若函数 $p(t)$ 在 $[0,\infty)$ 连续, 且当 $t\to+\infty$ 时, $p(t)=o(t^N)$ ($N$ 为正整数). 又 $\lm<0$, 证明: 当 $t\to\infty$ 时, $$\bex \int_t^\infty p(\tau)e^{\lm \tau}\rd \tau=o(t^{N+1})e^{\lm t}. \eex$$ (北京师范大学)

证明: 由题意, $\color{red}\forall\ \ve>0,\ \exists\ T> 1,\st$ $$\beex \bea {\color{red} t>T}&\ra \sev{\frac{p(t)}{t^N}}\leq \ve\\ &{\color{red}\ra \sev{\frac{e^{-\lm t}\int_t^\infty p(\tau)e^{\lm \tau}\rd \tau}{t^{N+1}}}} \leq \frac{e^{-\lm t}\int_t^\infty |p(\tau)|e^{\lm \tau}\rd\tau}{t^{N+1}} \leq \frac{e^{-\lm t}\int_t^\infty \ve \tau^N e^{\lm \tau}\rd \tau}{t^{N+1}}\\ &\leq \ve N! \sum_{i=1}^{N+1} \sex{\frac{1}{-\lm t}}^i <\ve N!e^{\frac{1}{-\lm t}} {\color{red}<\ve N! e^\frac{1}{-\lm}} \eea \eeex$$ 其中最后一步是因为 $$\beex \bea I_n&\equiv \int_t^\infty \tau^N e^{\lm \tau}\rd \tau\\ &=\frac{1}{\lm}\int_t^\infty \tau^N\rd e^{\lm \tau}\\ &=\frac{1}{\lm}\sez{-t^Ne^{\lm t} -\int_t^\infty N\tau^{N-1} e^{\lm \tau}\rd \tau}\\ &=-\frac{t^N}{\lm}e^{\lm t} -\frac{N}{\lm}I_{N-1}\\ &=\cdots\\ &=\sez{\sum_{i=1}^{N+1} (-1)^i \frac{N(N-1)\cdots (N-i+1)}{\lm^i} t^{N+1-i}}e^{\lm t}. \eea \eeex$$