Evaluate $\int_{0}^{1}\{1/x\}^2\,dx$

Question

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 $\displaystyle{x \in \left( {\frac{1}{{n + 1}},\frac{1}{n}} \right]}$

$$\displaystyle{n \leqslant \frac{1}{x} < n + 1 \Rightarrow \left\{ {\frac{1}{x}} \right\} = \frac{1}{x} - n \Rightarrow \int\limits_{1/\left( {n + 1} \right)}^{1/n} {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} = \int\limits_{1/\left( {n + 1} \right)}^{1/n} {{{\left( {\frac{1}{x} - n} \right)}^2}dx} = }$$

$$\displaystyle{ = \int\limits_{1/\left( {n + 1} \right)}^{1/n} {\left( {\frac{1}{{{x^2}}} - \frac{2}{x}n + {n^2}} \right)dx} = 1 + \frac{n}{{n + 1}} - 2n\ln \frac{{n + 1}}{n} \Rightarrow \boxed{\int\limits_0^1 {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} = \sum\limits_{n = 1}^\infty {\left( {1 + \frac{n}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} }}$$

$$\displaystyle{\sum\limits_{n = 1}^\infty {\left( {1 + \frac{n}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} = \sum\limits_{n = 1}^\infty {\left( {2 - \frac{1}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} = \mathop {\lim }\limits_{N \to \infty } \sum\limits_{n = 1}^N {\left( {2 - \frac{1}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} }$$

Answer 1

You made a good work, for sure.

Concerning the last summation $$S_N= \sum\limits_{n = 1}^N \left( {2 - \frac{1}{n + 1}} - 2n\log \left(\frac{n+1}{n}\right)\right) $$ it is given in terms of harmonic numbers, gamma function and the derivatives of the zeta function $$S_N=1+2N-H_{N+1}+2 \log (\Gamma (N+2))+$$ $$2 \left(\zeta ^{(1,0)}(-1,N+1)-\zeta ^{(1,0)}(-1,N+2)\right)$$ Using their asymptotics $$S_N=(\log (2 \pi )-\gamma-1)-\frac{1}{3 N}+\frac{5}{12 N^2}+O\left(\frac{1}{N^3}\right)$$ which is a very good approximation (for $N=10$, the relative error is $0.20$%).

Answer 2

\begin{align*} &amp; = 1 + 2\sum\limits_{n = 1}^N {\left( {\ln \frac{{e \cdot {n^n}}}{{{{\left( {n + 1} \right)}^n}}}} \right)} - \left( {{\gamma _N} + \ln N} \right) \\ &amp;= 1 - {\gamma _N} + 2\ln \left( {\prod\limits_{n = 1}^N {\frac{{e \cdot {n^n}}}{{{{\left( {n + 1} \right)}^n}}}} } \right) - \ln N \\ &amp;= 1 - {\gamma _N} + 2\ln \left( {\frac{{{e^N}}}{{\sqrt N }}\prod\limits_{n = 1}^N {\frac{{{n^n}}}{{{{\left( {n + 1} \right)}^n}}}} } \right) \\ &amp;= 1 - {\gamma _N} + 2\ln \left( {\frac{{{e^N}}}{{\sqrt N }}\frac{{N!}}{{{{\left( {N + 1} \right)}^N}}}} \right) \\ &amp;= 1 - {\gamma _N} + 2\ln \left( {\frac{{{e^N}}}{{\sqrt N }}\frac{{N!}}{{{{\left( {N + 1} \right)}^N}}}} \right). \end{align*}

However, we know that

$$\lim_{N \to \infty} \left( {\frac{{{e^N}}}{{\sqrt N }}\frac{{N!}}{{{{\left( {N + 1} \right)}^N}}}} \right) = \lim_{N \to \infty} \left( {\frac{{{e^N}}}{{{N^N}\sqrt N }}\frac{{N!}}{{{{\left( {\frac{{N + 1}}{N}} \right)}^N}}}} \right) = \frac{1}{e} \cdot \lim_{N \to \infty} \left( {\frac{{{e^N}}}{{{N^N}\sqrt N }}\sqrt{2\pi N} {{\left( {\frac{N}{e}} \right)}^N}} \right) = \frac{{\sqrt{2\pi}}}{e}$$

Therefore, we can conclude that

$$\int\limits_0^1 {{{\left\{ {\frac{1}{x}} \right\}}^2}dx} = \lim_{N \to \infty} \sum\limits_{n = 1}^N {\left( {2 - \frac{1}{{n + 1}} - 2n\ln \frac{{n + 1}}{n}} \right)} = 1 - \gamma + 2\ln \left( {\frac{{\sqrt{2\pi}}}{e}} \right) = - 1 - \gamma + \ln \left( {2\pi} \right)$$

where we have used the identities $$\sum\limits_{n = 1}^N {\frac{1}{n}} = {\gamma _N} + \ln N$$ with $$\lim_{N \to \infty} {\gamma _N} = \gamma$$ (Euler-Mascheroni constant) and $${\rm N}! \approx \sqrt{2\pi N} {\left( {\frac{N}{e}} \right)^N}$$

