lim_{n to infty} ln n - H_{n}+1 =1-gamma$$
    $endgroup$
    – reuns
    Jun 19 '16 at 2:39












  • $begingroup$
    @user1952009 you might want to write that as an answer because (1) it looks correct, and (2) it doesn't fit/format well inside of the comment
    $endgroup$
    – Chill2Macht
    Jun 19 '16 at 2:42










  • $begingroup$
    @user1952009 If you insist, although personally I think you are being modest Also it isn't necessarily one line, if you explained each of the 6/7 steps
    $endgroup$
    – Chill2Macht
    Jun 19 '16 at 2:48










  • $begingroup$
    This is also how you see that $$zeta(s) = sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx= frac{s}{s-1} - int_1^infty (x-lfloor x rfloor) x^{-s-1} dx = frac{1}{s-1} + gamma + o(1)$$ as $s to 1^+$ ($ sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx$ being some sort of integration by parts, see en.wikipedia.org/wiki/Abel%27s_summation_formula )
    $endgroup$
    – reuns
    Jun 19 '16 at 2:49


















1












$begingroup$


Does the integral exist? $displaystyleint_{0}^{1}{frac{1}{x}}dx,quad$ where {x} is the fractional part.



I have broken it into $$displaystyleint_{0}^{1}frac{1}{x}-lfloor frac{1}{x} rfloor dx = lim_{xto 0}-ln{x} -lim_{ntoinfty} H_n+1$$



This looks very close to a negative Euler-Mascheroni const + 1, although $gamma$ is defined as $$lim_{ntoinfty}quad H_n - ln{n}$$



Are $$lim_{xtoinfty}quad ln{x} text{ and } lim_{xto 0}quad -ln{x}$$ interchangeable in these cases? If so, it would seem then the integral would evaluate to $-gamma+1$, as is shown from an area approximation from .001 to 1 on my calculator.



Any help is appreciated, thanks!






definite-integrals improper-integrals fractional-part







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    I notice there is no tag for Euler-Mascheroni constant. Perhaps someone would add it?
    $endgroup$
    – Jasper Braun
    Jun 19 '16 at 2:30






  • 3




    $begingroup$
    $$int_0^1( frac{1}{x}- lfloor frac{1}{x}rfloor )dx = int_1^infty (x- lfloor x rfloor) frac{dx}{x^2} = lim_{n to infty} int_1^n ( x- lfloor x rfloor) frac{dx}{x^2} =lim_{n to infty} ln n - sum_{k=1}^{n-1} int_k^{k+1} frac{k}{x^2} dx = ln n - sum_{k=1}^{n-1} k (frac{1}{k}-frac{1}{k+1}) =lim_{n to infty} ln n - sum_{k=1}^{n-1} frac{1}{k+1} = lim_{n to infty} ln n - H_{n}+1 =1-gamma$$
    $endgroup$
    – reuns
    Jun 19 '16 at 2:39












  • $begingroup$
    @user1952009 you might want to write that as an answer because (1) it looks correct, and (2) it doesn't fit/format well inside of the comment
    $endgroup$
    – Chill2Macht
    Jun 19 '16 at 2:42










  • $begingroup$
    @user1952009 If you insist, although personally I think you are being modest Also it isn't necessarily one line, if you explained each of the 6/7 steps
    $endgroup$
    – Chill2Macht
    Jun 19 '16 at 2:48










  • $begingroup$
    This is also how you see that $$zeta(s) = sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx= frac{s}{s-1} - int_1^infty (x-lfloor x rfloor) x^{-s-1} dx = frac{1}{s-1} + gamma + o(1)$$ as $s to 1^+$ ($ sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx$ being some sort of integration by parts, see en.wikipedia.org/wiki/Abel%27s_summation_formula )
    $endgroup$
    – reuns
    Jun 19 '16 at 2:49
















1












1








1





$begingroup$


Does the integral exist? $displaystyleint_{0}^{1}{frac{1}{x}}dx,quad$ where {x} is the fractional part.



