$$$\sin{\left(\pi t^{2} \right)}$$$ 的积分

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

设$$$u=\sqrt{\pi} t$$$。

则$$$du=\left(\sqrt{\pi} t\right)^{\prime }dt = \sqrt{\pi} dt$$$ (步骤见»)，并有$$$dt = \frac{du}{\sqrt{\pi}}$$$。

因此，

$${\color{red}{\int{\sin{\left(\pi t^{2} \right)} d t}}} = {\color{red}{\int{\frac{\sin{\left(u^{2} \right)}}{\sqrt{\pi}} d u}}}$$

对 $$$c=\frac{1}{\sqrt{\pi}}$$$ 和 $$$f{\left(u \right)} = \sin{\left(u^{2} \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

该积分（菲涅耳正弦积分）没有闭式表达式：

$$\frac{{\color{red}{\int{\sin{\left(u^{2} \right)} d u}}}}{\sqrt{\pi}} = \frac{{\color{red}{\left(\frac{\sqrt{2} \sqrt{\pi} S\left(\frac{\sqrt{2} u}{\sqrt{\pi}}\right)}{2}\right)}}}{\sqrt{\pi}}$$

回忆一下 $$$u=\sqrt{\pi} t$$$:

$$\frac{\sqrt{2} S\left(\frac{\sqrt{2} {\color{red}{u}}}{\sqrt{\pi}}\right)}{2} = \frac{\sqrt{2} S\left(\frac{\sqrt{2} {\color{red}{\sqrt{\pi} t}}}{\sqrt{\pi}}\right)}{2}$$

因此，

$$\int{\sin{\left(\pi t^{2} \right)} d t} = \frac{\sqrt{2} S\left(\sqrt{2} t\right)}{2}$$

加上积分常数：

$$\int{\sin{\left(\pi t^{2} \right)} d t} = \frac{\sqrt{2} S\left(\sqrt{2} t\right)}{2}+C$$

$$$\int \sin{\left(\pi t^{2} \right)}\, dt = \frac{\sqrt{2} S\left(\sqrt{2} t\right)}{2} + C$$$A