Playing with the function $e^{t^2}$ I conjectured the following result :
Let $f\in C^2(\Bbb{R},\Bbb{R})$, assume that :
- $f'(x)\rightarrow_{x \rightarrow +\infty}+\infty$
- $\frac{f''(x)}{f'(x)}\int_1^{x}e^{f(t)-f(x)}dt\rightarrow_{x \rightarrow +\infty}0$
Then $\int_1^{x}e^{f(t)}dt\sim_{x \rightarrow +\infty}\frac{\exp(f(x))}{f'(x)}$
Detail for $\int_1^{x}e^{t^2}dt$
By integration by parts we have $\int_1^{x}e^{t^2}dt=xe^{x^2}-1-\int_1^x 2t^2e^{t^2}dt$. Now we can prove that $$\int_1^x 2t^2e^{t^2}dt\equiv_{+\infty} 2x^2\int_1^x e^{t^2}.$$ Proof. Denote $V(x)=\int_1^x 2t^2e^{t^2}dt$, the derivative gives us $$V'(x)=4x\int_1^xe^{t^2}dt+2x^2e^{x^2}=(2x^2e^{x^2})(1+\frac{4x\int_1^{x}e^{t^2}dt}{2x^2e^{x^2}}).$$ Moreover $\frac{4x\int_1^{x}e^{t^2}dt}{2x^2e^{x^2}}\rightarrow0$ it's easy to prove by using the fact that $\frac{4x\int_1^{x}e^{t^2}dt}{2x^2e^{x^2}}=\frac{2}{x}\int_1^x e^{t^2-x^2}dt$ and $0\le \frac{4x\int_1^{x}e^{t^2}dt}{2x^2e^{x^2}}\le 2\int_0^{1}e^{(1-s^2)x^2}ds$ and we conclude by Dominated convergence theorem. Finally, it is relatively easy to conclude that $\int_1^{x}e^{t^2}dt\equiv_{+\infty} \frac{e^{x^2}}{2x}$. Does anyone can prove the 'theorem' ?
We can prove something similar to your result with alternate (and in my opinion, simpler) assumptions. Depending on whether these assumptions are satisfactory, this might just be considered a partial answer.
Your result holds with the assumptions that:
1. $\int_a^\infty e^{f(x)}\ dx = \infty$
2. $f''(x) \in o\left(f'(x)^2\right)$
where $a$ is some fixed constant and where $o$ denotes the little-o. Your first assumption easily implies my first assumption, but I am not sure how our second assumptions compare.
First, note that we have $$\frac{d}{dx}\left(\frac{e^{f(x)}}{f'(x)}\right) = \frac{f'(x)^2e^{f(x)} - f''(x)e^{f(x)}}{f'(x)^2} = e^{f(x)}\left(1 - \frac{f''(x)}{f'(x)^2}\right)$$
Note that if $f''(x) \in o\left(f'(x)^2\right)$ then we have $$e^{f(x)} \sim e^{f(x)}\left(1 - \frac{f''(x)}{f'(x)^2}\right) = \frac{d}{dx}\left(\frac{e^{f(x)}}{f'(x)}\right)$$ Now as long as we have $$\int_a^\infty e^{f(x)}\ dx = \infty$$ then we may conclude (via the lemma below) that $$\int_a^x \frac{d}{dt}\left(\frac{e^{f(t)}}{f'(t)}\right)\ dt \sim \frac{e^{f(t)}}{f'(t)} \sim \int_a^x e^{f(t)}\ dt$$
Lemma: Let $f(x)$ and $g(x)$ be continuous. If $f(x) \sim g(x)$ and $\int_a^\infty f(t)\ dt = \infty$ then $$\int_a^x f(t)\ dt \sim \int_a^x g(t)\ dt$$ Proof: This is a consequence of L'Hoptial's rule. Since $\int_a^\infty f(t)\ dt = \infty$ (note that since $f(x) \sim g(x)$, this necessarily implies that the integral of $g$ also diverges to $\infty$), the limit $$\lim_{x\rightarrow \infty} \frac{\int_a^x f(t)\ dt}{\int_a^x g(t)\ dt}$$ is indeterminant of the form $\frac{\infty}{\infty}$. Applying L'Hopital's rule with the fundamental theorem of calculus gives $$\lim_{x\rightarrow \infty} \frac{\int_a^x f(t)\ dt}{\int_a^x g(t)\ dt} = \frac{f(x)}{g(x)} = 1$$ which is our desired result. $\square$
As a final note, knowing the above lemma allows us to pretty much guess this result with a bit of fiddling. We can make an educated guess that $$\int_a^x e^{f(t)} \ dt \sim g(t)e^{f(t)}$$ for some function $g(t)$. With the above lemma, it becomes natural to look at the functions $g$ which may satisfy $$e^{f(x)} \sim g'(x)e^{f(x)} + g(x)f'(x)e^{f(x)}$$ or equivalently $$g'(x) + g(x)f'(x) \sim 1$$ The most natural choice for this to occur is for $g(x) = 1/f'(x)$ with the condition that $g'(x) = -\frac{f''(x)}{f'(x)^2} \sim 0$.