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.

  • 2
    $\begingroup$ It is entire. Adding two or more non-entire functions may create an entire one. I will add a proof. $\endgroup$ Jan 15 '19 at 5:56
  • $\begingroup$ Sorry! I have deleted my comment and given an upvote for the question. $\endgroup$ Jan 15 '19 at 5:59
  • 2
    $\begingroup$ However I should thank you for making me notice the prove is non-trivial so that I can improve the post. $\endgroup$ Jan 15 '19 at 6:21
  • $\begingroup$ The question and answers are very interesting ! $\endgroup$
    – 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 $x\to\infty$.



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

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.


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