Class of functions whose integral behaves as an exponential function. 
I'm looking for increasing functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ satisfying $\int_1^n f(x)dx \to \infty$, as $n\to \infty$, but with the property that for all $a>1,$
$$
\lim_{n \to \infty}
 \frac{\left(\int_1^n f(x)^{a}dx\right)^{1/a}}{\int_1^n f(x)dx} \to \mathrm{constant} >0.
$$

Clearly one example is $f(x)=e^x$, but what I'm interested in is if there are other functions that grow more slowly than $e^x$ and satisfy this property. $e^x$ satisfies that $(\int_1^n f(x)dx)/f(n) \to Constant$, i.e. the integral is basically captured by the value of the function at $n$. For the purpose of a counterexample I wish to construct, I was hoping there would be an example where the property holds, but $(\int_1^n f(x)dx)/f(n) \to \infty$.
 A: No function as you would like does exist. To see this, note that your desired property $\int_1^n f(x) dx / f(n) \to \infty$ implies $f(n) / \int_1^n f(x) dx \to 0$. Therefore, since $f$ is increasing,
$$
\frac{(\int_1^n f(x)^a dx)^{1/a}}{\int_1^n f(x) dx}
= \bigg(
    \frac{\int_1^n f(x)^a dx}{(\int_1^n f(x) dx)^{a}}
  \bigg)^{1/a}
\leq \bigg(
    \frac{\int_1^n f(x) dx \cdot (f(n))^{a-1}}{\int_1^n f(x) dx \cdot (\int_1^n f(x) dx)^{a-1}}
  \bigg)^{1/a}
\xrightarrow{} 0,
$$
contradicting the convergence you would like.
A: I don't think such a function exists. Suppose that the limit in the OP exists and is equal to $C>0$. Then using properties of limits
$$\lim_{x\to\infty}\frac{\int_{1}^xf^{a}(t)dt}{(\int_{1}^x f(t)dt)^{a}}=C^a$$
We can apply L'Hopital's rule here if both integrals diverge (which they should, if we are seeking functions that are increasing and grow faster than constants) to obtain the equivalent limit- after raising to the power $1/(a-1)$ of course-
$$\lim_{x\to\infty}\frac{f(x)}{\int_{1}^x f(t)dt}=(aC^a)^{\frac{1}{a-1}}\equiv L$$
This limit tells us by its definition that
$$\forall ~x>x_0~,~|\frac{f(x)}{\int_1^xf(t)dt}-L|<\epsilon\Rightarrow F(x_0)e^{(L-\epsilon)(x-x_0)}<F(x)<F(x_0)e^{(L+\epsilon)(x-x_0)}$$
where the RHS has been obtained by integrating the inequality, for sufficiently small $\epsilon$, whereupon all the integrands are positive. Also $F(x)$ is the antiderivative of $f(x)$. This shows that there is not much wiggle room in the asymptotic behavior of $F(x)$ and no matter what one does for sufficiently large values, it should asymptote as an exponential. Also, if we assume as well that $\lim_{x\to\infty}f(x)=\infty$, and apply L'Hopital's rule again we can show the exact same result for f (substitute $F\rightarrow f$).
