Given a non-integral real $\alpha$, is there an entire (see <http://en.wikipedia.org/wiki/Entire_function>) function $h(x)$ such that $x^{-\alpha}h(x)\longrightarrow 1$
for $x\rightarrow+\infty$ (with $x$ real non-negative)?


Clearly, such a function if it exists is not unique since $h(x)+e^{-x}$ and similar functions
work also.



Asymptotic approximation of $x^\alpha$ by entire functions
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 $f$: $F(z)=\int\_1^z f(s)ds$.


We have $F(x)= ln(x)+C+O(1/x)$, with C some constant ($ \ C=\int\_1^\infty \ (f(x)-{1\over x})\ dx$ ).


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


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


