# 2024年考研数二第07题解析：积分敛散性的判别

## 一、题目

(1)若 $\int_{0}^{+\infty} f^{2}(x) \mathrm{~d} x$ 收敛, 则 $\int_{0}^{+\infty} f(x) \mathrm{~d} x$ 收敛.

(2)若存在 $p>1$, 使得 $\lim \limits_{x \rightarrow+\infty} x^{p} f(x)$ 存在, 则 $\int_{0}^{+\infty} f(x) \mathrm{~d} x$ 收敛.

(3)若 $\int_{0}^{+\infty} f(x) \mathrm{~d} x$ 收敛, 则存在 $p>1$, 使得 $\lim \limits_{x \rightarrow+\infty} x^{p} f(x)$ 存在.

(A) 0

(B) 1

(C) 2

(D) 3

## 二、解析

### 第 (1) 项

$$\int_{a}^{+\infty} \frac{1}{x^{p}} \mathrm{~d} x \ \begin{cases} 收敛, & p>1 \\ 发散, & p \leqslant 1 \end{cases}$$

$$f(x)=\frac{1}{x}$$

$$\int_{0}^{+\infty} f^{2}(x) \mathrm{d} x=\int_{0}^{+\infty} \frac{1}{x^{2}} \mathrm{~d} x$$

$$\int_{0}^{+\infty} f(x) \mathrm{d} x=\int_{0}^{+\infty} \frac{1}{x} \mathrm{~d} x$$

### 第 (2) 项

$$\int_{0}^{+\infty} \frac{1}{x^{p}} \mathrm{~d} x$$

\begin{aligned} & \lim \limits_{x \rightarrow+\infty} x^{p} f(x) = A \Rightarrow \\ & \lim \limits_{x \rightarrow+\infty} f(x) = \lim \limits_{x \rightarrow+\infty} \frac{A}{x^{p}} \end{aligned}

$$\int_{0}^{+\infty} f(x) \mathrm{~d} x = \int_{0}^{+\infty} \frac{A}{x^{p}} \mathrm{~d} x = A \int_{0}^{+\infty} \frac{1}{x^{p}} \mathrm{~d} x$$

$$\int_{0}^{+\infty} f(x) \mathrm{~d} x$$

### 第 (3) 项

$$\int_{0}^{+\infty} x^{p} f(x) \mathrm{~d} x$$

$$f(x)=\frac{1}{(x+1) \ln ^{2}(x+1)}$$

\begin{aligned} \int_{0}^{+\infty} f(x) \mathrm{d} x \\ \\ & = \int_{0}^{+\infty} \frac{1}{(x+1) \ln ^{2}(x+1)} \mathrm{d} x \\ \\ & = – \int_{0}^{+\infty} 1 \mathrm{d} \left[ \frac{1}{\ln (x+1)} \right] \\ \\ & = -\left.\frac{1}{\ln (x+1)}\right|_{0} ^{+\infty} \\ \\ & = \ln 2 \end{aligned}

\begin{aligned} \lim \limits_{x \rightarrow+\infty} x^{p} f(x) \\ \\ & = \lim \limits_{x \rightarrow+\infty} x^{p} \frac{1}{(x+2) \ln ^{2}(x+2)} \\ \\ & = \lim \limits_{x \rightarrow+\infty} \frac{x^{p}}{x \ln ^{2} x } \end{aligned}