I would like to simplify the following sum if possible: $$ e^{-x} \sum_{k=1}^\infty \frac{x^k}{k!} \sum_{j=1}^k \frac{\lambda^j}{j} $$ where $x \geq 0$ and $\lambda \in[0, 1]$.
When $\lambda = 1$ the inner sum is just the $k$th harmonic number, $H_k$, so the sum is known from the exponential generating function for the harmonic series: $$ e^{-x}\sum_{k=1}^\infty \frac{x^k}{k!} H_k = \operatorname{Ein}(x) = \Gamma(0,x) + \ln(x) + \gamma, $$ where $\gamma$ is Euler's constant and $\Gamma(\cdot)$ is the incomplete gamma function. And the case $\lambda = 0$ is, of course, easy.
But what about when $0 < \lambda < 1$? Any help appreciated.
(The sum arises in a problem involving the order statistics of a Poisson-distributed number of iid exponentially distributed random variables.)