Determine if $f(x)$ is an infinite or an infinitesimal as $x\to+\infty$, its order and the principal part $$f(x)=\, e^{x^2} \int_{x}^{+\infty}e^{-t^2}\,dt$$
I can write, and then, using L'Hôpital's rule:
$$\lim_{x \to +\infty}\dfrac{\displaystyle\int_{x}^{+\infty}e^{-t^2}\,dt}{e^{-x^2}}\,=\,\lim_{x \to +\infty}\dfrac{e^{-x^2}}{2xe^{-x^2}}\, = \, \lim_{x \to +\infty}\dfrac{1}{2x}\,=\,0$$
So, $f(x)$ is an infinitesimal as $x\to+\infty$.
I am not sure of what I have done in order to determine the order and the principal part:
$$\lim_{x \to +\infty}\dfrac{\displaystyle\int_{x}^{+\infty}e^{-t^2}\,dt}{e^{-x^2}}\setminus\dfrac{1}{x^{\alpha}}\,=\,\lim_{x \to +\infty}\dfrac{\displaystyle\int_{x}^{+\infty}e^{-t^2}\,dt}{x^{-\alpha}e^{-x^2}}\, = (H)\,= \, \lim_{x \to +\infty}\dfrac{e^{-x^2}}{e^{-x^2}x^{-\alpha-1}(\alpha+2x^2)}\,=\,\lim_{x \to +\infty}\dfrac{x^{\alpha+1}}{\alpha+2x^2}\,=\, L \iff \alpha=1$$
hence, the order of the infinitesimal is $1$ and the principal part of the infinitesimal is $\dfrac{1}{x}$
Is my solution correct or am I wrong? Thank you
As Szeto commented, you showed that $f(x)$ behaves as $\frac 1{2x}$; so order $1$ seems to be the answer.
Assuming that you already know the error function $$\int_{x}^{+\infty}e^{-t^2}\,dt=\frac{\sqrt{\pi } }{2}\, \text{erfc}(x)$$ then $$f(x)=\, e^{x^2} \int_{x}^{+\infty}e^{-t^2}\,dt=\frac{\sqrt{\pi } }{2}\, \text{erfc}(x)\, e^{x^2}$$ and, if you look here $$\operatorname{erfc}(x) = \frac{e^{-x^2}}{x\sqrt{\pi}}\sum_{n=0}^\infty (-1)^n \frac{(2n - 1)!!}{(2x^2)^n}$$ making $$f(x)=\frac 1 {2x}\sum_{n=0}^\infty (-1)^n \frac{(2n - 1)!!}{(2x^2)^n}=\frac{1}{2 x}-\frac{1}{4 x^3}+O\left(\frac{1}{x^5}\right)$$