一组关于{x}的积分

\[\Large\displaystyle \int_{0}^{1}\left \{ \frac{1}{x} \right \}\mathrm{d}x~，~\int_{0}^{1}\left \{ \frac{1}{x} \right \}^{2}\mathrm{d}x~，~\int_{0}^{1}\left \{ \frac{1}{x} \right \}^{3}\mathrm{d}x\]

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\(\Large\mathbf{Solution:}\)
1.
\[\begin{align*}
{\int_{0}^{1}\left \{ \frac{1}{x} \right \}\mathrm{d}x} &= \int_{1}^{\infty }\frac{\left \{ t \right \}}{t^{2}}\mathrm{d}t=\sum_{k=1}^{\infty }\int_{k}^{k+1}\frac{t-k}{t^{2}}\mathrm{d}t\\
&=\sum_{k=1}^{\infty }\left ( \ln\frac{k+1}{k}-\frac{1}{k+1} \right )\\
&=\Large\boxed{\color{blue}{1-\gamma}}
\end{align*}\]

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2.我们计算比较一般的情况
\[\int_{0}^{1}\left \{ \frac{k}{x} \right \}^{2}\mathrm{d}x = k\int_{k}^{\infty }\frac{\left \{ t \right \}}{t^{2}}\mathrm{d}t=k\sum_{m=k}^{\infty }\left ( 2-2m\ln \frac{m+1}{m}-\frac{1}{m+1}\right )\]
令\(\displaystyle S_{n}=\sum_{m=k}^{n }\left ( 2-2m\ln \frac{m+1}{m}-\frac{1}{m+1}\right )\)，我们有
\[\begin{align*}
S_{n}&=\sum_{m=k}^{n }\left ( 2-2m\ln \frac{m+1}{m}-\frac{1}{m+1}\right )\\
&=2\left ( n-k+1 \right )-\left ( \frac{1}{k+1}+\frac{1}{k+2}+\cdots +\frac{1}{k+n} \right )\\
&~~~~-2n\ln\left ( n+1 \right )+2k\ln k+2\ln n!-2\ln k!
\end{align*}\]
由\(2\ln n!\sim \ln\left ( 2\pi \right )+\left ( 2n+1 \right )\ln n-2n\)，得
\[\begin{align*}
S_{n} &\sim 2\left ( 1-k \right )+\ln\left ( 2\pi \right )+2k\ln k-2\ln k!-2n\ln\frac{n+1}{n}\\
&~~~~-\left ( \frac{1}{k+1}+\cdots +\frac{1}{n+1}-\ln n \right )
\end{align*}\]
所以
\[\lim_{n\rightarrow \infty }S_{n}=\ln\left ( 2\pi \right )-\gamma +1+\frac{1}{2}+\cdots +\frac{1}{k}+2k\ln k-2k-2\ln k!\]
因此，当\(k=1\)，则有
\[\Large\boxed{\displaystyle \int_{0}^{1}\left \{ \frac{1}{x} \right \}^{2}\mathrm{d}x=\color{Blue}{ \ln \left (2\pi \right ) -\gamma -1}}
\]

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3.
\[\begin{align*}
\int_{0}^{1}\left \{ \frac{1}{x} \right \}^{3}\mathrm{d}x &= \int_{1}^{\infty }\frac{\left \{ t \right \}^{3}}{t^{2}}\mathrm{d}t=\sum_{k=1}^{\infty }\int_{k}^{k+1}\frac{\left ( t-k \right )^{3}}{t^{2}}\mathrm{d}t\\
&=\sum_{k=1}^{\infty }\left ( 3k^2\ln\frac{k+1}{k}+\frac{3}{2}-3k-\frac{1}{k+1} \right )
\end{align*}\]
令\(\displaystyle S_{n}=\sum_{k=1}^{n }\left ( 3k^2\ln\frac{k+1}{k}+\frac{3}{2}-3k-\frac{1}{k+1} \right )\)，简单计算得
\[S_{n}=1-\left ( 1+\frac{1}{2}+\cdots +\frac{1}{n+1}-\ln n \right )-\frac{3}{2}n^{2}-\ln n+3\sum_{k=1}^{n}k^2\ln\frac{k+1}{k}\]
其中
\[\begin{align*}
&\sum_{k=1}^{n}k^2\ln\frac{k+1}{k} =\ln\prod_{k=1}^{n}\left ( \frac{k+1}{k} \right )^{k^{2}}=\ln\left [ \frac{\left ( n+1 \right )^{n^{2}}\cdot n!}{\left ( 2^{2}\cdot 3^{3}\cdots n^{n} \right )^{2}} \right ]\\
\Rightarrow &-\frac{3}{2}n^{2}-\ln n+3\sum_{k=1}^{n}k^2\ln\frac{k+1}{k}=\ln\left [ \frac{\left ( n+1 \right )^{3n^2}\cdot \left ( n! \right )^{3}}{\left ( 2^{2}\cdot 3^{3}\cdots n^{n} \right )^{6}\cdot e^{\frac{3n^{2}}{2}}\cdot n} \right ]
\end{align*}\]
令
\[a_{n}=\frac{\left ( n+1 \right )^{3n^2}\cdot \left ( n! \right )^{3}}{\left ( 2^{2}\cdot 3^{3}\cdots n^{n} \right )^{6}\cdot e^{\frac{3n^{2}}{2}}\cdot n}=\frac{n^{3n^2+3n+\frac{1}{2}}\cdot e^{-\frac{3n^{2}}{2}}}{\left ( 2^{2}\cdot 3^{3}\cdots n^{n} \right )^{6}}\cdot \frac{\left ( n+1 \right )^{3n^{2}}\cdot \left ( n! \right )^{3}}{n^{3n^2+3n+\frac{3}{2}}}\]
易知第一部分的极限为\(\dfrac{1}{\mathbf{A}^{6}}\)，对于第二部分，使用 Stirling's formula \(\displaystyle n!\sim \sqrt{2\pi n}\left ( \frac{n}{e} \right )^{n}\) 有
\[\frac{\left ( n+1 \right )^{3n^{2}}\cdot \left ( n! \right )^{3}}{n^{3n^2+3n+\frac{3}{2}}}\sim \left ( 2\pi \right )^{\frac{3}{2}}\left [ \left ( \frac{n+1}{n} \right )^{n}\frac{1}{e} \right ]^{3n}\rightarrow \left ( 2\pi \right )^{\frac{3}{2}}e^{-\frac{3}{2}}\]
所以 \(\displaystyle x_{n}\rightarrow \frac{\left ( 2\pi \right )^{\frac{3}{2}}e^{-\frac{3}{2}}}{\mathbf{A}^{6}}\)，由此可得
\[\Large\boxed{\displaystyle {\int_{0}^{1}\left \{ \frac{1}{x} \right \}^{3}\mathrm{d}x}=\color{blue}{-\frac{1}{2}-\gamma +\frac{3}{2}\ln\left ( 2\pi \right )-6\ln \mathbf{A}}}
\]