Integralen av $$$\sin{\left(\pi t^{2} \right)}$$$

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

Låt $$$u=\sqrt{\pi} t$$$ vara.

Då $$$du=\left(\sqrt{\pi} t\right)^{\prime }dt = \sqrt{\pi} dt$$$ (stegen kan ses »), och vi har att $$$dt = \frac{du}{\sqrt{\pi}}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=\frac{1}{\sqrt{\pi}}$$$ och $$$f{\left(u \right)} = \sin{\left(u^{2} \right)}$$$:

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

Denna integral (Fresnels sinusintegral) har ingen sluten form:

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

Kom ihåg att $$$u=\sqrt{\pi} t$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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