$$$e^{t^{2}}$$$ 的積分
您的輸入
求$$$\int e^{t^{2}}\, dt$$$。
解答
此積分(虛誤差函數)不存在閉式表示:
$${\color{red}{\int{e^{t^{2}} d t}}} = {\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erfi}{\left(t \right)}}{2}\right)}}$$
因此,
$$\int{e^{t^{2}} d t} = \frac{\sqrt{\pi} \operatorname{erfi}{\left(t \right)}}{2}$$
加上積分常數:
$$\int{e^{t^{2}} d t} = \frac{\sqrt{\pi} \operatorname{erfi}{\left(t \right)}}{2}+C$$
答案
$$$\int e^{t^{2}}\, dt = \frac{\sqrt{\pi} \operatorname{erfi}{\left(t \right)}}{2} + C$$$A
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