I'm trying to solve the following weighted probability
$$ \sum_{x=1}^\infty \operatorname{Prob}(x \mid x > 0) \frac{1}{x} $$
where $\operatorname{Prob}(x)$ is Poisson with rate $\lambda$.
I can rewrite
\begin{align} &\frac{e^{-\lambda}}{1-e^{-\lambda}}\sum_{x=1}^\infty \frac{\frac{\lambda^{x}}{x}}{x!} = \frac{e^{-\lambda}}{1-e^{-\lambda}} \sum_{x=1}^\infty \frac{\int_0^\lambda u^{x-1} \, du}{x!} \\[8pt] = {} & \frac{e^{-\lambda}}{1-e^{-\lambda}}\int_0^\lambda \sum_{x=1}^\infty \frac{u^{x-1}}{x!} \, du = \frac{e^{-\lambda}}{1-e^{-\lambda}} \int_0^\lambda \frac{1}{u} \sum_{x=1}^\infty \frac{u^{x}}{x!} \, du \\[8pt] = {} & \frac{e^{-\lambda}}{1-e^{-\lambda}}\int_0^\lambda \frac{1}{u}(e^u - 1 ) \, du \\[8pt] \end{align}
When integrated indefinitely, $\frac{e^u}{u} = Ei(u) + k$ (according to WolframAlpha). However, with the integration bounds, it doesnt give me an answer. How do I proceed?