一组关于{x}的积分

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


\(\Large\mathbf{Solution:}\)
1.

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


2.我们计算比较一般的情况

\[\int_{0}^{1}\left \{ \frac{k}{x} \right \}^{2}\mathrm{d}x = k\int_{k}^{\infty }\frac{\left \{ t \right \}}{t^{2}}\mathrm{d}t=k\sum_{m=k}^{\infty }\left ( 2-2m\ln \frac{m+1}{m}-\frac{1}{m+1}\right ) \]

\(\displaystyle S_{n}=\sum_{m=k}^{n }\left ( 2-2m\ln \frac{m+1}{m}-\frac{1}{m+1}\right )\),我们有

\[\begin{align*} S_{n}&=\sum_{m=k}^{n }\left ( 2-2m\ln \frac{m+1}{m}-\frac{1}{m+1}\right )\\ &=2\left ( n-k+1 \right )-\left ( \frac{1}{k+1}+\frac{1}{k+2}+\cdots +\frac{1}{k+n} \right )\\ &~~~~-2n\ln\left ( n+1 \right )+2k\ln k+2\ln n!-2\ln k! \end{align*}\]

\(2\ln n!\sim \ln\left ( 2\pi \right )+\left ( 2n+1 \right )\ln n-2n\),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

\[\Large\boxed{\displaystyle \int_{0}^{1}\left \{ \frac{1}{x} \right \}^{2}\mathrm{d}x=\color{Blue}{ \ln \left (2\pi \right ) -\gamma -1}} \]


3.

\[\begin{align*} \int_{0}^{1}\left \{ \frac{1}{x} \right \}^{3}\mathrm{d}x &= \int_{1}^{\infty }\frac{\left \{ t \right \}^{3}}{t^{2}}\mathrm{d}t=\sum_{k=1}^{\infty }\int_{k}^{k+1}\frac{\left ( t-k \right )^{3}}{t^{2}}\mathrm{d}t\\ &=\sum_{k=1}^{\infty }\left ( 3k^2\ln\frac{k+1}{k}+\frac{3}{2}-3k-\frac{1}{k+1} \right ) \end{align*}\]

\(\displaystyle S_{n}=\sum_{k=1}^{n }\left ( 3k^2\ln\frac{k+1}{k}+\frac{3}{2}-3k-\frac{1}{k+1} \right )\),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

\[\begin{align*} &\sum_{k=1}^{n}k^2\ln\frac{k+1}{k} =\ln\prod_{k=1}^{n}\left ( \frac{k+1}{k} \right )^{k^{2}}=\ln\left [ \frac{\left ( n+1 \right )^{n^{2}}\cdot n!}{\left ( 2^{2}\cdot 3^{3}\cdots n^{n} \right )^{2}} \right ]\\ \Rightarrow &-\frac{3}{2}n^{2}-\ln n+3\sum_{k=1}^{n}k^2\ln\frac{k+1}{k}=\ln\left [ \frac{\left ( n+1 \right )^{3n^2}\cdot \left ( n! \right )^{3}}{\left ( 2^{2}\cdot 3^{3}\cdots n^{n} \right )^{6}\cdot e^{\frac{3n^{2}}{2}}\cdot n} \right ] \end{align*}\]

\[a_{n}=\frac{\left ( n+1 \right )^{3n^2}\cdot \left ( n! \right )^{3}}{\left ( 2^{2}\cdot 3^{3}\cdots n^{n} \right )^{6}\cdot e^{\frac{3n^{2}}{2}}\cdot n}=\frac{n^{3n^2+3n+\frac{1}{2}}\cdot e^{-\frac{3n^{2}}{2}}}{\left ( 2^{2}\cdot 3^{3}\cdots n^{n} \right )^{6}}\cdot \frac{\left ( n+1 \right )^{3n^{2}}\cdot \left ( n! \right )^{3}}{n^{3n^2+3n+\frac{3}{2}}} \]

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 \(\displaystyle n!\sim \sqrt{2\pi n}\left ( \frac{n}{e} \right )^{n}\)

\[\frac{\left ( n+1 \right )^{3n^{2}}\cdot \left ( n! \right )^{3}}{n^{3n^2+3n+\frac{3}{2}}}\sim \left ( 2\pi \right )^{\frac{3}{2}}\left [ \left ( \frac{n+1}{n} \right )^{n}\frac{1}{e} \right ]^{3n}\rightarrow \left ( 2\pi \right )^{\frac{3}{2}}e^{-\frac{3}{2}} \]

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

\[\Large\boxed{\displaystyle {\int_{0}^{1}\left \{ \frac{1}{x} \right \}^{3}\mathrm{d}x}=\color{blue}{-\frac{1}{2}-\gamma +\frac{3}{2}\ln\left ( 2\pi \right )-6\ln \mathbf{A}}} \]

posted @ 2016-05-09 20:04  Renascence_5  阅读(1303)  评论(0编辑  收藏  举报