Evaluate $$\displaystyle{\int_{0}^{1}\{1/x\}^2\,dx}$$ Where {•} is fractional part
My work
$$\displaystyle{\int\limits_0^1 {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} = \sum\limits_{n = 1}^\infty {\left( {\int\limits_{1/\left( {n + 1} \right)}^{1/n} {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} } \right)} }$$
For $\displaystyle{x \in \left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]}$
$$\displaystyle{n \leqslant \frac{1}{x} < n + 1 \Rightarrow \left\{ {\frac{1}{x}} \right\} = \frac{1}{x} - n \Rightarrow \int\limits_{1/\left( {n + 1} \right)}^{1/n} {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} = \int\limits_{1/\left( {n + 1} \right)}^{1/n} {{{\left( {\frac{1}{x} - n} \right)}^2}dx} = }$$
$$\displaystyle{ = \int\limits_{1/\left( {n + 1} \right)}^{1/n} {\left( {\frac{1}{{{x^2}}} - \frac{2}{x}n + {n^2}} \right)dx} = 1 + \frac{n}{{n + 1}} - 2n\ln \frac{{n + 1}}{n} \Rightarrow \boxed{\int\limits_0^1 {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} = \sum\limits_{n = 1}^\infty {\left( {1 + \frac{n}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} }}$$
$$\displaystyle{\sum\limits_{n = 1}^\infty {\left( {1 + \frac{n}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} = \sum\limits_{n = 1}^\infty {\left( {2 - \frac{1}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} = \mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {\left( {2 - \frac{1}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} }$$