Integral de $$$e^{\frac{p^{2}}{4}}$$$

Encontre $$$\int e^{\frac{p^{2}}{4}}\, dp$$$.

Seja $$$u=\frac{p}{2}$$$.

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

Assim,

$${\color{red}{\int{e^{\frac{p^{2}}{4}} d p}}} = {\color{red}{\int{2 e^{u^{2}} d u}}}$$

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

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

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

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

Recorde que $$$u=\frac{p}{2}$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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