Convergence of mean value of entire function bounded on the real line Let $f(z)$ be entire analytic function which is bounded for $z \in \mathbb{R}$. Is it true that the following limit  always exists 
$$\lim_{T\rightarrow \infty} \frac{1}{T} \int_0^T f(t) \, d t\text{ ?}$$
We have
$f(z) = \sum_{n=0}^\infty a_n z^n$, where the sequence is absolutely convergent, i.e. $\sum_{n=0}^\infty |a_n| r^n < \infty$ for all real $r$. 
We can write
$$\frac{1}{T} \int_0^T f(t) \, d t = \sum_{n=0}^\infty \frac{a_n}{n+1} T^n$$
Because of the following inequality 
$$\sum_{n=0}^\infty \frac{|a_n|}{n+1} T^n \leq \sum_{n=0}^\infty |a_n|T^n \leq \infty\text{ for all }T$$
the sequence $\sum_{n=0}^\infty \frac{a_n}{n+1} T^n$ is absolutely convergent for all $T$. We also know that $\sup_T \sum_{n=0}^\infty a_n T^n \leq \infty$. Can one use these facts to prove existence of $\frac{1}{T} \int_0^T f(t) \, d t $?
 A: No, the limit $\frac{1}{T} \int_0^T f(t)\,dt$ need not exist. A family of counterexamples can be obtained using the following  observation:
Fact. Let $0<\lambda<1$. If $g$ is an entire function such that $g(0)=0$, then $ f(z)=\sum_{k=0}^\infty g(\lambda^k z)$ is entire.
Indeed, on every disk $|z|\le R$ we have a bound $|g(z)|\le C|z|$ (with $C$ depending on $R$) which  implies that the series converges uniformly on the disk. $\quad \Box$
Let's take $g(z)=z \exp(-z^2)$ above. The function $f(z)= \sum_{k=0}^\infty g(\lambda^k z)$ is bounded on $\mathbb R$ by 
$$\sum_{n\in \mathbb Z} \sup_{ \lambda^{n+1}\le t\le \lambda^n}|g(t)|<\infty$$
(The series converges because its terms decay exponentially  for positive $n$, and super-exponentially for negative $n$). If we let 
 $$G(z)= \frac{1}{z}\int_0^z g(t)\,dt =   \frac{1}{2z}  (1-\exp(-z^2)) $$  the average of $f$ takes the form 
 $$m(T) := \frac{1}{T}\int_0^T f(t)\,dt = \sum_{k=1}^\infty 
 G(\lambda^k T) $$
As $T$ runs through the sequence  $T=\lambda^{-n}a$, the average has a limit: 
$$m(\lambda^{-n}a)  = \sum_{k=1}^\infty 
 G(\lambda^{k-n} a)  \to \sum_{k=-\infty}^\infty  G(\lambda^{k}a),\quad n\to\infty$$
To show that $m(T)$ does not have a limit as $T\to\infty$, it remains to show that the expression $$H(a,\lambda):=\sum_{k=-\infty}^\infty  G(\lambda^{k}a)$$ depends on $a$, at least for some $\lambda\in(0,1)$. One way to do it is to observe that $G$ vanishes at $0$ and $\infty$, which implies that for every fixed $a$, 
$$\lim_{\lambda\to 0} H(a,\lambda)=G(a)  $$
Since $G$ is nonconstant, you can pick $a_1,a_2$ with  $G(a_1)\ne G(a_2)$, and let $\lambda$ be small enough so that $H(a_1,\lambda)\ne H(a_2,\lambda)$.
