Integral of $$$e^{\frac{y^{2}}{2}}$$$

Find $$$\int e^{\frac{y^{2}}{2}}\, dy$$$.

Let $$$u=\frac{\sqrt{2} y}{2}$$$.

Then $$$du=\left(\frac{\sqrt{2} y}{2}\right)^{\prime }dy = \frac{\sqrt{2}}{2} dy$$$ (steps can be seen »), and we have that $$$dy = \sqrt{2} du$$$.

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\sqrt{2}$$$ and $$$f{\left(u \right)} = e^{u^{2}}$$$:

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

This integral (Imaginary Error Function) does not have a closed form:

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

Recall that $$$u=\frac{\sqrt{2} y}{2}$$$:

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

Therefore,

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

Add the constant of integration:

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

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