Let $\alpha>0$ and $F:\mathbb{R}^2 \to \mathbb{R}$ the CDF with:

$F(x,y) = \left\{\begin{array}{ll} 0, & \min\{x,y\} < 0 \\
\int_{0}^{\infty}\left(\sum_{k=0}^{\lfloor x \rfloor}\mathrm{e}^{-\theta} \frac{\theta^k}{k!}\right)\left(\sum_{k=0}^{\lfloor y \rfloor}\mathrm{e}^{-\theta} \frac{\theta^k}{k!}\right)\alpha \mathrm{e}^{-\alpha \theta} \mathrm{d}\theta, & \mathrm{otherwise}\end{array}\right.$

Calculate the marginal distribution of $f_X$ and $f_Y$ as well as the CDF $F_X$ and $F_Y$. Additionally calculate the expected value vector as well as the covariance matrix.

So I started and tried to calculate the marginal distribution $f_X$ via:

$\int_{0}^{\infty} \limits \left(\sum_{k=0}^{\lfloor x \rfloor} \limits \mathrm{e}^{-\theta} \frac{\theta^k}{k!}\right)\alpha \mathrm{e}^{-\alpha \theta} \int_{0}^{\infty} \limits \left(\sum_{k=0}^{\lfloor y \rfloor} \limits \mathrm{e}^{-\theta} \frac{\theta^k}{k!} \ \mathrm{d}y \right) \mathrm{d}\theta$

My problem is to evaluate the integral: $\int_{0}^{\infty} \limits \left(\sum_{k=0}^{\lfloor y \rfloor} \limits \mathrm{e}^{-\theta} \frac{\theta^k}{k!} \ \mathrm{d}y \right)$ Does anyone know a trick or sees an error in my thinking, any help is appreciated.