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

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

Cambiar la base:

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

Sea $$$u=x \sqrt{\ln{\left(2 \right)}}$$$.

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

La integral puede reescribirse como

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

Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=\frac{1}{\sqrt{\ln{\left(2 \right)}}}$$$ y $$$f{\left(u \right)} = e^{- u^{2}}$$$:

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

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

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

Recordemos que $$$u=x \sqrt{\ln{\left(2 \right)}}$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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