I have broken it into $$displaystyleint_{0}^{1}frac{1}{x}-lfloor frac{1}{x} rfloor dx = lim_{xto 0}-ln{x} -lim_{ntoinfty} H_n+1$$



This looks very close to a negative Euler-Mascheroni const + 1, although $gamma$ is defined as $$lim_{ntoinfty}quad H_n - ln{n}$$



Are $$lim_{xtoinfty}quad ln{x} text{ and } lim_{xto 0}quad -ln{x}$$ interchangeable in these cases? If so, it would seem then the integral would evaluate to $-gamma+1$, as is shown from an area approximation from .001 to 1 on my calculator.



Any help is appreciated, thanks!






definite-integrals improper-integrals fractional-part







share|cite|improve this question









$endgroup$




Does the integral exist? $displaystyleint_{0}^{1}{frac{1}{x}}dx,quad$ where {x} is the fractional part.



I have broken it into $$displaystyleint_{0}^{1}frac{1}{x}-lfloor frac{1}{x} rfloor dx = lim_{xto 0}-ln{x} -lim_{ntoinfty} H_n+1$$



This looks very close to a negative Euler-Mascheroni const + 1, although $gamma$ is defined as $$lim_{ntoinfty}quad H_n - ln{n}$$



Are $$lim_{xtoinfty}quad ln{x} text{ and } lim_{xto 0}quad -ln{x}$$ interchangeable in these cases? If so, it would seem then the integral would evaluate to $-gamma+1$, as is shown from an area approximation from .001 to 1 on my calculator.



Any help is appreciated, thanks!






definite-integrals improper-integrals fractional-part




definite-integrals improper-integrals fractional-part






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jun 19 '16 at 2:28









Jasper BraunJasper Braun

385




385








  • 3




    $begingroup$
    I notice there is no tag for Euler-Mascheroni constant. Perhaps someone would add it?
    $endgroup$
    – Jasper Braun
    Jun 19 '16 at 2:30






  • 3




    $begingroup$
    $$int_0^1( frac{1}{x}- lfloor frac{1}{x}rfloor )dx = int_1^infty (x- lfloor x rfloor) frac{dx}{x^2} = lim_{n to infty} int_1^n ( x- lfloor x rfloor) frac{dx}{x^2} =lim_{n to infty} ln n - sum_{k=1}^{n-1} int_k^{k+1} frac{k}{x^2} dx = ln n - sum_{k=1}^{n-1} k (frac{1}{k}-frac{1}{k+1}) =lim_{n to infty} ln n - sum_{k=1}^{n-1} frac{1}{k+1} = lim_{n to infty} ln n - H_{n}+1 =1-gamma$$
    $endgroup$
    – reuns
    Jun 19 '16 at 2:39












  • $begingroup$
    @user1952009 you might want to write that as an answer because (1) it looks correct, and (2) it doesn't fit/format well inside of the comment
    $endgroup$
    – Chill2Macht
    Jun 19 '16 at 2:42










  • $begingroup$
    @user1952009 If you insist, although personally I think you are being modest Also it isn't necessarily one line, if you explained each of the 6/7 steps
    $endgroup$
    – Chill2Macht
    Jun 19 '16 at 2:48










  • $begingroup$
    This is also how you see that $$zeta(s) = sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx= frac{s}{s-1} - int_1^infty (x-lfloor x rfloor) x^{-s-1} dx = frac{1}{s-1} + gamma + o(1)$$ as $s to 1^+$ ($ sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx$ being some sort of integration by parts, see en.wikipedia.org/wiki/Abel%27s_summation_formula )
    $endgroup$
    – reuns
    Jun 19 '16 at 2:49
















  • 3




    $begingroup$
    I notice there is no tag for Euler-Mascheroni constant. Perhaps someone would add it?
    $endgroup$
    – Jasper Braun
    Jun 19 '16 at 2:30






  • 3




    $begingroup$
    $$int_0^1( frac{1}{x}- lfloor frac{1}{x}rfloor )dx = int_1^infty (x- lfloor x rfloor) frac{dx}{x^2} = lim_{n to infty} int_1^n ( x- lfloor x rfloor) frac{dx}{x^2} =lim_{n to infty} ln n - sum_{k=1}^{n-1} int_k^{k+1} frac{k}{x^2} dx = ln n - sum_{k=1}^{n-1} k (frac{1}{k}-frac{1}{k+1}) =lim_{n to infty} ln n - sum_{k=1}^{n-1} frac{1}{k+1} = lim_{n to infty} ln n - H_{n}+1 =1-gamma$$
    $endgroup$
    – reuns
    Jun 19 '16 at 2:39












