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

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

Seja $$$u=\frac{1}{x}$$$.

Então $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (veja os passos »), e obtemos $$$\frac{dx}{x^{2}} = - du$$$.

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=-1$$$ e $$$f{\left(u \right)} = e^{- \frac{1}{u^{2}}}$$$:

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

Para a integral $$$\int{e^{- \frac{1}{u^{2}}} d u}$$$, use integração por partes $$$\int \operatorname{g} \operatorname{dv} = \operatorname{g}\operatorname{v} - \int \operatorname{v} \operatorname{dg}$$$.

Sejam $$$\operatorname{g}=e^{- \frac{1}{u^{2}}}$$$ e $$$\operatorname{dv}=du$$$.

Então $$$\operatorname{dg}=\left(e^{- \frac{1}{u^{2}}}\right)^{\prime }du=\frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}} du$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{1 d u}=u$$$ (os passos podem ser vistos »).

A integral torna-se

$$- {\color{red}{\int{e^{- \frac{1}{u^{2}}} d u}}}=- {\color{red}{\left(e^{- \frac{1}{u^{2}}} \cdot u-\int{u \cdot \frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}} d u}\right)}}=- {\color{red}{\left(u e^{- \frac{1}{u^{2}}} - \int{\frac{2 e^{- \frac{1}{u^{2}}}}{u^{2}} d u}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=2$$$ e $$$f{\left(u \right)} = \frac{e^{- \frac{1}{u^{2}}}}{u^{2}}$$$:

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

Seja $$$v=\frac{1}{u}$$$.

Então $$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (veja os passos »), e obtemos $$$\frac{du}{u^{2}} = - dv$$$.

Portanto,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ usando $$$c=-1$$$ e $$$f{\left(v \right)} = e^{- v^{2}}$$$:

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

Esta integral (Função erro) não possui forma fechada:

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

Recorde que $$$v=\frac{1}{u}$$$:

$$- u e^{- \frac{1}{u^{2}}} - \sqrt{\pi} \operatorname{erf}{\left({\color{red}{v}} \right)} = - u e^{- \frac{1}{u^{2}}} - \sqrt{\pi} \operatorname{erf}{\left({\color{red}{\frac{1}{u}}} \right)}$$

Recorde que $$$u=\frac{1}{x}$$$:

$$- \sqrt{\pi} \operatorname{erf}{\left({\color{red}{u}}^{-1} \right)} - {\color{red}{u}} e^{- {\color{red}{u}}^{-2}} = - \sqrt{\pi} \operatorname{erf}{\left({\color{red}{\frac{1}{x}}}^{-1} \right)} - {\color{red}{\frac{1}{x}}} e^{- {\color{red}{\frac{1}{x}}}^{-2}}$$

Portanto,

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

Adicione a constante de integração:

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

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