Given a smooth function $f : \mathbb{R} \to \mathbb{R}$, I am interessed in the behavior of $$I_{\epsilon} = \int_{0}^{\epsilon} f(x) dx \sim ~ ?\quad, \quad \text{when }\epsilon \to 0$$
When $f(0) \neq 0$, $I_{\epsilon} \sim \epsilon f(0)$, when $\epsilon \to 0$. When $f(0) = 0$ but $f'(0) \neq 0$, then $I_{\epsilon} \sim \epsilon^2 f'(0)$. More generally, if $f^{(i)}(0) = 0$ for $i \leq k-1$ and $f^{(k)}(0) \neq 0$ then $I_{\epsilon} \sim \epsilon^k f^{(k)}(0)$.
My question is : what happen if $f^{(i)}(0) = 0$ for all $i \geq 0$, but $f$ is not identically zero ? Is there a canonical way to renormalize $I_{\epsilon}$ by $u_{\epsilon}$ to obtain a finite non zero limit for $\dfrac{I_{\epsilon}}{u_{\epsilon}}$ ? For the previous case it was $u_{\epsilon} = \epsilon^k$, with $k$ linked to the nullity of the derivatives of $f$.
If a canonical renormalization is not possible, I have the following question : given a sequence $u_{\epsilon} \to 0$, can we construct $f$ such that $\dfrac{I_{\epsilon}}{u_{\epsilon}}$ converges to a finite, non zero limit ?
Thanks.
There is no such "canonical renormalization".
With your notations, take any smooth function $\epsilon \mapsto u(\epsilon)$ defined for $\epsilon \in [0,1]$ with $u(0) = 0$. Then simply set $f(x) := u'(x)$. Hence $I_\epsilon / u_\epsilon = 1$ for every $\epsilon \in (0,1]$.
For example, with $u(\epsilon) := e^{-1/\epsilon}$ you indeed get a "flat" function with all derivatives vanishing at $0$. But you could choose any other $u$ to get any other rate of convergence to $0$.