# Entire function $f(z)$ grows like $\exp(x^\pi)$ as $x\to+\infty$

Does there exists an entire function $$f(z)$$ such that $$\lim_{x\to+\infty}f(x)/\exp(x^\pi)=1$$ (along the real axis)?

I have successfully constructed $$f(z)$$ when $$\pi$$ is replaced by a rational number $$\frac pq$$.
For $$\lim_{x\to+\infty}f(x)/\exp(x^{p/q})=1$$, take $$f(z)=\exp(z^{p/q})+\exp(z^{p/q}e^{2/q\pi i})+\exp(z^{p/q}e^{4/q\pi i})+\cdots+\exp(z^{p/q}e^{2(q-1)/q\pi i})$$ It is easy to verify $$\lim_{x\to+\infty}f(x)/\exp(x^{p/q})=1$$.

Proof of $$f(z)$$ is entire
It is easy to see $$f(z^q)$$ is entire. Denote $$g(z)=\exp(z^p)+\exp(z^pe^{2/q\pi i})+\exp(z^pe^{4/q\pi i})+\cdots+\exp(z^pe^{2(q-1)/q\pi i}),$$ $$g$$ has property $$g(z)=g(ze^{2/q\pi i})$$ and $$f(z)=g(z^{1/q})$$.
Let $$g(x)=a_0+a_1x+\cdots$$, substituting $$g(z)=g(ze^{2/q\pi i})$$ repeatedly and solving the simultaneous equation gives $$g(x)=a_0+a_qx^q+a_{2q}x^{2q}+\cdots$$. Hence the entirety of $$f$$.

But for $$\pi$$? I can't take the limit with respect to $$p/q$$. I have no idea how to proceed.

• It is entire. Adding two or more non-entire functions may create an entire one. I will add a proof. Jan 15 '19 at 5:56
• Sorry! I have deleted my comment and given an upvote for the question. Jan 15 '19 at 5:59
• However I should thank you for making me notice the prove is non-trivial so that I can improve the post. Jan 15 '19 at 6:21
• The question and answers are very interesting !
– mick
Aug 19 '21 at 19:29

Let's start with the entire function $$f_1(z) = \dfrac{1-e^{-z}}{z}$$ and let $$f_2(z) = \int_1^z f_1(w) \,\mathrm{d}w + \int_1^\infty\dfrac{e^{-t}}{t} \,\mathrm{d}t.$$ (The last term is a real improper integral with the aim to cancel out the integral of $$e^{-w}/w$$.)

For real $$x>1$$, we have \begin{align*} f_2(x) &= \int_1^x\left(\frac1t-\frac{e^{-t}}t\right) \,\mathrm{d}t + \int_1^\infty\dfrac{e^{-t}}{t} \,\mathrm{d}t \\ &= \log x + \int_x^\infty \dfrac{e^{-t}}{t} \,\mathrm{d}t \\ &= \log x + O(e^{-x}/x). \end{align*}

Then consider $$f_3(z) = \exp \big(\pi f_2(z)\big)$$ and $$f(z) = \exp f_3(z)$$; we obtain \begin{align*} f_3(x) &= \exp\Big(\pi\log x + O(e^{-x}/x)\Big) \\ &= x^\pi\cdot\exp\Big(O(e^{-x}/x)\Big) \\ &= x^\pi\Big(1+O(e^{-x}/x)\Big) \\ &= x^\pi+O\big(x^{\pi-1}e^{-x}\big) \end{align*} so \begin{align*} f(x) &= \exp(f_3(x)) = \exp(x^\pi) \cdot \exp\Big(O\big(x^{\pi-1}e^{-x}\big)\Big) =\\ &= \exp(x^\pi) \cdot\Big(1+O\big(x^{\pi-1}e^{-x}\big)\Big), \end{align*} if $$x\to+\infty$$.

Therefore, $$f(z) = \exp \exp \Bigg( \pi\cdot \bigg( \int_1^z \frac{1-e^{-w}}w \,\mathrm{d}w + \int_1^\infty\dfrac{e^{-t}}{t} \,\mathrm{d}t \bigg)\Bigg)$$ is an entire function, satisfying $$f(x)\sim e^{x^\pi}$$ if $$x$$ is real and $$x\to\infty$$.

Define

$$f(z)=\sum_{n=0}^\infty\left(\frac{z}{n^{1/\pi}}\right)^{\lceil n\pi\rceil} \frac{n^n}{n!}$$

where $$\lceil n\pi\rceil$$ means the smallest integer greater than or equal to $$n\pi,$$ and the $$n=0$$ term is $$1$$ by convention. Then $$f$$ is entire: it converges faster than the usual power series for $$\exp(z^4).$$

We need to show that $$\exp(-x^\pi)f(x)\to 1$$ as $$x\to\infty.$$ Let $$N$$ be a Poisson distributed random variable with mean $$x^\pi,$$ and let $$Y=(xN^{-1/\pi})^{\lceil N\pi\rceil-\pi N}$$ (with $$Y=1$$ for $$N=0$$). Then

$$\exp(-x^\pi)f(x) =\exp(-x^\pi)\sum_{n=0}^\infty\left(\frac{x}{n^{1/\pi}}\right)^{\lceil n\pi\rceil-\pi n} \frac{x^{\pi n}}{n!} =\mathbb E[Y].$$

From here the proof is just routine probabilistic estimates. Consider a fixed $$0<\epsilon<1$$ and large $$x$$ (how large to be determined later, depending on $$\epsilon$$). Let $$E$$ be the event $$x/N^{1/\pi}\in (\exp(-\epsilon),\exp(\epsilon))$$ or in other words $$N\in (x^\pi e^{-\epsilon\pi},x^\pi e^{\epsilon\pi}).$$ The variance of $$N$$ is $$x^\pi,$$ so by Chebyshev's inequality, $$\mathbb P[E]$$ is at least $$1-O(x^{-\pi}\epsilon^{-2}).$$ This means it is possible to pick $$x$$ large enough such that $$\mathbb P[E]\geq 1-\epsilon/x.$$ Then as $$\epsilon\to 0$$ we get $$\mathbb E[Y1_E]\to 1,$$ and $$\mathbb E[Y(1-1_E)]\leq x(1-\mathbb P[E])\to 0.$$ This gives $$\mathbb E[Y]\to 1$$ as required.