So I was bored, and decided to try and evaluate$$\int e^{-1/x^2}dx$$in as many ways as I could. After a while, I came up with this method for my first method, but I'm unsure if it's correct:
Using IBP, we get
| D | I | |
|---|---|---|
| + | $e^{-1/x^2}$ | $1$ |
| - | $2x^{-3}e^{-1/x^2}$ | $x$ |
So we get from[1]$$\text R1\text C1\cdot\text R2\text C2-\int\text R2\text C1\cdot\text R2\text C2\,dx=xe^{-1/x^2}-2\int\dfrac{e^{-1/x^2}}{x^2}dx\because IBP$$
So now we let our integral now be$$I=xe^{-1/x^2}+I_1,I_1=-2\int\dfrac{e^{-1/x^2}}{x^2}dx\overset{u=1/x,du=(-1/x^2)dx}=-2\int\dfrac{e^{-u^2}}{\require{cancel}\cancel{x^2}}\cancel{-x^2}du$$So now our integral is just$$I_1=2\int e^{-u^2}du=\sqrt\pi\operatorname{erf}(u)+c$$so we can add this to $xe^{-1/x^2}$ to get that our antiderivative is in simplest form$$I=\int e^{-1/x^2}dx=xe^{-1/x^2}+\sqrt\pi\operatorname{erf}(1/x)+c$$
My question
Did I evaluate this correctly, or is there another, more simple way to evaluate $\displaystyle\int e^{-1/x^2}dx$?
Notes
[1]If anyone is confused about the $\text RX_{\text{#}_1}\text CX_{\text{#}_2}\cdot\cdots$ stuff, what I'm basically trying to say here is that what is in row X column Y is being multiplied by row Z column A or something like that for IBP