  • $begingroup$
    @user1952009 you might want to write that as an answer because (1) it looks correct, and (2) it doesn't fit/format well inside of the comment
    $endgroup$
    – Chill2Macht
    Jun 19 '16 at 2:42










  • $begingroup$
    @user1952009 If you insist, although personally I think you are being modest Also it isn't necessarily one line, if you explained each of the 6/7 steps
    $endgroup$
    – Chill2Macht
    Jun 19 '16 at 2:48










  • $begingroup$
    This is also how you see that $$zeta(s) = sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx= frac{s}{s-1} - int_1^infty (x-lfloor x rfloor) x^{-s-1} dx = frac{1}{s-1} + gamma + o(1)$$ as $s to 1^+$ ($ sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx$ being some sort of integration by parts, see en.wikipedia.org/wiki/Abel%27s_summation_formula )
    $endgroup$
    – reuns
    Jun 19 '16 at 2:49










3




3




$begingroup$
I notice there is no tag for Euler-Mascheroni constant. Perhaps someone would add it?
$endgroup$
– Jasper Braun
Jun 19 '16 at 2:30




$begingroup$
I notice there is no tag for Euler-Mascheroni constant. Perhaps someone would add it?
$endgroup$
– Jasper Braun
Jun 19 '16 at 2:30




3




3




$begingroup$
$$int_0^1( frac{1}{x}- lfloor frac{1}{x}rfloor )dx = int_1^infty (x- lfloor x rfloor) frac{dx}{x^2} = lim_{n to infty} int_1^n ( x- lfloor x rfloor) frac{dx}{x^2} =lim_{n to infty} ln n - sum_{k=1}^{n-1} int_k^{k+1} frac{k}{x^2} dx = ln n - sum_{k=1}^{n-1} k (frac{1}{k}-frac{1}{k+1}) =lim_{n to infty} ln n - sum_{k=1}^{n-1} frac{1}{k+1} = lim_{n to infty} ln n - H_{n}+1 =1-gamma$$
$endgroup$
– reuns
Jun 19 '16 at 2:39






$begingroup$
$$int_0^1( frac{1}{x}- lfloor frac{1}{x}rfloor )dx = int_1^infty (x- lfloor x rfloor) frac{dx}{x^2} = lim_{n to infty} int_1^n ( x- lfloor x rfloor) frac{dx}{x^2} =lim_{n to infty} ln n - sum_{k=1}^{n-1} int_k^{k+1} frac{k}{x^2} dx = ln n - sum_{k=1}^{n-1} k (frac{1}{k}-frac{1}{k+1}) =lim_{n to infty} ln n - sum_{k=1}^{n-1} frac{1}{k+1} = lim_{n to infty} ln n - H_{n}+1 =1-gamma$$
$endgroup$
– reuns
Jun 19 '16 at 2:39














$begingroup$
@user1952009 you might want to write that as an answer because (1) it looks correct, and (2) it doesn't fit/format well inside of the comment
$endgroup$
– Chill2Macht
Jun 19 '16 at 2:42




$begingroup$
@user1952009 you might want to write that as an answer because (1) it looks correct, and (2) it doesn't fit/format well inside of the comment
$endgroup$
– Chill2Macht
Jun 19 '16 at 2:42












$begingroup$
@user1952009 If you insist, although personally I think you are being modest Also it isn't necessarily one line, if you explained each of the 6/7 steps
$endgroup$
– Chill2Macht
Jun 19 '16 at 2:48




$begingroup$
@user1952009 If you insist, although personally I think you are being modest Also it isn't necessarily one line, if you explained each of the 6/7 steps
$endgroup$
– Chill2Macht
Jun 19 '16 at 2:48












$begingroup$
This is also how you see that $$zeta(s) = sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx= frac{s}{s-1} - int_1^infty (x-lfloor x rfloor) x^{-s-1} dx = frac{1}{s-1} + gamma + o(1)$$ as $s to 1^+$ ($ sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx$ being some sort of integration by parts, see en.wikipedia.org/wiki/Abel%27s_summation_formula )
$endgroup$
– reuns
Jun 19 '16 at 2:49






