$$$3^{x^{2}}$$$ 的積分
您的輸入
求$$$\int 3^{x^{2}}\, dx$$$。
解答
換底:
$${\color{red}{\int{3^{x^{2}} d x}}} = {\color{red}{\int{e^{x^{2} \ln{\left(3 \right)}} d x}}}$$
令 $$$u=x \sqrt{\ln{\left(3 \right)}}$$$。
則 $$$du=\left(x \sqrt{\ln{\left(3 \right)}}\right)^{\prime }dx = \sqrt{\ln{\left(3 \right)}} dx$$$ (步驟見»),並可得 $$$dx = \frac{du}{\sqrt{\ln{\left(3 \right)}}}$$$。
該積分可改寫為
$${\color{red}{\int{e^{x^{2} \ln{\left(3 \right)}} d x}}} = {\color{red}{\int{\frac{e^{u^{2}}}{\sqrt{\ln{\left(3 \right)}}} d u}}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=\frac{1}{\sqrt{\ln{\left(3 \right)}}}$$$ 與 $$$f{\left(u \right)} = e^{u^{2}}$$$:
$${\color{red}{\int{\frac{e^{u^{2}}}{\sqrt{\ln{\left(3 \right)}}} d u}}} = {\color{red}{\frac{\int{e^{u^{2}} d u}}{\sqrt{\ln{\left(3 \right)}}}}}$$
此積分(虛誤差函數)不存在閉式表示:
$$\frac{{\color{red}{\int{e^{u^{2}} d u}}}}{\sqrt{\ln{\left(3 \right)}}} = \frac{{\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erfi}{\left(u \right)}}{2}\right)}}}{\sqrt{\ln{\left(3 \right)}}}$$
回顧一下 $$$u=x \sqrt{\ln{\left(3 \right)}}$$$:
$$\frac{\sqrt{\pi} \operatorname{erfi}{\left({\color{red}{u}} \right)}}{2 \sqrt{\ln{\left(3 \right)}}} = \frac{\sqrt{\pi} \operatorname{erfi}{\left({\color{red}{x \sqrt{\ln{\left(3 \right)}}}} \right)}}{2 \sqrt{\ln{\left(3 \right)}}}$$
因此,
$$\int{3^{x^{2}} d x} = \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \sqrt{\ln{\left(3 \right)}} \right)}}{2 \sqrt{\ln{\left(3 \right)}}}$$
加上積分常數:
$$\int{3^{x^{2}} d x} = \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \sqrt{\ln{\left(3 \right)}} \right)}}{2 \sqrt{\ln{\left(3 \right)}}}+C$$
答案
$$$\int 3^{x^{2}}\, dx = \frac{\sqrt{\pi} \operatorname{erfi}{\left(x \sqrt{\ln\left(3\right)} \right)}}{2 \sqrt{\ln\left(3\right)}} + C$$$A