Asymptotic Equivalence $e^{x^2+x} \sim e^{x^2}$? I've a doubt regarding the asymptotic equivalence of two functions.
From the definition
$f(x)\sim_{x\to x_0} g(x) \iff \lim_{x\to x_0} \frac{f(x)}{g(x)} =1$
While I was trying to determine the convergence of an improper integral (to $\infty$) I had to manipulate this expression
$e^{x^2+x}$
Following the definition
 $e^{x^2+x} \not \sim e^{x^2}$ since of course $\lim_{x\to \infty} \frac{e^{x^2+x}}{e^{x^2}} =\infty$
But it is also true that, considering the exponents,
 $x^2+x\sim_{x\to \infty} x^2$
And since in general $ \lim_{x\to x_0} e^{f(x)}=e^{ \lim_{x\to x_0} f(x)}$
Isn't it possible to say at least that $e^{x^2+x}$ "behaves" as $e^{x^2}$ when $x\to \infty$ and therefore, if I'm dealing with an improper integral for example, I can "replace"  $e^{x^2+x}$ with $e^{x^2}$?
Thanks a lot for your help
Just to make an example the improper integral I'm dealing with is 
$\int_1^\infty \frac{1}{x^2e^{x^2+x} }$
The question is the same for $e^{x^2+x}$ and $x^2$.
Can I substitute $x^2e^{x^2+x}$ with  $e^{x^2+x}$ or just with  $e^{x^2}$, even if  $x^2e^{x^2+x}  \not \sim e^{x^2}$ ?
 A: I want to challenge your thought by asking some questions and hope that it helps you to figure it out.
What is your definition of behaving? I mean how do you define two functions behaving like each other at the neighborhood of some point $x_0$. Can you give a definition for it like the one for asymptotic equivalence? We should be elaborate and exact when we say something in math! Just saying a word does not suffice!
1) We say two functions $f(x)$ and $g(x)$ are asymptoticaly equivalent at the neighborhood of $x_0$ if and only if 
$$ \lim_{x \to x_0} \frac{f(x)}{g(x)}=1$$
What do you say about two function $f(x)$ and $g(x)$ behaving like each other at the neighborhood of $x_0$?
2) Although it is meaningless as we don't have a precise definition yet, assume that you could say $e^{x^2+x}$ and $e^{x^2}$ behave like each other at $\infty$. However, can you do the replacement in an integral $\int_1^\infty \frac{1}{x^2e^{x^2+x}}$? I mean they behave like each other at $\infty$ but not at $7$ for example! I leave the final conclusion to yourself! :)

By the way we can obtain some inequalities for your integral which may help. Noting that $x\in[1,\infty)$, it is obvious $x^2>1,\,e^{x^2}>1,\,e^x>1$. Also, $x^2 > x$ which implies $e^{-x^2}<e^{-x}$. Then we have
$$\frac{1}{x^2 e^{x^2+x}}=\frac{1}{x^2 e^{x^2}e^x}<\frac{1}{x e^{x^2} e^x}< \frac{1}{x e^x}=\frac{e^{-x}}{x}.$$
Since the integral is order preserving we obtain
$$\int_{1}^{\infty}\frac{1}{x^2 e^{x^2+x}}\text{d}x < \int_{1}^{\infty}\frac{e^{-x}}{x}\text{d}x.$$
The upper bound we found is a famous integral, known as the exponential integral.
A: $\int_1^\infty \frac{dx}{x^2e^{x^2+x}}$ is a finite  integral.
we have $\int_1^\infty \frac{dx}{x^2e^{x^2+x} } <\int_1^\infty \frac{e^{-x^2}}{x^2}dx <  \int_1^\infty \frac{e^{-x}}{x^2}dx$ ...
A: Do not substitute any of them because they will change the result drastically. If I'm allowed to do so, I'd suggest you make a substitution $y = x + 1/2$ and your integral will become $$\int_1^\infty \frac{dx}{x^2e^{x^2+x} } = \int_{3/2}^\infty \frac {e^{1/4} dy} {(y - 1/2)^2 e^{y^2}}$$ Hope I've been of help.
