Let $F:[0,1]\to\mathbb{R}$ be a measurable function such that
$$
\mu\_n(F)=\int\_0^1F(t)t^ndt\sim\frac1{(\ln n)^n}\quad\mbox{as}\quad n\to\infty.
$$
More precisely, 
$$
0<c<|\mu\_n(F)|(\ln n)^n<C<\infty,\quad\forall n\in\mathbb{N}.
$$
Note that in this case the series
$$
\sum\_{n=0}^\infty\mu\_n(F)z^n
$$
represents an entire function. The generating function
$$
\sum\_{n=0}^\infty\frac{\mu\_n(F)z^n}{n!}
$$
is an entire function of minimal exponential type. I am not aware of any explicit description of such a function, so this is already a sign.


**Question:** Can we find an explicitly given such function $F$? Here by "explicitly given" I mean in a "closed form", e.g., no series.


I think that by convexity arguments it can be shown that $F$ cannot be non-negative (non-positive). In fact, I do not need $F$ be a function; $F(x)dx$ may be a signed measure on $[0,1]$, just given explicitly.


Thank you.



Find a function $F$ on $[0,1]$ with moments decaying as $(\ln n)^{-n}$
$\newcommand{\R}{\mathbb{R}}
\newcommand{\si}{\sigma}
\newcommand{\supp}{\operatorname{\mathrm supp}}
\newcommand{\cch}{\operatorname{\mathrm cch}}
$
If $F\in L^2$, then the condition 
\begin{equation\*}
 |\mu\_n(F)|(\ln n)^n<C<\infty\quad\forall n\in\mathbb{N} \tag{1}
\end{equation\*}
implies that $F=0$ almost everywhere (a.e.) on $[0,1]$. 


Indeed, let 
\begin{equation\*}
 f(z):=\sum\_{n=0}^\infty\frac{\mu\_n(F)(iz)^n}{n!}
 =\int\_0^1 F(t)\,dt \sum\_{n=0}^\infty\frac{(itz)^n}{n!}
 =\int\_0^1 F(t)e^{itz}\,dt
\end{equation\*}
for all complex $z$. 
By (1), for any natural $k\ge2$
\begin{equation\*}
 |f(z)|\le O(1+|z|^{k-1})+C\sum\_{n=k}^\infty\frac{(|z|/\ln n)^n}{n!}
\le O(1+|z|^{k-1})+C\sum\_{n=0}^\infty\frac{(|z|/\ln k)^n}{n!}
\le c\_k e^{|z|/\ln k}
\end{equation\*}
for some real $c\_k>0$ and all complex $z$. 
So, $f(z)$ is an entire function of exponential type $a$ for any real $a>0$. 


Hence, by a [Paley--Wiener theorem](https://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem) (more specifically, see e.g. [Theorem 19.3 on page 375](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwidramz0vHhAhUHQq0KHVAcCicQFjAAegQIARAC&url=https%3A%2F%2F59clc.files.wordpress.com%2F2011%2F01%2Freal-and-complex-analysis.pdf&usg=AOvVaw3_ZfMwIfUWx3Lrt4d_zS5e)), for each real $a>0$ there is an $L^2$ function $F\_a$ such that for all complex $z$
\begin{equation\*}
 f(z)=\int\_{-a}^a F\_a(t)e^{itz}\,dt
\end{equation\*}
Taking the inverse Fourier transform, we see that for each $a\in(0,1)$ and all complex $z$
\begin{equation\*}
 f(z)=\int\_0^a F(t)e^{itz}\,dt. 
\end{equation\*}
Thus, for all complex $z$
\begin{equation\*}
 0=f(z)=\int\_0^1 F(t)e^{itz}\,dt. 
\end{equation\*}
So, indeed $F=0$ a.e. on $[0,1]$. 




---


Consider now the more general setting when $F(t)\,dt$ is replaced by $\rho(dt)$, where $\rho$ is a signed measure over $\R$ with support $\supp\rho\subseteq[0,1]$. Then condition (1) is replaced by 
\begin{equation\*}
 |\mu\_n(\si)|(\ln(n+2))^n<C<\infty\quad\forall n\in\mathbb{N}\_0:=\{0,1,\dots\}, \tag{1a}
\end{equation\*}
where 
\begin{equation\*}
 \mu\_n(\si):=\int\_0^1 t^n\si(dt)=\int\_0^1 t^{n+2}\rho(dt)\quad\text{and}\quad\si(dt):=t^2\rho(dt). 
\end{equation\*}
So, $\supp\si\subseteq[0,1]$. Also, (1a) implies $\mu\_0(\si)<\infty$, that is, the signed measure $\si$ is finite. So, by reasoning quite similar to that in the above $L^2$ case, we see that under condition (1a) 
\begin{equation\*}
 f(z):=\int\_\R e^{itz}\,\si(dt)
\end{equation\*}
is an entire function of exponential type $a$ for any real $a>0$. 


