$$$e^{\frac{y^{2}}{2}}$$$ 的积分

求$$$\int e^{\frac{y^{2}}{2}}\, dy$$$。

设$$$u=\frac{\sqrt{2} y}{2}$$$。

则$$$du=\left(\frac{\sqrt{2} y}{2}\right)^{\prime }dy = \frac{\sqrt{2}}{2} dy$$$ (步骤见»)，并有$$$dy = \sqrt{2} du$$$。

积分变为

$${\color{red}{\int{e^{\frac{y^{2}}{2}} d y}}} = {\color{red}{\int{\sqrt{2} e^{u^{2}} d u}}}$$

对 $$$c=\sqrt{2}$$$ 和 $$$f{\left(u \right)} = e^{u^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\sqrt{2} e^{u^{2}} d u}}} = {\color{red}{\sqrt{2} \int{e^{u^{2}} d u}}}$$

该积分（虚误差函数）没有闭式表达式：

$$\sqrt{2} {\color{red}{\int{e^{u^{2}} d u}}} = \sqrt{2} {\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erfi}{\left(u \right)}}{2}\right)}}$$

回忆一下 $$$u=\frac{\sqrt{2} y}{2}$$$:

$$\frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left({\color{red}{u}} \right)}}{2} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left({\color{red}{\left(\frac{\sqrt{2} y}{2}\right)}} \right)}}{2}$$

因此，

$$\int{e^{\frac{y^{2}}{2}} d y} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} y}{2} \right)}}{2}$$

加上积分常数：

$$\int{e^{\frac{y^{2}}{2}} d y} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} y}{2} \right)}}{2}+C$$

$$$\int e^{\frac{y^{2}}{2}}\, dy = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} y}{2} \right)}}{2} + C$$$A