I'm trying to calculate the cumulative function of a Laplace distribution whose PDF is
$f_X(x)=\frac{\lambda}{2}e^{-\lambda|x|},\forall x \in \mathbb{R}$
I proved that, if $x<0$:
$\int_{-\infty}^{x}\frac{\lambda}{2}e^{-\lambda|x|}dx=[\frac{\lambda}{2}e^{\lambda x}]_{-\infty}^{x} =\frac{\lambda}{2}e^{\lambda x}$
But I have difficulties in the other case. I obtained, if $x\geq0$:
$\mathbb{P}(X\geq 0)=\mathbb{P}(X=0 \bigcup X>0)=\mathbb{P}(X=0)+\mathbb{P}(X>0)=\frac{\lambda}{2}+\int_{0}^{x}\frac{\lambda}{2}e^{-\lambda|x|}dx=\frac{\lambda}{2}+[\frac{\lambda}{2}e^{-\lambda x}]_{0}^{x}=\frac{\lambda}{2}e^{-\lambda x}$
but I should have gotten $1-\frac{\lambda}{2}e^{-\lambda x}$. Where i wrong?
Thanks for any help!
For $x < 0$,
$$ \int_{-\infty}^{x}{\rm d}t~ \frac{\lambda}{2}e^{-\lambda |t|} = \frac{e^{\lambda x}}{2} \tag{1} $$
For $x \ge 0$
\begin{eqnarray} \int_{-\infty}^{x}{\rm d}t ~ \frac{\lambda}{2}e^{-\lambda |t|} &=& \color{blue}{\int_{-\infty}^{0}{\rm d}t ~ \frac{\lambda}{2}e^{-\lambda |t|}} + \color{red}{\int_{0}^{x}{\rm d}t ~ \frac{\lambda}{2}e^{-\lambda |t|}} \\ &=& \color{blue}{\frac{1}{2}e^{\lambda\cdot 0}} + \color{red}{\frac{1}{2} - \frac{1}{2}e^{-\lambda x}} \\ &=& 1 - \frac{1}{2}e^{-\lambda x} \tag{2} \end{eqnarray}