$$$s^{2} \sin{\left(x^{2} \right)}$$$ の $$$x$$$ に関する積分

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

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=s^{2}$$$ と $$$f{\left(x \right)} = \sin{\left(x^{2} \right)}$$$ に対して適用する:

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

この積分（フレネル正弦積分）には閉形式はありません:

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

したがって、

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

積分定数を加える:

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

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