expected value of a $e^{-2|x|}$ Let $f(x)=e^{-2|x|}$

Find: 


$E(X)$
$E(|X|)$
$E(x')$ where $x'$ denotes the largest integer not greater than $x$.


I'm stuck on this question and am confused about how to use the modulus sign. I feel I need to show it separately for cases when $x\ge0$ and $x\le0$ but I'm not sure.

Any help is appreciated
By definition, 
$$E(X)=\int_{-\infty}^\infty xe^{-2|x|}\,dx.\tag{1}$$

To evaluate this integral is easy. First, a technical thin, the integral does converge. The function $xe^{-2|x|}$ is an odd function, so the integral is $0$.

We have
$$E(|X|)=\int_{-\infty}^\infty |x|e^{-2|x|}\,dx.\tag{2}$$
Now the integrand is symmetric about the $y$-axis, so we find the integral from $0$ to $\infty$, and double.

To find the integral from $0$ to $\infty$, we need
$$\int_{-\infty}^\infty xe^{-2x}\,dx.$$
We can use integration by parts, or recall that the exponential with parameter $2$ has mean $\frac{1}{2}$. That tells us that $\int_0^\infty (x)2e^{-2x}=\frac{1}{2}$.

The greatest integer one takes a bit more woek to analyze. The integral breaks up into parts. From $0$ to $1$ we are integrating $0\cdot e^{-2x}$. From $1$ to $2$ we are integrating $1\cdot e^{-2x}$, and so on. 

So we get 
$$(1)\frac{1}{2}(e^{-2}-e^{-4})+(2)\frac{1}{2}(e^{-4}-e^{-6})+(3)\frac{1}{2}(e^{-6}-e^{-8})+\cdots.$$
This is a geometric series, not hard to evaluate. 

However, we can be tricky, and do the negative part, hoping for cancellation. We get
$$(-1)\frac{1}{2}(e^{0}-e^{2})+(-2)\frac{1}{2}(e^{-2}-e^{-4})+(-3)\frac{1}{2}(e^{-4}-e^{-6})+\cdots.$$
Add the two, simplify.
Hint: 


  $$\operatorname{E}[X] = \int_{-\infty}^\infty x f(x)\, \mathrm{d}x,  $$


and more generally


  $$\operatorname{E}[g(X)] = \int_{-\infty}^\infty g(x) f(x)\, \mathrm{d}x. $$


To deal with integrals contain the floor function $\lfloor x\rfloor$ see here.