Help with Exponential Distribution Exercise: Struggling to Eliminate -1
$Let ( X_i \sim \text{Exp}(\theta_1) and\ ( X_j \sim \text{Exp}(\theta_2) ), determinate ( X_i + X_j = U ).$
$\int_0^u \theta_1 e^{-\theta_1*k} \cdot \theta_2e^{-\theta_2u+\theta_2k} dk $
$\theta_1\theta_2e^{-\theta_{2}u}\int_0^u e^{\theta_2k-\theta_1k}dk $
after of evaluate
$\theta_1\theta_2e^{-\theta_{2}u} \cdot \dfrac{1}{\theta_1\theta_2}(e^{(\theta_1-\theta_2)u}-{e^0}) $
$\theta_1\theta_2e^{-\theta_{2}u} \cdot \dfrac{1}{\theta_1\theta_2}(e^{(\theta_1-\theta_2)u}-{1}) $
The principal problem is that answer is:
$\theta_1\theta_2e^{-\theta_{2}u} \cdot \dfrac{e^{(\theta_1-\theta_2)u}}{\theta_1\theta_2} $
I need help, please
If $\theta _2>\theta_1$ (say) then for $s$ small enough $$E(e^{s U})=\frac{\theta_1}{\theta_1-s}\frac{\theta_2}{\theta_2-s}=\frac{1}{\theta_2-\theta_1}\left(\theta_2\frac{\theta_1}{\theta_1-s}-\theta_1\frac{\theta_2}{\theta_2-s}\right)$$$$=\frac{\theta_1\theta_2}{\theta_2-\theta_1}\int_{0}^{\infty}e^{su}\left(e^{-\theta_1 u}-e^{-\theta_1 u}\right)du$$ The density of $U$ is $$\frac{\theta_1\theta_2}{\theta_2-\theta_1}\times \left(e^{-\theta_1 u}-e^{-\theta_2 u}\right)$$