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Let $\displaystyle \gamma$ denote the Euler constant defined by $\displaystyle \gamma := \lim\limits_{n \to \infty} \left(\frac11+\frac12+\cdots+\frac1n- \log n\right)$.

Here is an integral for $\gamma$.

$$ \gamma = \int_0^1 \frac{ \left\{1/x\right\}1/x}{\lfloor1/x\rfloor} \mathrm{d}x $$

where $\displaystyle \left\{x\right\}$ and $\displaystyle \lfloor x \rfloor$ denote the fractional part and the integer part of $x$ respectively.

We may generalize the previous result.

Let $z$ be a complex number such that $\Re{z}>-1$. Then

$$ \psi(z+1) =\int_{0}^{1} \frac{z-\left\{1/x\right\}}{z+\lfloor1/x\rfloor} \: \frac{\mathrm{d}x }{x} $$ where $\displaystyle \psi:= \Gamma'/\Gamma.$

Could you prove it?

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    $\begingroup$ Why do you say it is new? $\endgroup$ – Thomas Andrews Jul 23 '14 at 23:57
  • $\begingroup$ @Thomas Andrews As far as I know, it is new :)! $\endgroup$ – Olivier Oloa Jul 24 '14 at 0:00
  • $\begingroup$ Why do you think it is true? $\endgroup$ – Hoseyn Heydari Jul 24 '14 at 0:00
  • $\begingroup$ @Hoseyn Heydari This is a (small) challenging question. Thanks. $\endgroup$ – Olivier Oloa Jul 24 '14 at 0:01
  • $\begingroup$ Just usually we try to avoid bragging in questions, and just because you haven't seen a trivial result... $\endgroup$ – Thomas Andrews Jul 24 '14 at 0:03
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Substitute $x=\frac{1}{u}$. Then $\frac{1}{x}dx = \frac{1}{x}\frac{-1}{u^2}du = -\frac{1}{u}du$.

So $$\int_{1}^\infty \frac{\{u\}}{u\lfloor u\rfloor}du = \int_{1}^\infty\left(\frac{1}{\lfloor u\rfloor}-\frac{1}{u}\right) du = \gamma$$

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  • $\begingroup$ Ok, it is not new at all! :) $\endgroup$ – Olivier Oloa Jul 24 '14 at 0:04
  • $\begingroup$ I think it is fair to accept this answer which was appropriate to the initial question! Tell me... Thanks! $\endgroup$ – Olivier Oloa Jul 24 '14 at 2:15

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