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

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=s^{2}$$$ 與 $$$f{\left(x \right)} = \sin{\left(x^{2} \right)}$$$：

$${\color{red}{\int{s^{2} \sin{\left(x^{2} \right)} d x}}} = {\color{red}{s^{2} \int{\sin{\left(x^{2} \right)} d x}}}$$

此積分（菲涅耳正弦積分）不存在閉式表示：

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

因此，

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

加上積分常數：

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

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