I am trying to reconstruct this probabilistic model, \begin{equation} \begin{split} \frac{1}{\mu}\int^{\infty}_{0}P(N \geq n\, |\, L=l, T=t)\,e^{-\frac{l}{\mu}} dl &= \frac{1}{(n-1)!\mu}\int^{\infty}_{0} \left\{ \int^{tl}_{0}e^{-y}y^{n-1}dy \right\}e^{-\frac{l}{\mu}}dl \\ &= \frac{1}{(n-1)!}\int^{\infty}_{0} e^{-\{1+1/(t\mu)\}y}y^{n-1}dy \\ &= \frac{1}{(1+1/(t\mu))^{n}} = \left(\frac{t\mu}{1+t\mu}\right)^{n} \end{split} \end{equation} We know that $N$ follows a Poisson distribution. Such that, \begin{equation} P(N \geq n) = \frac{1}{(n-1)!}\int^{\lambda}_{0}e^{-x}x^{n-1}dx. \end{equation} and also that $L$ is gamma distribution parameter.
I know how to get (using integration by parts), \begin{equation} \frac{1}{(n-1)!}\int^{\infty}_{0} e^{-\{1+1/(t\mu)\}y}y^{n-1}dy = \frac{1}{(1+1/(t\mu))^{n}} = \left(\frac{t\mu}{1+t\mu}\right)^{n} \end{equation} But I have no idea how to get: \begin{equation} \frac{1}{(n-1)!\mu}\int^{\infty}_{0} \left\{ \int^{tl}_{0}e^{-y}y^{n-1}dy \right\}e^{-\frac{l}{\mu}}dl = \frac{1}{(n-1)!}\int^{\infty}_{0} e^{-\{1+1/(t\mu)\}y}y^{n-1}dy \end{equation} I have tried different methods of approach, by parts, substitution, but to no avail. Any help with this would be greatly appreciated!
The left hand side of your expression is the Laplace transform of an integral from 0 to $tl$.
$\begin{align} & \int_{0}^{\infty }{\left\{ \int_{0}^{tl}{{{e}^{-y}}}{{y}^{n-1}}dy \right\}}{{e}^{-\frac{l}{\mu }}}dl= \int_{0}^{\infty }{{{e}^{-\frac{l}{\mu }}}}\left\{ \int_{0}^{l}{{{e}^{-tp}}}{{(tp)}^{n-1}}tdp \right\}dl =\mu \mathfrak{L}(f)\left( \frac{1}{\mu } \right) \end{align}$
where $\mathfrak{L}(f)$ is the Laplace transform of $f(y)={{e}^{-ty}}{{(ty)}^{n-1}}t$ and we have used the result that $\mathfrak{L}\left( \int_{0}^{l}{f(y)dy} \right)(s)=\frac{\mathfrak{L}\left( f \right)(s)}{s}$ for s = $\frac{1}{\mu}$.
Then
$\mathfrak{L}\left( f \right)\left( \frac{1}{\mu } \right)=\int_{0}^{\infty }{{{e}^{-\tfrac{y}{\mu }}}{{e}^{-ty}}{{(ty)}^{n-1}}tdy=}\int_{0}^{\infty }{{{e}^{-\tfrac{p}{t\mu }}}{{e}^{-p}}{{(p)}^{n-1}}dp=}\int_{0}^{\infty }{{{e}^{-\left( 1+\tfrac{1}{t\mu } \right)p}}{{p}^{n-1}}dp}$
and replace p with y to recover the right hand side of your integral equality.