Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
What is the question? –  Per Alexandersson May 28 '10 at 9:05
Existence, yes or no, and if yes, an example, of an entire function $h$ such that $x^{-\alpha}h(x)\rightarrow 1$ for $x\rightarrow+\infty$ with $x$ real. –  Roland Bacher May 28 '10 at 9:20
Looking at the example of $1/\Gamma(1+x)\sim (e/x)^x(2\pi x)^{-1/2}$ as $x\to+\infty$, I would say "yes" but definitely a maitre in complex analysis is wanted. :) –  Wadim Zudilin May 28 '10 at 10:11
Casorati-Weierstrass: If $f$ has an essential singularity at $a$, then the image under $f$ of any punctured disk around $a$ is dense in $\mathbb C$. Use for $a=\infty$. Take an entire function $f(z)$ and consider the entire $g(z)=zf(z)^2$; there exists a direction $\lambda$ along which $g(z)\to C\ne0$ as $z\to\infty$. Then $f(x)=c_0\sqrt{x}g(x/\lambda)$ will give an example with $\alpha=-1/2$. –  Wadim Zudilin May 28 '10 at 10:28

2 Answers 2

up vote 13 down vote accepted

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

share|improve this answer
Yes, it works. –  Wadim Zudilin May 28 '10 at 10:33
It works up to a constant (which is easily dealt with) picked up by the integral. –  Roland Bacher May 28 '10 at 15:20

As a matter of fact, real entire functions (that is, entire functions that map the real line into itself, or equivalently, functions represented by a power series centered in 0, with real coefficients and radius of convergence infinite) are dense in $C^0({\mathbb R}, \mathbb{R})$ in the sense of the order, that is:

Theorem (T.Carleman, 1927). For any two continuous real valued functions f < g there exists a real entire function $\phi$ in between:

$f(x)<\phi(x) < g(x)$ for all $x\in\mathbb{R}$.

So in particular, an entire function may be asymptotic to any continuous real function, and also, it may grow as fast as any continuous function.

share|improve this answer
There seems to be an issue with the latex and markdown. The last line of Pietro's box should read: $f(x)<\phi(x)<g(x)$ for all $x\in\mathbb{R}$. –  j.c. May 28 '10 at 18:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.