2023年浙江大学强基计划数学试题（回忆版）#21
已知 $C, L \in \mathbb{R}$ 且 $L \neq 0$，有 \[\lim _{n \rightarrow \infty} \dfrac{n^c \displaystyle\int_0^{\frac{\pi}{2}} x^n \sin x \mathrm{~d} x}{\displaystyle\int_0^{\frac{\pi}{2}} x^n \cos x \mathrm{~d} x}=L,\]则 $L=$ _____．
答案 $\dfrac{2}{\pi}$．
解析 设\[a_n=\int_0^{\frac{\pi}{2}} x^n \sin x \mathrm{~d} x,\quad b_n=\int_0^{\frac{\pi}{2}} x^n \cos x \mathrm{~d} x,\]其中 $n\in\mathbb N$，则\[ a_n =\int_0^{\frac{\pi}{2}} x^n \sin x \mathrm{~d} x =\int_0^{\frac{\pi}{2}} x^n \mathrm{~d}(-\cos x) =-x^n \cos x\Bigg|_0 ^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}(-\cos x) \mathrm{d} x^n =n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \mathrm{~d} x =n b_{n-1},\]且\[ b_n =\int_0^{\frac{\pi}{2}} x^n \mathrm{~d} \sin x =x^n\sin x\Bigg|_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \sin x \mathrm{~d} x^n=\left(\dfrac {\pi}2\right)^n-n\int_{0}^{\frac{\pi}2}x^{n-1}\sin x\mathrm{~d}x=\left(\dfrac{\pi}2\right)^n-na_{n-1},\]因此对任意 $n \geqslant 2$ 且 $n \in \mathbb{N}$，有\[\begin{split} a_n & =n b_{n-1}=n \cdot\left(\frac{\pi}{2}\right)^{n-1}-n(n-1) a_{n-2}, \\ b_n & =\left(\frac{\pi}{2}\right)^n-n a_{n-1}=\left(\frac{\pi}{2}\right)^n-n(n-1) b_{n-2} \end{split}\]从而\[ a_n =n \cdot\left(\frac{\pi}{2}\right)^{n-1}-n(n-1)(n-2) \cdot\left(\frac{\pi}{2}\right)^{n-3}+\cdots=\sum_{i=0}^{\infty} \frac{(-1)^i \cdot n!}{(n-2 i-1)!} \cdot\left(\frac{\pi}{2}\right)^{n-2 i-1} ,\]且\[ b_n =\left(\frac{\pi}{2}\right)^n-n(n-1)\left(\frac{\pi}{2}\right)^{n-2}+\cdots =\sum_{i=0}^{\infty} \frac{(-1)^i \cdot n!}{(n-2 i)!} \cdot\left(\frac{\pi}{2}\right)^{n-2 i},\]因此\[ L=\lim _{n \to \infty} n^{C+1} \cdot \frac{n^{-1} a_n}{b_n}=\frac{2}{\pi} \cdot \lim _{n \to \infty} n^{C+1}=\begin{cases} 0,&C<-1\\ +\infty,&C>-1,\\ \dfrac{2}{\pi},&C=-1.\end{cases}\]