Vtyjkyuk

Determine if $f(x)$ is an infinite or an infinitesimal as $xto+infty$, its order and the principal part
$$f(x)=, e^x^2 int_x^+inftye^-t^2,dt$$

I can write, and then, using L'Hôpital's rule:

$$lim_x to +inftydfracdisplaystyleint_x^+inftye^-t^2,dte^-x^2,=,lim_x to +inftydfrace^-x^22xe^-x^2, = , lim_x to +inftydfrac12x,=,0$$

So, $f(x)$ is an infinitesimal as $xto+infty$.

I am not sure of what I have done in order to determine the order and the principal part:

$$lim_x to +inftydfracdisplaystyleint_x^+inftye^-t^2,dte^-x^2setminusdfrac1x^alpha,=,lim_x to +inftydfracdisplaystyleint_x^+inftye^-t^2,dtx^-alphae^-x^2, = (H),= , lim_x to +inftydfrace^-x^2e^-x^2x^-alpha-1(alpha+2x^2),=,lim_x to +inftydfracx^alpha+1alpha+2x^2,=, L iff alpha=1$$

hence, the order of the infinitesimal is $1$ and the principal part of the infinitesimal is $dfrac1x$

Is my solution correct or am I wrong? Thank you

As Szeto commented, you showed that $f(x)$ behaves as $frac 12x$; so order $1$ seems to be the answer.

Assuming that you already know the error function
$$int_x^+inftye^-t^2,dt=fracsqrtpi 2, texterfc(x)$$ then
$$f(x)=, e^x^2 int_x^+inftye^-t^2,dt=fracsqrtpi 2, texterfc(x), e^x^2$$ and, if you look here
$$operatornameerfc(x) = frace^-x^2xsqrtpisum_n=0^infty (-1)^n frac(2n - 1)!!(2x^2)^n$$ making
$$f(x)=frac 1 2xsum_n=0^infty (-1)^n frac(2n - 1)!!(2x^2)^n=frac12 x-frac14 x^3+Oleft(frac1x^5right)$$