How does one deal with improper integrals with absolute value bars?
I need to show that a given function is a density function, which means I need to show that $\int_{-\infty}^\infty p(|x|) dx = 1$ yet what do to with those bars?
The function is of the form $$p(x) = \frac{1}{2}e^{-|x|}$$
When you don't have any other insight as to what to do with absolute values, you revert to the definition and consider those cases. As $|x| = x$ when $x$ is positive and $-x$ when negative, your integral can be written as
$$\int_{-\infty}^\infty \frac{1}{2}e^{-|x|} dx = \frac{1}{2} \int_{-\infty}^0 e^x \ dx + \frac{1}{2} \int_0^\infty e^{-x} \ dx$$
Use the definition of the absolute value: $$|x| = \cases{x & $x \ge 0$ \\ -x & else}$$ So $$\int_{-\infty}^\infty p(|x|) \;\mathrm dx = \int_{-\infty}^0 p(-x)\;\mathrm dx + \int_0^\infty p(x)\;\mathrm dx = 2\int_0^\infty p(x)\;\mathrm dx$$
Split the integral into a sum of two integrals, one from 0 to infinity where $|x|=x$, and one from -infinity to 0 where $|x|=-x$.