For real $b>0$ and $t\in\R$, let 
\begin{equation\*}
 G\_b(t):=(g\_b\*d\si)(t):=\int\_\R g\_b(t-s)\,\si(ds)
 =\frac{\si([t-b,t+b])}{2b}, 
\end{equation\*}
where $g\_b:=\frac1{2b}\,1\_{[-b,b]}$. Then $G\_b\in L^2(\R)$, since the signed measure $\si$ is finite with $\supp\si\subseteq[0,1]$. Also, 
\begin{equation\*}
 \hat g\_b(z):=\int\_\R e^{itz}g\_b(t)\,dt=\frac{\sin bz}{bz}
\end{equation\*}
for $z\ne0$, and hence 
\begin{equation\*}
 f\_b(z):=\int\_\R e^{itz}\,G\_b(t)\,dt
 =\int\_\R e^{itz}\,(g\_b\*d\si)(t)\,dt=\hat g\_b(z)f(z)
\end{equation\*}
is of exponential type $b+a$ for all real $a>0$. Thus, by the cited Paley--Wiener theorem, $\supp(g\_b\*d\si)\subseteq[-b-a,b+a]$ for all real $a>0$ and hence 
\begin{equation\*}
 \supp(g\_b\*d\si)\subseteq[-b,b]. 
\end{equation\*}


On the other hand, because $\supp g\_b$ and $\supp\si$ are both compact, by [Theorem 4.3.3 on page 117](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&uact=8&ved=2ahUKEwigxZCtvvXhAhUR16wKHcMWCMkQFjAEegQIBxAC&url=http%3A%2F%2Fwww.lama.univ-savoie.fr%2F~raibaut%2FHormander-LPDOI.pdf&usg=AOvVaw3ozf7r2YsA4HKPwTiAFpZ_), $\cch\supp(g\_b\*d\si)=\cch\supp g\_b+\cch\supp\si$, where $\cch$ denotes the closed convex hull. Since $\cch\supp g\_b\ni0$, we conclude that $\cch\supp\si\subseteq\cch\supp(g\_b\*d\si)\subseteq[-b,b]$, whence 
$\supp\rho\subseteq\supp\si\cup\{0\}\subseteq[-b,b]$, for any real $b>0$. Thus, condition (1a) implies 
\begin{equation\*}
\supp\rho\subseteq\{0\}. \tag{2} 
\end{equation\*}


Vice versa, trivially (2) implies (1a). Thus, (1a) holds iff $\supp\rho\subseteq\{0\}$. 




---


**Response to the second comment by the OP:**
Let us try to sort all this out. 


1. Your condition 
\begin{equation}
0<c<|\mu\_n(F)|(\ln n)^n<C<\infty \quad\forall n\in\mathbb{N} \tag{3} 
\end{equation} 
can actually never hold as stated, because the inequalities $0<c<|\mu\_n(F)|(\ln n)^n$ will always be false for $n=1$. So, of course, I assumed (3) with $\mathbb{N}$ replaced by $\{2,3,\dots\}$.
2. In your post here, you did not even mention "analytical distributions" or Ehrenpreis. Instead, you wrote: "Let $F:[0,1]\to\mathbb{R}$ be a measurable function" and then you also wrote "In fact, I do not need $F$ be a function; $F(x)dx$ may be a signed measure on $[0,1]$".
3. If $F(x)dx$ is replaced by a signed measure $\rho(dx)$, then the corrected version of your condition (3) can be rewritten as the conjunction of my condition (1a) and the condition 
\begin{equation}
0<c<|\mu\_n(\si)|(\ln n)^n \quad\forall n\in\{2,3,\dots\}. \tag{4}
\end{equation}
4. It was shown in this answer that, under condition (1a), the support of the signed measure $\rho$ must be contained in the set $\{0\}$. (Method-wise, this was first done in the case when $\rho(dx)=F(x)dx$ with $F\in L^2$, and then extended to the general case of any signed measure $\rho$ on $[0,1]$.)
5. It then immediately follows that there is no signed measure $\rho$ on $[0,1]$ satisfying (1a) and (4) -- or, equivalently, satisfying the corrected version of your condition (3). In other words, your conditions do result in a contradiction. This completely answers your posted question.