Answer 3

$$t=\frac{1}{x} , dx =\frac{-dt}{x^2}$$

$$\mathcal{I}:= \int_0^1 \left\{ \frac{1}{x}\right\}^2dx = \int_1^{\infty}\frac{(t- \lfloor t\rfloor)^2}{t^2} dt = \lim\limits_{n\to \infty} \sum_{k=1}^n\int_k^{k+1}\frac{t^2 -2tk +k^2}{t^2} dt $$

$$=\lim\limits_{n\to \infty} \sum_{k=1}^n 2k \ln(k)- 2k\ln(k+1)+2-\frac{1}{k+1} $$

$$S_N:= \sum_{k=1}^N2k \ln(k)- 2k\ln(k+1)+2-\frac{1}{k+1}= 2\ln(N!)-2N\ln(N+1) -H_{N+1}+2N+1 $$

$$= 2\ln\left(\frac{N!}{N^N}\right) -2N\ln\left(1+\frac{1}{N}\right)+2N +1 - \ln(N)+ (\ln(N)- H_{N+1})$$

$$ =2\ln\left(\frac{\sqrt{2 \pi N}}{e^N}\right)+ \ln\left(\frac{N!}{\sqrt{2\pi N} \left(\frac{n^n}{e^n} \right)}\right) -2N\ln\left(1+\frac{1}{N}\right)+2N +1 - \ln(N)+ (\ln(N)- H_{N+1}) $$

$$=\ln\left({2 \pi }\right) + \ln(N) - 2N -2N\ln\left(1+\frac{1}{N}\right)+2N +1 - \ln(N)+ (\ln(N)- H_{N+1})+ \ln\left(\frac{N!}{\sqrt{2\pi N} \left(\frac{n^n}{e^n} \right)}\right) $$

$$=\ln\left({2 \pi }\right) + -2N\ln\left(1+\frac{1}{N}\right)+1 + (\ln(N)- H_{N+1})+ \ln\left(\frac{N!}{\sqrt{2\pi N} \left(\frac{n^n}{e^n} \right)}\right) $$

$$\mathcal{I}= \lim_{N\to \infty} S_N= -2 +1 -\gamma + \ln(2 \pi) =-1-\gamma + \ln(2 \pi) $$

where $\gamma$ is Euler–Mascheroni constant

Answer 4

\begin{align*} \int_0^1\left\{\frac1x\right\}^2\;dx &= \int_0^1\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right)^2\;dx \\ &= \sum_{n=1}^{\infty}\int_{\frac1{n+1}}^{\frac1n}\left(\frac1x-\left\lfloor\frac1x\right\rfloor\right)^2\;dx \\ &= \sum_{n=1}^{\infty}\int_{\frac1{n+1}}^{\frac1n}\left(\frac1x-n\right)^2\;dx \\ &= \sum_{n=1}^{\infty}\int_{\frac1{n+1}}^{\frac1n}\left(\frac1{x^2}-\frac{2n}x+n^2\right)\;dx \\ &= \sum_{n=1}^{\infty}\left[-\frac1x-2n\log x+n^2x\right]_{x=\frac1{n+1}}^{\frac1n} \\ &= \sum_{n=1}^{\infty}\left[-n+(n+1)+2n\log(n)-2n\log(n+1)+n-\frac{n^2}{n+1}\right] \\ &= \sum_{n=1}^{\infty}\left[\frac{2n+1}{n+1}+2n\log(n)-2n\log(n+1)\right] \\ \end{align*}\begin{align*} &= \lim_{N\rightarrow\infty}\sum_{n=1}^N\left[2-\frac1{n+1}+2n\log(n)-2n\log(n+1)\right] \\ &= \lim_{N\rightarrow\infty}\left( 2N + 1 - H_{N+1} + 2\log(N!) - 2N\log(N+1) \right) \\ &= \lim_{N\rightarrow\infty}\left( 2N + 1 - H_{N+1} + 2\log\left(\sqrt{2\pi N}\left(\frac{N}e\right)^N\right) - 2\log\left((N+1)^N\right) \right) \\ &= \lim_{N\rightarrow\infty}\left( 2N + 1 - H_{N+1} + \log(2\pi) + \log N + 2\log\left(N^N\right) - 2N - 2\log\left((N+1)^N\right) \right) \\ &= \log(2\pi) + 1 + \lim_{N\rightarrow\infty}\left(-H_{N+1} + \log N - 2\log\left(\left(1+\frac1N\right)^N\right) \right) \\ &= \log(2\pi) + 1 - 2\log(e) - \lim_{N\rightarrow\infty}\left(H_{N+1} - \log N\right) \\ &= \log(2\pi) - 1 - \lim_{N\rightarrow\infty}\left(H_N - \log N + \frac1{N+1}\right) \\ &= \log(2\pi) - 1 - \gamma \;\blacksquare \end{align*}