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Today there is a new integral with fractional part function… At least, this is the last integral and ufficially recognised by the author himself has the most difficult on his book. We have $$\int_0^1 \sqrt{\frac{\left\{\frac1x\right\}}{1-\left\{\frac1x\right\}}}\frac{\mathrm{d}x}{1-x}$$ and the hint given by the book is to consider the integral definition of $\Gamma(x)$ but it doesn't help me anyway… Have you some ideas, also without Gamma function? The result is instead a very simple $\pi$ :)

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Substitute $x=1/t$ and then \begin{align*} &\int_1^{ + \infty } {\sqrt {\frac{{\left\{ t \right\}}}{{1 - \left\{ t \right\}}}} \frac{1}{{t(t - 1)}}dt} = \sum\limits_{n = 1}^\infty {\int_n^{n + 1} {\sqrt {\frac{{\left\{ t \right\}}}{{1 - \left\{ t \right\}}}} \frac{1}{{t(t - 1)}}dt} } \\ & = \sum\limits_{n = 1}^\infty {\int_0^1 {\sqrt {\frac{{\left\{ {n + s} \right\}}}{{1 - \left\{ {n + s} \right\}}}} \frac{1}{{(n + s)(n + s - 1)}}ds} } \\ &= \int_0^1 {\sqrt {\frac{s}{{1 - s}}} \left( {\sum\limits_{n = 1}^\infty {\frac{1}{{(n + s)(n + s - 1)}}} } \right)ds} \\ &= \int_0^1 {\sqrt {\frac{s}{{1 - s}}} \left( {\sum\limits_{n = 1}^\infty {\left( \frac{1}{{n + s - 1}}-\frac{1}{{n + s}}\right)} } \right)ds} \\ & = \int_0^1 {\frac{1}{{\sqrt {1 - s} }}\frac{1}{{\sqrt s }}ds} = B\left( {\tfrac{1}{2},\tfrac{1}{2}} \right) = \frac{{\Gamma \left( {\frac{1}{2}} \right)\Gamma \left( {\frac{1}{2}} \right)}}{{\Gamma (1)}} = \pi . \end{align*} So we used the integral representation of the beta function.

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