Integral de $$$e^{\frac{y^{2}}{2}}$$$

Halla $$$\int e^{\frac{y^{2}}{2}}\, dy$$$.

Sea $$$u=\frac{\sqrt{2} y}{2}$$$.

Entonces $$$du=\left(\frac{\sqrt{2} y}{2}\right)^{\prime }dy = \frac{\sqrt{2}}{2} dy$$$ (los pasos pueden verse »), y obtenemos que $$$dy = \sqrt{2} du$$$.

La integral se convierte en

$${\color{red}{\int{e^{\frac{y^{2}}{2}} d y}}} = {\color{red}{\int{\sqrt{2} e^{u^{2}} d u}}}$$

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\sqrt{2}$$$ y $$$f{\left(u \right)} = e^{u^{2}}$$$:

$${\color{red}{\int{\sqrt{2} e^{u^{2}} d u}}} = {\color{red}{\sqrt{2} \int{e^{u^{2}} d u}}}$$

Esta integral (Función error imaginaria) no tiene una forma cerrada:

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

Recordemos que $$$u=\frac{\sqrt{2} y}{2}$$$:

$$\frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left({\color{red}{u}} \right)}}{2} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left({\color{red}{\left(\frac{\sqrt{2} y}{2}\right)}} \right)}}{2}$$

Por lo tanto,

$$\int{e^{\frac{y^{2}}{2}} d y} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} y}{2} \right)}}{2}$$

Añade la constante de integración:

$$\int{e^{\frac{y^{2}}{2}} d y} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} y}{2} \right)}}{2}+C$$

$$$\int e^{\frac{y^{2}}{2}}\, dy = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} y}{2} \right)}}{2} + C$$$A