Proof of existence of the Laplace distribution's first moment

Question

Given the Laplace distribution's PDF:

$$f_X(x|\mu,\sigma) = \dfrac{1}{2 \sigma}\,e^{-\frac{|x-\mu |}\sigma}$$

I'm trying to compute $\mathbb{E}[\lvert X \vert]$, or at the very least to show that it's finite.

$$\mathbb{E}[\lvert X \vert] = \frac{1}{2 \sigma} \int_{\mathbb{R}} \lvert x \rvert e^{-\frac{|x-\mu |}\sigma} dx = \ ?$$

What integration tactics should I use?

Answer 1

WLOG, assume $\mu>0$. Then we see that \begin{align} \int^\infty_{-\infty}|x|e^{-\frac{|x-\mu|}{\sigma}}\ dx =&\ \int^\infty_\mu xe^{-\frac{x-\mu}{\sigma}}\ dx+\int^\mu_0 xe^{-\frac{\mu-x}{\sigma}}\ dx-\int^0_{-\infty}x e^{-\frac{\mu-x}{\sigma}}\ dx \\ =&\ I_1 + I_2 +I_3. \end{align} Observe \begin{align} I_1 = \sigma \int^\infty_0 (\sigma y+\mu)e^{-y}\ dy = \sigma(\sigma+\mu) \end{align} and \begin{align} I_2 = \sigma\int^{\mu/\sigma}_0 (\mu-\sigma y)e^{-y}\ dy = \sigma^2\left(e^{-\mu/\sigma}-1 \right)+\mu\sigma \end{align} and lastly \begin{align} I_3 = \sigma^2e^{-\mu/\sigma}. \end{align} Thus, we have \begin{align} \mathbb{E}[|X|] = \frac{2\mu\sigma +2\sigma^2e^{-\mu/\sigma}}{2\sigma} = \mu + \sigma e^{-\mu/\sigma}. \end{align}