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

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

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

Then $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{x^{2}} = - du$$$.

Therefore,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=-1$$$ and $$$f{\left(u \right)} = e^{- \frac{1}{u^{2}}}$$$:

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

For the integral $$$\int{e^{- \frac{1}{u^{2}}} d u}$$$, use integration by parts $$$\int \operatorname{o} \operatorname{dv} = \operatorname{o}\operatorname{v} - \int \operatorname{v} \operatorname{do}$$$.

Let $$$\operatorname{o}=e^{- \frac{1}{u^{2}}}$$$ and $$$\operatorname{dv}=du$$$.

Then $$$\operatorname{do}=\left(e^{- \frac{1}{u^{2}}}\right)^{\prime }du=\frac{2 e^{- \frac{1}{u^{2}}}}{u^{3}} du$$$ (steps can be seen ») and $$$\operatorname{v}=\int{1 d u}=u$$$ (steps can be seen »).

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=2$$$ and $$$f{\left(u \right)} = \frac{e^{- \frac{1}{u^{2}}}}{u^{2}}$$$:

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

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

Then $$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (steps can be seen »), and we have that $$$\frac{du}{u^{2}} = - dv$$$.

Thus,

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

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=-1$$$ and $$$f{\left(v \right)} = e^{- v^{2}}$$$:

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

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

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

Recall that $$$v=\frac{1}{u}$$$:

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

Recall that $$$u=\frac{1}{x}$$$:

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

Therefore,

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

Add the constant of integration:

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

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