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.
Best Answer
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 $x\to\infty$.