Other way to express $e^{|x|+|y|}$ I have a joint PDF with $e^{|x|+|y|}$. I know I can separate the function in two functions, $e^{|x|}$ and $e^{|y|}$. The limits for $x$ and $y$ are from $-\infty$ to $\infty$. Can I integrate from $0$ to $\infty$ and multiply the integral by $2$:
$$
2 \int_{0}^{\infty} e^x \,dx = \int_{-\infty}^{\infty} e^x \,dx \quad ?
$$
The complete problem is 
I have a pdf $f(x,Y)=(1/4)2 \exp[-1/2 (|x| + |y|)].$ the goal is to find the joint PDF f(W,Z) taking in consideration this $W=XY$ and $Z=Y/X$
 A: Use this fact.
$$\int_{\mathbb{R}^2} e^{-(|x| + |y|)}\, dx\, dy = \left(\int_{-\infty}^\infty e^{-|x|}\, dx\right)^ 2 =  4$$
A: The question in the title does not seem to match what is being insisted on in the comments.
Cleaning up the question, it seems that the joint pdf is
$$f_{X,Y}(x,y) = \left(\frac{1}{4}\right)^2 \exp\left(-\frac{|x|+|y|}{2}\right),
-\infty < x, y < \infty$$
which is the product of the two LaPlacian pdfs $\frac{1}{4}e^{-|x|/2}$ and 
$\frac{1}{4}e^{-|y|/2}$ of the independent random variables $X$ and $Y$.  The
OP desires to find the joint pdf of $W = XY$ and $Z = Y/X$.  
This is a standard problem in transformation of random variables
and the calculations involve Jacobians etc. However, if these
are not allowed, we can proceed from first principles as follows.
Suppose that $\alpha$ and $\beta$ are two positive numbers. Then,
$$F_{W,Z}(\alpha, \beta)= P\{W\leq \alpha, Z \leq \beta\}$$
can be found by integrating $f_{X,Y}(x,y)$ over two regions:


*

*the fourth quadrant and the region in the first quadrant
below the line $y = \beta x$ and the hyperbola $xy = \alpha$

*the second quadrant and the region in the third quadrant
above the line $y = \beta x$ and the hyperbola $xy = \alpha$
By the symmetry of the joint pdf, the integrals over the
two regions have the same
value, the integrals over the quadrants have value $\frac{1}{4}$ 
each, and the integral over the region in the first quadrant
below the line $y = \beta x$ and the hyperbola $xy = \alpha$
is just
$$\int_{y=0}^{\sqrt{\alpha\beta}} \int_{x=y/\beta}^{\alpha/ y}
\frac{1}{4} e^{-(x+y)/2}\,\mathrm dx\,\mathrm dy.$$
Note the absence of absolute value signs since we know we
are in the first quadrant and so $x$ and $y$ are positive.
One can evaluate this integral and then take the partial
derivative with respect to $\alpha$ and $\beta$ or
one can use the method of differentiating under the
integral sign (see for example, the comment following
this answer)
to get $f_{W,Z}(\alpha, \beta)$ without actually computing
the integral.
Finally, this gives the joint pdf in the first quadrant.
You need to figure out its value in other quadrants too,
and in each case, you can make the absolute value signs
go away by using known restrictions on $x$ and $y$ to
replace $|x|$ by $x$ or $-x$ etc as the case may be.
