# 2023年考研数二第06题解析：换元积分、指数函数的求导法则

## 一、题目

A. $-\frac{1}{\ln (\ln 2)}$

C. $\frac{1}{\ln 2}$

B. $-\ln (\ln 2)$

D. $\ln 2$

## 二、解析

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

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

$$f(\alpha) = \int_{2}^{+\infty}(\ln x)^{-(\alpha+1)} \mathrm{~d} (\ln x) \Rightarrow$$

$$f(\alpha) = \left.\frac{1}{-\alpha}(\ln x)^{-\alpha}\right|_{2} ^{+\infty}=\frac{-1}{\alpha}\left[0-\frac{1}{(\ln 2)^{\alpha}}\right] \Rightarrow$$

$$f(\alpha) = \frac{1}{\alpha (\ln 2)^{\alpha}} = \frac{1}{\alpha} \cdot \frac{1}{(\ln 2)^{\alpha}}$$

$$f^{\prime}(\alpha)=\frac{-1}{\alpha^{2}} \cdot \frac{1}{(\ln 2)^{\alpha}}+\frac{1}{\alpha} \cdot(\ln 2)^{-\alpha} \cdot \ln (\ln 2) \Rightarrow$$

$$f^{\prime}(\alpha)=\frac{1}{(\ln 2)^{\alpha}}\left[\frac{-1}{\alpha^{2}}+\frac{\ln (\ln 2)}{\alpha}\right] \Rightarrow$$

$$f^{\prime}(\alpha)=0 \Rightarrow$$

$$\alpha \ln (\ln 2)=-1 \Rightarrow \alpha=\frac{-1}{\ln (\ln 2)}$$

$$g(\alpha)=\alpha(\ln 2)^{\alpha}$$

$$g^{\prime}(\alpha)=(\ln 2)^{\alpha}+\alpha \cdot(\ln 2)^{\alpha} \cdot \ln (\ln 2) \Rightarrow$$

$$g^{\prime}(\alpha)=(\ln 2)^{\alpha} \cdot(1+\alpha \ln (\ln 2)) \Rightarrow$$

$$g^{\prime}(\alpha)=0 \Rightarrow$$

$$1+\alpha(\ln (\ln 2))=0 \Rightarrow$$

$$\alpha(\ln (\ln 2))=-1 \Rightarrow \alpha=\frac{-1}{\ln (\ln 2)}$$