Integral de $$$e^{- 2 x^{2}}$$$

Encontre $$$\int e^{- 2 x^{2}}\, dx$$$.

Seja $$$u=\sqrt{2} x$$$.

Então $$$du=\left(\sqrt{2} x\right)^{\prime }dx = \sqrt{2} dx$$$ (veja os passos »), e obtemos $$$dx = \frac{\sqrt{2} du}{2}$$$.

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{\sqrt{2}}{2}$$$ e $$$f{\left(u \right)} = e^{- u^{2}}$$$:

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

Esta integral (Função erro) não possui forma fechada:

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

Recorde que $$$u=\sqrt{2} x$$$:

$$\frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left({\color{red}{u}} \right)}}{4} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left({\color{red}{\sqrt{2} x}} \right)}}{4}$$

Portanto,

$$\int{e^{- 2 x^{2}} d x} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\sqrt{2} x \right)}}{4}$$

Adicione a constante de integração:

$$\int{e^{- 2 x^{2}} d x} = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\sqrt{2} x \right)}}{4}+C$$

$$$\int e^{- 2 x^{2}}\, dx = \frac{\sqrt{2} \sqrt{\pi} \operatorname{erf}{\left(\sqrt{2} x \right)}}{4} + C$$$A