$$$\sin{\left(\pi x^{2} \right)}$$$ 的積分

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

令 $$$u=\sqrt{\pi} x$$$。

則 $$$du=\left(\sqrt{\pi} x\right)^{\prime }dx = \sqrt{\pi} dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{\sqrt{\pi}}$$$。

所以，

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{\sqrt{\pi}}$$$ 與 $$$f{\left(u \right)} = \sin{\left(u^{2} \right)}$$$：

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

此積分（菲涅耳正弦積分）不存在閉式表示：

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

回顧一下 $$$u=\sqrt{\pi} x$$$：

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

因此，

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

加上積分常數：

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

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