$begingroup$
This is also how you see that $$zeta(s) = sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx= frac{s}{s-1} - int_1^infty (x-lfloor x rfloor) x^{-s-1} dx = frac{1}{s-1} + gamma + o(1)$$ as $s to 1^+$ ($ sum_{n=1}^infty n^{-s} = s int_1^infty lfloor x rfloor x^{-s-1} dx$ being some sort of integration by parts, see en.wikipedia.org/wiki/Abel%27s_summation_formula )
$endgroup$
– reuns
Jun 19 '16 at 2:49












1 Answer
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$begingroup$

As explained by @user1952009, use the substitution $u = dfrac{1}{x}$, $du = dfrac{-dx}{x^2}$. So
$$int_{0}^{1}{ dfrac{1}{x}} = int_{1}^{infty}dfrac{{u}}{u^2}du = $$
(Now use ${u} = u - lfloor urfloor$)
$$lim_{ntoinfty} text{ }int_{1}^{n}dfrac{u - lfloor urfloor}{u^2}du = lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}int_{k}^{k+1}dfrac{k}{x^2}dx =$$
$$lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}dfrac{1}{k+1} = lim_{ntoinfty}ln{n}-H_n+1 = Largeboxed{-gamma + 1}$$






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    $begingroup$

    As explained by @user1952009, use the substitution $u = dfrac{1}{x}$, $du = dfrac{-dx}{x^2}$. So
    $$int_{0}^{1}{ dfrac{1}{x}} = int_{1}^{infty}dfrac{{u}}{u^2}du = $$
    (Now use ${u} = u - lfloor urfloor$)
    $$lim_{ntoinfty} text{ }int_{1}^{n}dfrac{u - lfloor urfloor}{u^2}du = lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}int_{k}^{k+1}dfrac{k}{x^2}dx =$$
    $$lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}dfrac{1}{k+1} = lim_{ntoinfty}ln{n}-H_n+1 = Largeboxed{-gamma + 1}$$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      As explained by @user1952009, use the substitution $u = dfrac{1}{x}$, $du = dfrac{-dx}{x^2}$. So
      $$int_{0}^{1}{ dfrac{1}{x}} = int_{1}^{infty}dfrac{{u}}{u^2}du = $$
      (Now use ${u} = u - lfloor urfloor$)
      $$lim_{ntoinfty} text{ }int_{1}^{n}dfrac{u - lfloor urfloor}{u^2}du = lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}int_{k}^{k+1}dfrac{k}{x^2}dx =$$
      $$lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}dfrac{1}{k+1} = lim_{ntoinfty}ln{n}-H_n+1 = Largeboxed{-gamma + 1}$$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        As explained by @user1952009, use the substitution $u = dfrac{1}{x}$, $du = dfrac{-dx}{x^2}$. So
        $$int_{0}^{1}{ dfrac{1}{x}} = int_{1}^{infty}dfrac{{u}}{u^2}du = $$
        (Now use ${u} = u - lfloor urfloor$)
        $$lim_{ntoinfty} text{ }int_{1}^{n}dfrac{u - lfloor urfloor}{u^2}du = lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}int_{k}^{k+1}dfrac{k}{x^2}dx =$$
        $$lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}dfrac{1}{k+1} = lim_{ntoinfty}ln{n}-H_n+1 = Largeboxed{-gamma + 1}$$






        share|cite|improve this answer











        $endgroup$



        As explained by @user1952009, use the substitution $u = dfrac{1}{x}$, $du = dfrac{-dx}{x^2}$. So
        $$int_{0}^{1}{ dfrac{1}{x}} = int_{1}^{infty}dfrac{{u}}{u^2}du = $$
        (Now use ${u} = u - lfloor urfloor$)
        $$lim_{ntoinfty} text{ }int_{1}^{n}dfrac{u - lfloor urfloor}{u^2}du = lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}int_{k}^{k+1}dfrac{k}{x^2}dx =$$
        $$lim_{ntoinfty}ln{n}-sum_{k=1}^{n-1}dfrac{1}{k+1} = lim_{ntoinfty}ln{n}-H_n+1 = Largeboxed{-gamma + 1}$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 9:13









        Von Neumann

        16.3k72543




        16.3k72543










        answered Jun 19 '16 at 16:24









        Jasper BraunJasper Braun

        385




        385






























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            VLT Carioca

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            VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
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