Integral de $$$1 - e^{- \frac{y^{2}}{2}}$$$

Halla $$$\int \left(1 - e^{- \frac{y^{2}}{2}}\right)\, dy$$$.

Integra término a término:

$${\color{red}{\int{\left(1 - e^{- \frac{y^{2}}{2}}\right)d y}}} = {\color{red}{\left(\int{1 d y} - \int{e^{- \frac{y^{2}}{2}} d y}\right)}}$$

Aplica la regla de la constante $$$\int c\, dy = c y$$$ con $$$c=1$$$:

$$- \int{e^{- \frac{y^{2}}{2}} d y} + {\color{red}{\int{1 d y}}} = - \int{e^{- \frac{y^{2}}{2}} d y} + {\color{red}{y}}$$

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 $$$dy = \sqrt{2} du$$$.

Por lo tanto,

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

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 $$$f{\left(u \right)} = e^{- u^{2}}$$$:

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

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

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

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

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

Por lo tanto,

$$\int{\left(1 - e^{- \frac{y^{2}}{2}}\right)d y} = y - \frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\frac{\sqrt{2} y}{2} \right)}}{2}$$

Añade la constante de integración:

$$\int{\left(1 - e^{- \frac{y^{2}}{2}}\right)d y} = y - \frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\frac{\sqrt{2} y}{2} \right)}}{2}+C$$

$$$\int \left(1 - e^{- \frac{y^{2}}{2}}\right)\, dy = \left(y - \frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\frac{\sqrt{2} y}{2} \right)}}{2}\right) + C$$$A