\[\displaystyle\lim\limits_{n\to\infty}\frac{1}{(2n-1)^{2011}}\sum_{k=0}^{n-1}\int_{2k\pi}^{(2k+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x\]

(tian_275461)
解:
根据推广的积分第一中值定理，对每个正整数~$n~\exists\qquad \theta_n\in(0,1)$~使得
\begin{equation*}
    \int_{2n\pi}^{(2n+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x=
    ((2n+\theta_n)\pi)^{2010}\int_{2n\pi}^{(2n+1)\pi}\sin^3 x\cos^2 x \mathrm{d}x
\end{equation*}
由此得
\begin{align*}
&\quad\int_{2n\pi}^{(2n+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x\\
&=\left((2n\pi)^{2010}+o(n^{2010})\right)\int_{2n\pi}^{2n\pi+\pi}\sin^3 x\cos^2 x\mathrm{d}x\\
&=\left((2n\pi)^{2010}+o(n^{2010})\right)\left(\frac{\cos 5x}{80}-\frac{\cos 3x}{48}-\frac{\cos x}{8}\right)\Bigg|_{2n\pi}^{(2n+1)\pi}\\
&=\frac{4}{15}((2n\pi)^{2010}+o(n^{2010}))\qquad (n\to\infty)
\end{align*}
另外
\begin{equation*}
    (2n+1)^{2011}-(2n-1)^{2011}=4022(2n)^{2010}+o(n^{2010})\qquad (n\to\infty)
\end{equation*}
根据~Stolz~定理
\begin{align*}
&\quad\lim\limits_{n\to\infty}\frac{1}{(2n-1)^{2011}}\sum_{k=0}^{n-1}\int_{2k\pi}^{(2k+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x\\
&=\lim_{n\to\infty}\frac{\displaystyle\int_{2n\pi}^{(2n+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x}{ (2n+1)^{2011}-(2n-1)^{2011}}\\
&=\frac{2}{30165}\lim_{n\to \infty}\frac{(2n\pi)^{2010}+o(n^{2010})}{(2n)^{2010}+o(n^{2010})}\\
&=\frac{2\pi^{2010}}{30165}
\end{align*}

此题的更一般结果为
\begin{equation*}
    \lim_{n\to\infty}\frac{1}{(2n-1)^{p+1}}\sum_{k=0}^{n-1}\int_{2k\pi}^{(2k+1)\pi}x^p\sin^3 x\cos^2 x \mathrm{d}x
=\frac{2\pi^p}{15(p+1)}(p>0)
\end{equation*}