# 当变限积分和无穷限反常积分在一起会碰撞出什么火花？

## 二、解析

$$t=\sqrt{b} k \Rightarrow k^{2}=\frac{t^{2}}{b} \Rightarrow$$

$$t \in(0, x) \Rightarrow k \in\left(0, \frac{x}{\sqrt{b}}\right) \Rightarrow$$

$$\mathrm{~ d} t=\sqrt{b} \mathrm{~ d} k$$

$$\varphi(x)=\frac{2}{\sqrt{\pi b}} \int_{0}^{x} e^{-\frac{t^{2}}{b}} \mathrm{~ d} t \Rightarrow$$

$$\varphi(x)=\frac{2}{\sqrt{\pi b}} \cdot \sqrt{b} \int_{0}^{\frac{x}{\sqrt{b}}} e^{-k^{2}} \mathrm{~ d} k \Rightarrow$$

$$\varphi(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{\frac{x}{\sqrt{b}}} e^{-k^{2}} \mathrm{~ d} k.$$

$$\lim \limits_{x \rightarrow+\infty} \varphi(x)=\frac{2}{\sqrt{\pi}} \int_{0}^{+\infty} e^{-k^{2}} \mathrm{~ d} k=\frac{2}{\sqrt{\pi}} \cdot \frac{\sqrt{\pi}}{2}=1.$$

$$\int_{0}^{+\infty}[1-\varphi(x)] \mathrm{~ d} x=$$

$$\left.x[1-\varphi(x)]\right|_{0} ^{+\infty}-\int_{0}^{+\infty} x \mathrm{~d} [1-\varphi(x)].$$

$$\lim \limits_{x \rightarrow+\infty} x[1-\varphi(x)]=\lim \limits_{x \rightarrow+\infty} \frac{1-\varphi(x)}{\frac{1}{x}} \Rightarrow$$

$\frac{0}{0}$ 型极限 $\Rightarrow$ 洛必达法则 $\Rightarrow$

$$\lim \limits_{x \rightarrow+\infty} \frac{-\varphi^{\prime}(x)}{\frac{-x^{2}}{b}}=\lim \limits_{x \rightarrow+\infty} \frac{-\frac{2}{\sqrt{\pi b}} e^{\frac{-x^{2}}{b}}}{\frac{-1}{x^{2}}}=$$

$$\frac{2}{\sqrt{\pi b}} x^{2} e^{\frac{-x^{2}}{b}}=\frac{2}{\sqrt{\pi b}} \frac{e^{\frac{-x^{2}}{b}}}{x^{-2}}=\frac{2}{\sqrt{\pi b}} \cdot 0=0$$

$$\left.x[1-\varphi(x)]\right|_{0} ^{+\infty} \textcolor{red}{-} \int_{0}^{+\infty} x \mathrm{~d} [1-\varphi(x)]=$$

$$(0-0) \textcolor{red}{+} \int_{0}^{+\infty} x \cdot \frac{2}{\sqrt{\pi b}} e^{\frac{-x^{2}}{b}} \mathrm{~ d} x=$$

$$\frac{2}{\sqrt{\pi b}} \int_{0}^{+\infty} x e^{\frac{-x^{2}}{b}} \mathrm{~ d} x=$$

$$\frac{2}{\sqrt{\pi b}} \cdot \frac{1}{2} \int_{0}^{+\infty} e^{\frac{-x^{2}}{b}} \mathrm{~d} \left(x^{2}\right) \Rightarrow$$

$$\frac{1}{\sqrt{\pi b}} \int_{0}^{+\infty} e^{\frac{-x}{b}} \mathrm{~ d} x=$$

$$\left.\frac{1}{\sqrt{\pi b}} \cdot \frac{-b}{1} \cdot e^{\frac{-x}{b}}\right|_{0} ^{+\infty}=$$

$$\frac{-b}{\sqrt{\pi b}}(0-1)=\frac{(\sqrt{b})^{2}}{\sqrt{\pi} \cdot \sqrt{b}}=\frac{\sqrt{b}}{\sqrt{\pi}}=\sqrt{\frac{b}{\pi}}.$$