Method of moments estimate: Laplace distribution 
Consider an i.i.d sample of $n$ random variables with density function $$f(x\mid\mu,\sigma) = \frac{1}{2\sigma}\cdot e^{-{|x-\mu|}/{\sigma}}$$
  for $-\infty <  x < \infty$. Use the method of moments to find estimates of $\mu$  and $\sigma$.

Im stuck at the evaluation of $E[X]$ and $E[X^2]$. Can someone help me with this . Thanks

$$E[X] = \frac{1}{2\sigma} \int x e^{-|x - \mu|/\sigma} \, dx \stackrel{(1)}{=} \frac{1}{2\sigma} \int\sigma (\sigma t + \mu)e^{-|t|} \, dt \stackrel{(2)}{=} \frac{\mu}{2} \int e^{-|t|} \, dt \stackrel{(3)}{=} \mu \int_0^\infty e^{-t} \, dt = \mu$$
Here, (1) follows from a substitution and (2) resp. (3) follow from symmetry ($t \mapsto t e^{-|t|}$ is odd and $t \mapsto e^{-|t|}$ is even).
A similar calculation shows $E[X^2] = 2\sigma^2 + \mu^2$. This distribution is called Laplace distribution.
