Let $X$ be Poisson r.v. with $\lambda$ find $f(x)$ such that $E[f(X)]=\lambda \log (\lambda)$ I am looking for a function $f(x)$ such that 
\begin{align}
E[f(X)]=\lambda \log (\lambda) \quad  \text{ for all } \lambda \ge 0 \tag{$*$}
\end{align}
where $X$ is a Poisson random varaible with parameter $\lambda$. Note, we are looking for a function idenpendent of $\lambda$. 
Here are some thoughts:
\begin{align}
\lambda \log (\lambda) = E[f(X)]= \sum_{k=0}^\infty f(k) \frac{\lambda^k e^{-\lambda}}{k!} 
\end{align}
Therefore, we have that 
\begin{align}
e^{\lambda} \lambda \log (\lambda) =  \sum_{k=0}^\infty f(k) \frac{\lambda^k }{k!}= \sum_{k=0}^\infty a_k \frac{\lambda^k }{k!} \tag{$**$}
\end{align}
where in the last step I defined $f(k)=a_k$. 
Now, this is the point where I am a bit stuck.  
It doesn't look like the expression in $(**)$ can hold as I don't think the function $g(x)=e^x x \log(x)$ can be written as a Maclaurin series. 
These lets me to conclude that there is no function $f(x)$ such that $(*)$ holds.  
 A: To be precise, we want to find $f:\mathbb{N}_0\to\mathbb{C}$ such that
$
E\left[|f(X)|\right]<\infty,
$ and $$
E[f(X)]=\lambda\log\lambda,\quad\forall \lambda\ge 0.
$$
Now, we can write
$$
\sum_{k=0}^\infty f(k)\frac{\lambda^k e^{-\lambda}}{k!}=\lambda\log\lambda,
$$or equivalently
$$
\sum_{k=0}^\infty f(k)\frac{\lambda^k}{k!}=\lambda e^\lambda\log\lambda .
$$ The LHS is by the assumption an absolutely convergent power series for all $\lambda\ge 0$. Then this implies that it can be extended uniquely to some entire function $F(\lambda)$. Since $F(0) = \lim\limits_{\lambda\to 0^+}\lambda e^\lambda \log \lambda =0$, $G(\lambda)=\frac{F(\lambda)} {\lambda}$ also defines an entire function. Now, we have
$$
e^{-\lambda}G(\lambda) =\log\lambda
$$ for all $\lambda>0$, but $\lim\limits_{\lambda\to 0^+}e^{-\lambda}G(\lambda)=G(0)\neq \lim\limits_{\lambda\to 0^+}\log\lambda=-\infty.$ This leads to a contradiction. Therefore, there does not exist $f$ such that $E[|f(X)|]<\infty$ and $E[f(X)]=\lambda \log \lambda$ for all $\lambda\ge 0$.
