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

I have broken it into $$\displaystyle\int_{0}^{1}\frac{1}{x}-\lfloor \frac{1}{x} \rfloor dx = \lim_{x\to 0}-\ln{x} -\lim_{n\to\infty} H_n+1$$

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

Are $$\lim_{x\to\infty}\quad \ln{x} \text{ and } \lim_{x\to 0}\quad -\ln{x}$$ 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!

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    I notice there is no tag for Euler-Mascheroni constant. Perhaps someone would add it? –  Jun 19 '16 at 02:30
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    $$\int_0^1( \frac{1}{x}- \lfloor \frac{1}{x}\rfloor )dx = \int_1^\infty (x- \lfloor x \rfloor) \frac{dx}{x^2} = \lim_{n \to \infty} \int_1^n ( x- \lfloor x \rfloor) \frac{dx}{x^2} =\lim_{n \to \infty} \ln n - \sum_{k=1}^{n-1} \int_k^{k+1} \frac{k}{x^2} dx = \ln n - \sum_{k=1}^{n-1} k (\frac{1}{k}-\frac{1}{k+1}) =\lim_{n \to \infty} \ln n - \sum_{k=1}^{n-1} \frac{1}{k+1} = \lim_{n \to \infty} \ln n - H_{n}+1 =1-\gamma$$ – reuns Jun 19 '16 at 02:39
  • @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 – Chill2Macht Jun 19 '16 at 02:42
  • @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

     – Chill2Macht Jun 19 '16 at 02:48
  • 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 https://en.wikipedia.org/wiki/Abel%27s_summation_formula ) – reuns Jun 19 '16 at 02:49
  • @user1952009 if you make your comment an answer, you can get the points associated with a correct answer, as well as the reasons William mentioned. –  Jun 19 '16 at 16:06
  • ? If you want an answer why don't you write it ? – reuns Jun 19 '16 at 16:11
  • @user1952009 Alright. –  Jun 19 '16 at 16:12

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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 $\{u\} = u - \lfloor u\rfloor$) $$\lim_{n\to\infty} \text{ }\int_{1}^{n}\dfrac{u - \lfloor u\rfloor}{u^2}du = \lim_{n\to\infty}\ln{n}-\sum_{k=1}^{n-1}\int_{k}^{k+1}\dfrac{k}{x^2}dx =$$ $$\lim_{n\to\infty}\ln{n}-\sum_{k=1}^{n-1}\dfrac{1}{k+1} = \lim_{n\to\infty}\ln{n}-H_n+1 = \Large\boxed{-\gamma + 1}$$

Enrico M.
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