I would like a precision concerning the Tauberian theorem for the Laplace transform. Actually the statement is not clear to me and depends on the books. Assume we have a function $f : (0,+\infty) \to \mathbb{C}.$ And assume that $f \in o(t^{-\alpha})$ with $\alpha<1$ for $t\to 0^+$. (This is enough for the integrability at $0$.) Then $$z \mapsto \int_{0}^{+\infty} e^{-zt} f(t)dt$$ is in $o(???)$ when $|z|\to +\infty.$ What should I put instead of ??? In some books I found $|z|^{\alpha}$ in other $|z|^{\alpha+1}.$ Thanks for any help.
PS : We, of course, assume that there is at least one $z\in \mathbb{C}$ such that this integral exists. Moreover, by the standard theory we know that it exists on a halfplane. I let $|z|$ go to infinity in this halfplane.
PS2 : I would like to have a reversible condition. That is, if $F$ is holomorphic on the halfplane and $F\in o(???)$ when $|z|\to +\infty$ then $\mathcal{L}^{-1}(F) \in o(t^{-\alpha})$ when $t\to 0^+.$
No Tauberian theorem here.
If $f\in L^1(0,\infty)$ and $|f(t)| \le C t^{-a}$ as $t \to 0$ for some $a \in (0,1)$
then $|\int_b^\infty e^{-xt}f(t)dt| \le \|f\|_{L^1}e^{-bx}$ and $|\int_0^b f(t) e^{-xt} dt| \le C \int_0^b t^{-a}dt = C \frac{b^{1-a}}{1-a}$.
Letting $b = (1-a)\frac{\log x}{x}$ you get $$|\int_0^\infty f(t) e^{-xt}dt| \le x^{a-1}(\|f\|_{L^1} + C \frac{((1-a)\log x)^{1-a}}{1-a})$$