problem

\[\text { 求極限: } \lim _{x \rightarrow+\infty} \frac{\int_{0}^{x} \frac{|\sin t|}{t} d t}{\ln x} \text {. } \]

solution

解: 利用不等式: \(\ln (1+n) \leqslant \sum_{k=1}^{n} \frac{1}{k} \leqslant 1+\ln n\)
對任意 \(x>0\), 存在 \(n \in N\), 使得: \(n \pi \leqslant x \leqslant(n+1) \pi\), 於是有:

\[\begin{aligned} &\int_{0}^{n \pi} \frac{|\sin t|}{t} d t \leqslant \int_{0}^{x} \frac{|\sin t|}{t} d t \leqslant \int_{0}^{(n+1) \pi} \frac{|\sin t|}{t} d t \\ &\int_{0}^{n \pi} \frac{|\sin t|}{t} d t=\sum_{k=1}^{n} \int_{(k-1) \pi}^{k \pi} \frac{|\sin t|}{t} d t=\int_{(k-1) \pi}^{k \pi} \sum_{k=1}^{n} \frac{|\sin t|}{t} d t \geqslant \int_{(k-1) \pi}^{k \pi} \sum_{k=1}^{n} \frac{|\sin t|}{k \pi} d t \\ &=\int_{(k-1) \pi}^{k \pi}|\sin t| d t \cdot \frac{1}{\pi} \sum_{k=1}^{n} \frac{1}{k} \geqslant 2 \cdot \frac{1}{\pi} \cdot \ln (1+n)=\frac{2}{\pi} \ln (1+n) \end{aligned} \]

於是: \(\int_{0}^{n \pi} \frac{|\sin t|}{t} d t \geqslant \frac{2}{\pi} \ln (1+n)\)

;
\(\int_{0}^{(n+1) \pi} \frac{|\sin t|}{t} d t=\int_{0}^{\pi} \frac{|\sin t|}{t} d t+\int_{\pi}^{(n+1) \pi} \frac{|\sin t|}{t} d t\); 研究第二項積分有:
\(\int_{\pi}^{(n+1) \pi} \frac{|\sin t|}{t} d t=\sum_{k=1}^{n} \int_{k \pi}^{(k+1) \pi} \frac{|\sin t|}{t} d t=\int_{k \pi}^{(k+1) \pi} \sum_{k=1}^{n} \frac{|\sin t|}{t} d t \leqslant \int_{k \pi}^{(k+1) \pi} \sum_{k=1}^{n} \frac{|\sin t|}{k \pi} d t\)
\(=\int_{k \pi}^{(k+1) \pi}|\sin t| d t \cdot \frac{1}{\pi} \sum_{k=1}^{n} \frac{1}{k} \leqslant 2 \cdot \frac{1}{\pi} \cdot(1+\ln n)=\frac{2}{\pi}(1+\ln n)\)
於是: \(\int_{0}^{(n+1) \pi} \frac{|\sin t|}{t} d t \leqslant \int_{0}^{\pi} \frac{|\sin t|}{t} d t+\frac{2}{\pi}(1+\ln n)\);
於是當 \(n \pi \leqslant x \leqslant(n+1) \pi\) 時, 滿足下列不等式組:

\[\left\{\begin{array}{l} \int_{0}^{n \pi} \frac{|\sin t|}{t} d t \\ \frac{\ln [(n+1) \pi]}{\int_{0}^{n \pi} \frac{|\sin t|}{t} d t \geqslant \frac{\int_{0}^{x} \frac{|\sin t|}{t} d t}{\pi} \ln (1+n)} \leqslant \frac{\int_{0}^{(n+1) \pi} \frac{|\sin t|}{t} d t}{\ln (n \pi)} \\ \int_{0}^{(n+1) \pi} \frac{|\sin t|}{t} d t \leqslant \int_{0}^{\pi} \frac{|\sin t|}{t} d t+\frac{2}{\pi}(1+\ln n) \\ \frac{2}{\pi} \ln (1+n) \\ {\frac{\int_{0}^{n}[(n+1) \pi]}{\int_{0}^{n} \frac{|\sin t|}{t} d t} \int_{0}^{\pi} \frac{|\sin t|}{t} d t+\frac{2}{\pi}(1+\ln n)}{\ln x} \leqslant \frac{\ln (n \pi)}{} \end{array}\right. \]

當 \(x \rightarrow+\infty\) 時, \(n \rightarrow \infty\), 對上述不等式取極限有:

\[\begin{aligned} &\lim _{n \rightarrow \infty} \frac{\frac{2}{\pi} \ln (1+n)}{\ln [(n+1) \pi]}=\frac{2}{\pi} \lim _{n \rightarrow \infty} \frac{\ln (1+n)}{\ln (1+n)+\ln \pi}=\frac{2}{\pi} \\ &\lim _{n \rightarrow \infty} \frac{\int_{0}^{\pi} \frac{|\sin t|}{t} d t+\frac{2}{\pi}(1+\ln n)}{\ln (n \pi)}=\lim _{n \rightarrow \infty} \frac{\int_{0}^{\pi} \frac{|\sin t|}{t} d t}{\ln (n \pi)}+\lim _{n \rightarrow \infty} \frac{\frac{2}{\pi}(1+\ln n)}{\ln (n \pi)} \\ &=0+\frac{2}{\pi} \lim _{n \rightarrow \infty} \frac{\ln n+1}{\ln n+\ln \pi}=\frac{2}{\pi} \end{aligned} \]

由夾逼準則可得: \(\lim _{x \rightarrow+\infty} \frac{\int_{0}^{x} \frac{|\sin t|}{t} d t}{\ln x}=\frac{2}{\pi}\).