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

I have successfully constructed $f(z)$ when $pi$ is replaced by a rational number $frac pq$.

For $lim_{xto+infty}f(x)/exp(x^{p/q})=1$, take $$f(z)=exp(z^{p/q})+exp(z^{p/q}e^{2/qpi i})+exp(z^{p/q}e^{4/qpi i})+cdots+exp(z^{p/q}e^{2(q-1)/qpi i})$$
It is easy to verify $lim_{xto+infty}f(x)/exp(x^{p/q})=1$.

Proof of $f(z)$ is entire

It is easy to see $f(z^q)$ is entire. Denote $$g(z)=exp(z^p)+exp(z^pe^{2/qpi i})+exp(z^pe^{4/qpi i})+cdots+exp(z^pe^{2(q-1)/qpi i}),$$ $g$ has property $g(z)=g(ze^{2/qpi i})$ and $f(z)=g(z^{1/q})$.

Let $g(x)=a_0+a_1x+cdots$, substituting $g(z)=g(ze^{2/qpi i})$ 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.

Let's start with the entire function
$$ f_1(z) = dfrac{1-e^{-z}}{z} $$
and let
$$
f_2(z) = int_1^z f_1(w) ,mathrm{d}w + int_1^inftydfrac{e^{-t}}{t} ,mathrm{d}t.
$$
(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^xleft(frac1t-frac{e^{-t}}tright) ,mathrm{d}t
+ int_1^inftydfrac{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) &= expBig(pilog x + O(e^{-x}/x)Big) \
&= x^picdotexpBig(O(e^{-x}/x)Big) \
&= x^piBig(1+O(e^{-x}/x)Big) \
&= x^pi+Obig(x^{pi-1}e^{-x}big)
end{align*}
so
begin{align*}
f(x) &= exp(f_3(x))
= exp(x^pi) cdot expBig(Obig(x^{pi-1}e^{-x}big)Big) =\
&= exp(x^pi) cdotBig(1+Obig(x^{pi-1}e^{-x}big)Big),
end{align*}
if $xto+infty$.

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

Define

$$f(z)=sum_{n=0}^inftyleft(frac{z}{n^{1/pi}}right)^{lceil npirceil} frac{n^n}{n!}$$

where $lceil npirceil$ means the smallest integer greater than or equal to $npi,$ 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 $xtoinfty.$
Let $N$ be a Poisson distributed random variable with mean $x^pi,$ and let $Y=(xN^{-1/pi})^{lceil Npirceil-pi N}$ (with $Y=1$ for $N=0$). Then

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

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 $Nin (x^pi e^{-epsilonpi},x^pi e^{epsilonpi}).$
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 $epsilonto 0$ 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.