Integral de $$$e^{\frac{x^{2}}{8}}$$$

Encontre $$$\int e^{\frac{x^{2}}{8}}\, dx$$$.

Seja $$$u=\frac{\sqrt{2} x}{4}$$$.

Então $$$du=\left(\frac{\sqrt{2} x}{4}\right)^{\prime }dx = \frac{\sqrt{2}}{4} dx$$$ (veja os passos »), e obtemos $$$dx = 2 \sqrt{2} du$$$.

A integral torna-se

$${\color{red}{\int{e^{\frac{x^{2}}{8}} d x}}} = {\color{red}{\int{2 \sqrt{2} e^{u^{2}} d u}}}$$

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

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

Esta integral (Função erro imaginária) não possui forma fechada:

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

Recorde que $$$u=\frac{\sqrt{2} x}{4}$$$:

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

Portanto,

$$\int{e^{\frac{x^{2}}{8}} d x} = \sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} x}{4} \right)}$$

Adicione a constante de integração:

$$\int{e^{\frac{x^{2}}{8}} d x} = \sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} x}{4} \right)}+C$$

$$$\int e^{\frac{x^{2}}{8}}\, dx = \sqrt{2} \sqrt{\pi} \operatorname{erfi}{\left(\frac{\sqrt{2} x}{4} \right)} + C$$$A