The Laplace functional of the Poisson Point Process (PPP) $\Phi$ with intensity measure $\Lambda$ on $\mathbb{R}^d$ for non-negative function $f(x)$ is: $$ \mathcal{L}_\Phi(f) = \exp\bigg\{-\int_{\mathbb{R}^d} (1-e^{-f(x)})\Lambda(dx)\bigg\}.$$ The corresponding Laplace functional of an independently marked PPP $\tilde{\Phi}$ with intensity measure $\Lambda$ on $\mathbb{R}^d$ and marks with distributions $F_x(dm)$ on $\mathbb{R}^l$ for non-negative function $\tilde{f}(x,m)$ is: $$ \mathcal{L}_\tilde{\Phi}(\tilde{f}) = \exp\bigg\{-\int_{\mathbb{R}^d} \bigg(1-\int_{\mathbb{R}^l}e^{-\tilde{f}(x,m)F_x(dm)}\bigg)\Lambda(dx)\bigg\}.$$
I am a beginner in this field and have three simple questions.
1). What is the intuition of the expression $\int_{\mathbb{R}^l}e^{-\tilde{f}(x,m) F_x(dm)}$ in above equation that makes it different than PPP?
The intensity measure of independently marked PPP $\tilde{\Phi}$ is: $$ \tilde{\Lambda}(A \times K) = \int_A \tilde{p}(x,K) \Lambda(dx), \quad A \subset \mathbb{R}^d, \quad K \subset \mathbb{R}^l,$$ where $\tilde{p}(x,K) = \int_K F_x(dm)$.
2). What is the difference between $\int_K F_x(\color{red}{dm})$ and $\int_K \color{red}{d}F_x(\color{red}{m})$?
For non-negative random variable $X$ and corresponding CDF $F_X$, we have: $$E(X) = \int_0^\infty (1-F_X (x)) \, dx$$
3). Can we derive such relation for $\int_K F_x(dm)$?
Thanks in advance.