$$$e^{\frac{x^{2}}{8}}$$$ 的積分

求$$$\int e^{\frac{x^{2}}{8}}\, dx$$$。

令 $$$u=\frac{\sqrt{2} x}{4}$$$。

則 $$$du=\left(\frac{\sqrt{2} x}{4}\right)^{\prime }dx = \frac{\sqrt{2}}{4} dx$$$ (步驟見»)，並可得 $$$dx = 2 \sqrt{2} du$$$。

因此，

$${\color{red}{\int{e^{\frac{x^{2}}{8}} d x}}} = {\color{red}{\int{2 \sqrt{2} e^{u^{2}} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=2 \sqrt{2}$$$ 與 $$$f{\left(u \right)} = e^{u^{2}}$$$：

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

此積分（虛誤差函數）不存在閉式表示：

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

回顧一下 $$$u=\frac{\sqrt{2} x}{4}$$$：

$$\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left({\color{red}{u}} \right)} = \sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left({\color{red}{\left(\frac{\sqrt{2} x}{4}\right)}} \right)}$$

因此，

$$\int{e^{\frac{x^{2}}{8}} d x} = \sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} x}{4} \right)}$$

加上積分常數：

$$\int{e^{\frac{x^{2}}{8}} d x} = \sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} x}{4} \right)}+C$$

$$$\int e^{\frac{x^{2}}{8}}\, dx = \sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} x}{4} \right)} + C$$$A