4 forgot the constant of integration.

Start with an entire function $f$ such that $f(x)=1/x + O(1/x^2)$ for $x>0$, $x\rightarrow\infty$. For example $f(z)= (1-e^{-z})/z$.

Let F be some primitive for $f$: $F(z)=\int_0^z F(z)=\int_1^z f(s)ds$.

We have $F(x)= ln(x)+O(1/x)$ln(x)+C+O(1/x)$, with C some constant ($ \ C=\int_1^\infty \ (f(x)-{1\over x})\ dx$). Then consider$h(x)=exp(\alpha F(x))$F(x)-\alpha C)$.

We get ${h(x)\over x^\alpha} = exp(O(1/x))\rightarrow 1$.

3 typo

Start with an entire function $f$ such that $f(x)=1/x + O(1/x^2)$ for $x>0$, $x\rightarrow\infty$. For example $f(z)= (1-e^{-z})/z$.

Let F be some primitive for $f$: $F(z)=\int_0^z f(s)ds$. We have $F(x)= ln(x)+O(1/x)$.

Then consider $h(x)=exp(-\alpha h(x)=exp(\alpha F(x))$.

We get ${h(x)\over x^\alpha} = exp(O(1/x))\rightarrow 1$.

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