Let $X \sim \mathsf{Poisson}(\lambda)$ for some $\lambda>0$. Let $Y=X \cdot \log X$, where we define $0 \cdot \log 0 = \lim_{x \to 0^+} x \cdot \log x = 0$.
I would like an expression for $$\mathbb{E}[Y] = \sum_{n=1}^\infty \frac{\lambda^n \cdot n \cdot \log n}{e^\lambda \cdot n!}$$ and $$\mathbb{E}[Y^2] = \sum_{n=1}^\infty \frac{\lambda^n \cdot ( n \cdot \log n)^2}{e^\lambda \cdot n!}.$$
An exact closed-form expression would be ideal. However, even a good approximation would be very useful.
Note that $$\mathbb{E}[Y] = \sum_{n=1}^\infty \frac{\lambda^n \cdot n \cdot \log n}{e^\lambda \cdot n!} = \lambda \cdot \sum_{n=1}^\infty \frac{\lambda^{n-1} \cdot \log n}{e^\lambda \cdot (n-1)!} = \lambda \cdot \sum_{m=0}^\infty \frac{\lambda^m \cdot \log(m+1)}{e^\lambda \cdot m!} = \lambda \cdot \mathbb{E}[\log(X+1)].$$
We also have the following alternate expression for the expectation in terms of fractional moments: $$\mathbb{E}[Y] = \mathbb{E}[X \cdot \log X] = \mathbb{E}\left[ \frac{\mathrm{d}}{\mathrm{d}t} X^t |_{t=1}\right] = \frac{\mathrm{d}}{\mathrm{d}t} \mathbb{E}\left[X^t\right] |_{t=1} = \lim_{t \to 1} \frac{\mathbb{E}[X^t]-\lambda}{t-1}.$$
Since $x \mapsto x \cdot \log x$ is convex, we have $\mathbb{E}[X \cdot \log X] \ge \mathbb{E}[X] \cdot \log \mathbb{E}[X] = \lambda \cdot \log \lambda$. Since $x \cdot \log x \le x \cdot (x-1)$, we have $\mathbb{E}[X \cdot \log X] \le \mathbb{E}[X \cdot (X-1)] = \mathsf{Var}[X] + \mathbb{E}[X]^2 - \mathbb{E}[X] = \lambda^2$.