求一个一元函数的极限

试证：极限$$\lim\limits_{n\to+\infty}\int_0^1\left(\frac{1}{n}\left[\frac{n}{x}\right]-\left[\frac{1}{x}\right]\right)dx$$一定存在有限，并求这个极限。
注：$[x]$表示Gauss取整函数。

易知
\[\int_0^1\left(\frac{1}{n}\left[\frac{n}{x}\right]-\left[\frac{1}{x}\right]\right)dx=\sum
_{k=1}^\infty\int_{\frac{1}{k+1}}^{\frac{1}{k}}\left(\frac{1}{n}\left[\frac{n}{x}\right]-\left[\frac{1}{x}\right]\right)dx,\]
其中
\[\int_{\frac{1}{k+1}}^{\frac{1}{k}}\frac{1}{n}\left[\frac{n}{x}\right]dx=\frac{1}{n}\int_k^{k+1}\frac{[nt]}{t^2}dt=
\frac{1}{n}\sum_{i=1}^n\int_{k+\frac{i-1}{n}}^{k+\frac{i}{n}}\frac{[nt]}{t^2}dt=\sum_{i=1}^n\frac{1}{nk+i},\]
\[\int_{\frac{1}{k+1}}^{\frac{1}{k}}\left[\frac{1}{x}\right]dx=\frac{1}{k+1}.\]
因此
\[\int_{\frac{1}{k+1}}^{\frac{1}{k}}\left(\frac{1}{n}\left[\frac{n}{x}\right]-\left[\frac{1}{x}\right]\right)dx=\sum_{i=1}^n\frac{1}{nk+i}-\frac{1}{k+1}.\]
注意到
\[0<\ln\frac{k+1}{k}-\sum_{i=1}^n\frac{1}{nk+i}=\sum_{i=1}^n\int_{i-1}^i\left(\frac{1}{nk+t}-\frac{1}{nk+i}\right)dt<\sum_{i=1}^n\frac{1}{n^2k^2}
=\frac{1}{nk^2}.\] 由此可得
\[\ln\frac{k+1}{k}-\frac{1}{nk^2}<\sum_{i=1}^n\frac{1}{nk+i}<\ln\frac{k+1}{k},\]
故
\[\ln\frac{k+1}{k}-\frac{1}{k+1}-\frac{1}{nk^2}<\int_{\frac{1}{k+1}}^{\frac{1}{k}}\left(\frac{1}{n}\left[\frac{n}{x}\right]-\left[\frac{1}{x}\right]\right)dx<\ln\frac{k+1}{k}-\frac{1}{k+1},\]
进而得到
\[\sum_{k=1}^\infty\left(\ln\frac{k+1}{k}-\frac{1}{k+1}\right)-\sum_{k=1}^\infty\frac{1}{nk^2}
<\int_0^1\left(\frac{1}{n}\left[\frac{n}{x}\right]-\left[\frac{1}{x}\right]\right)dx<\sum_{k=1}^\infty\left(\ln\frac{k+1}{k}-\frac{1}{k+1}\right),\]
即\[1-\gamma-\frac{\pi^2}{6n}<\int_0^1\left(\frac{1}{n}\left[\frac{n}{x}\right]-\left[\frac{1}{x}\right]\right)dx<1-\gamma,\]
其中$\gamma$是Euler常数. 因此有
\[\lim_{n\to\infty}\int_0^1\left(\frac{1}{n}\left[\frac{n}{x}\right]-\left[\frac{1}{x}\right]\right)dx=1-\gamma.\]