Equivalent to the non-convergent integral Excuse me, please, for the initial post, I did not know that someone would apprehend this rudely, and also excuse my English ...
Task:
Find the equivalent to a function, when $ t \to +\infty $
$$ f(t) = \int \limits_{t}^{2t} \frac{x^2}{e^{x^2}} dx $$
My ideas:
1) I tried to find such a function to integrate by parts, I would get some function in a closed form and an asymptotically small function.
The problem is that it is not so easy to find it by choosing it, but in general terms it is not clear how to do it.
2) One can try to use the mean value theorem if we find an equivalent function for the integrand
The problem is that there seems to be no equivalent ...
3) You can try to guess the answer if you use the rule of l'Hospital for the original function and function, which we do not yet know.
The problem is that in an adequate form I could not do it.
P.S. I came up with this task myself based on some examples given to me at the lecture. I really do not understand why people do not like my questions ...
 A: We are looking for a nice easy function $g(t)$ such that
$$\lim_{t\to \infty}\frac{f(t)}{g(t)}=1.$$
Since $\lim_{t\to \infty}f(t)=0,$ the same will have to be true of $g.$ That means the ratio will have the form $0/0$ at $\infty,$ suggesting L'Hopital may be the way to go.
Let $F(x)$ be an antiderivative for $x^2e^{-x^2}.$ Then $f(t)= F(2t)-F(t),$ which implies
$$f'(t) = 2F'(2t)-F'(t) = 8t^2e^{-4t^2}- t^2e^{-t^2}.$$
Since $e^{-4t^2}$ is much smaller than $e^{-t^2}$ for large $t,$ the term $- t^2e^{-t^2}$ above will dominate in this expression. So if we can find $g$ such that $g'(t) = - t^2e^{-t^2}+h(t),$ where $h(t)= o(t^2e^{-t^2}),$ then we'll be in good shape.
I just played around with this. The first thing I tried was $g(t)=te^{-t^2}.$ Here $g'(t)= e^{-t^2}-2t^2e^{-t^2}.$ Lucky guess! I just need to divide by $2.$ Thus, taking $g(t)= (te^{-t^2})/2,$ we will have $f(t)/g(t) \to 1$ at $\infty$ by L'Hopital, and we're done.
A: Integration by parts will work. We have
$$\int_t^{2 t} x^2 e^{-x^2} dx =
\int_t^{2 t} \left( -\frac x 2 \right) d(e^{-x^2}) = \\
-\frac x 2 e^{-x^2} \bigg\rvert_t^{2 t} -
 \int_t^{2 t} \frac 1 {4 x} d(e^{-x^2}) = \dots \,.$$
All terms evaluated at $x = 2t$ will produce faster decaying exponentials and can be discarded. In other words, we could have extended the upper limit of integration to infinity.
Repeating the same procedure, we obtain a full asymptotic series:
$$\int_t^{2 t} x^2 e^{-x^2} dx \sim
e^{-t^2} \sum_{k = 0}^\infty
 \frac {(-1)^{k + 1} \Gamma(k - 1/2)} {4 \sqrt \pi} t^{-2 k + 1}.$$
Essentially, we have computed an asymptotic expansion for $\operatorname{erf}$.
