$$$e^{t^{2}} - e^{- t^{2}}$$$の積分

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

項別に積分せよ:

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

この積分（誤差関数）には閉形式はありません:

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

この積分（虚誤差関数）には閉形式はありません:

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

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