Queremos calcular, para \(t \geq 0\) y \(\lambda > 0\), la integral doble: \[\begin{equation} F(t) = \int_{0}^{t} \int_{0}^{t - x_{2}} \lambda^2 e^{-\lambda (x_{1} + x_{2})} \, dx_{1} \, dx_{2} \end{equation}\]
Fijamos \(x_{2}\): \[\begin{equation} I_{1}(x_{2}) = \int_{0}^{t - x_{2}} \lambda^2 e^{-\lambda (x_{1} + x_{2})} \, dx_{1} \end{equation}\]
Factorizamos lo que no depende de \(x_{1}\): \[\begin{equation} I_{1}(x_{2}) = \lambda^2 e^{-\lambda x_{2}} \int_{0}^{t - x_{2}} e^{-\lambda x_{1}} \, dx_{1} \end{equation}\]
La integral es: \[\begin{equation} \int_{0}^{t - x_{2}} e^{-\lambda x_{1}} \, dx_{1} = \left[ -\frac{1}{\lambda} e^{-\lambda x_{1}} \right]_{0}^{t - x_{2}} = \frac{1}{\lambda} \left( 1 - e^{-\lambda (t - x_{2})} \right) \end{equation}\]
Por lo tanto: \[\begin{equation} I_{1}(x_{2}) = \lambda e^{-\lambda x_{2}} \left( 1 - e^{-\lambda (t - x_{2})} \right) \end{equation}\]
Ahora integramos: \[\begin{equation} F(t) = \int_{0}^{t} \lambda e^{-\lambda x_{2}} \left( 1 - e^{-\lambda (t - x_{2})} \right) dx_{2} \end{equation}\]
Separamos en dos integrales: \[\begin{equation} F(t) = \lambda \int_{0}^{t} e^{-\lambda x_{2}} dx_{2} - \lambda \int_{0}^{t} e^{-\lambda x_{2}} e^{-\lambda (t - x_{2})} dx_{2} \end{equation}\]
Notamos que: \[\begin{equation} e^{-\lambda x_{2}} e^{-\lambda (t - x_{2})} = e^{-\lambda t} \end{equation}\]
\[\begin{equation} \lambda \int_{0}^{t} e^{-\lambda x_{2}} dx_{2} = \lambda \left[ -\frac{1}{\lambda} e^{-\lambda x_{2}} \right]_{0}^{t} = 1 - e^{-\lambda t} \end{equation}\]
\[\begin{equation} \lambda \int_{0}^{t} e^{-\lambda t} dx_{2} = \lambda e^{-\lambda t} \cdot t \end{equation}\]
Por lo tanto: \[\begin{equation} F(t) = (1 - e^{-\lambda t}) - \lambda t e^{-\lambda t} = 1 - e^{-\lambda t}(1 + \lambda t) \end{equation}\]
\[\begin{equation} F(t) = 1 - e^{-\lambda t}(1 + \lambda t) \end{equation}\]