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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 ...

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  • $\begingroup$ "The equivalent to a function". What do you really mean? Are you concerned to find a function which behaves similarly to $f(t)$ as $t \to +\infty$? $\endgroup$ May 1 '18 at 20:57
  • $\begingroup$ You'll find that simple "Here's the statement of my exercise, solve it for me" posts will be poorly received. What is better is for you to add context: What you understand about the problem, what you've tried so far, etc. Something to both show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. $\endgroup$
    – Shaun
    May 1 '18 at 21:01
  • $\begingroup$ OP, great job on improving your question! I've voted to reopen as a result of your changes. $\endgroup$
    – B. Mehta
    May 2 '18 at 0:12
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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.

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  • $\begingroup$ So how do you guess this function? The answer is to check, I'm certainly able to $\endgroup$ May 2 '18 at 8:51
  • $\begingroup$ Anyway, thanks you! $\endgroup$ May 2 '18 at 8:52
  • $\begingroup$ "So how do you guess this function?" Good question. It wasn't even a good hint since I just pulled the answer out of a hat. I edited it and hopefully improved it. $\endgroup$
    – zhw.
    May 2 '18 at 15:44
  • $\begingroup$ Yes, it looks interesting enough. And is there any tool that allows you to find this assimptotics without guessing? But in general, your method is very good for not very complex functions. $\endgroup$ May 2 '18 at 16:12
  • $\begingroup$ P.S. I am a first-year student in theoretical mathematics at St. Petersburg State University (St. Petersburg, Russia) $\endgroup$ May 2 '18 at 16:14
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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 $\operatorname{erf}$.

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