$$$e^{t^{2}} - e^{- t^{2}}$$$ 的積分

求$$$\int \left(e^{t^{2}} - e^{- t^{2}}\right)\, dt$$$。

逐項積分：

$${\color{red}{\int{\left(e^{t^{2}} - e^{- t^{2}}\right)d t}}} = {\color{red}{\left(- \int{e^{- t^{2}} d t} + \int{e^{t^{2}} d t}\right)}}$$

此積分（誤差函數）不存在閉式表示：

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

此積分（虛誤差函數）不存在閉式表示：

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

因此，

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

化簡：

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

加上積分常數：

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

$$$\int \left(e^{t^{2}} - e^{- t^{2}}\right)\, dt = \frac{\sqrt{\pi} \left(- \operatorname{erf}{\left(t \right)} + \operatorname{erfi}{\left(t \right)}\right)}{2} + C$$$A