Notation Concerning Improper Integral and the Absolute Value of an Integrand I have the following integral
$$\int_{- \infty}^\infty e^{-|x|} dx$$
Since the preimages $x$ determine the the images $e^{-|x|}$ for nonnegative and negatives preimages, I believe the integral must be evaluated in the following manner
$$\lim \limits_{c \rightarrow 0} \enspace \int_{- \infty}^{c} e^{-|x|} \enspace dx \quad + \quad \int_{0}^{\infty} e^{- |x|} \enspace dx$$
where $c$ obviously approaches $0$ without necessarily equalling $0$, so that $e^{- |x|} = e{-(-x)}$. So,
$$\lim \limits_{c \rightarrow 0} \enspace \int_{- \infty}^{c} e^{-|x|} \enspace dx \quad + \quad \int_{0}^{\infty} e^{- |x|} \enspace dx $$
$$= \lim \limits_{c \rightarrow 0} \enspace \int_{- \infty}^{c} e^{x} \enspace dx \quad + \quad \int_{0}^{\infty} e^{-x} \enspace dx $$
$$= 2$$
Also, I think it is wrong to write 
$$\int_{- \infty}^{\infty} e^{-|x|} \enspace dx \quad + \quad \int_{0}^{\infty} e^{- |x|} \enspace dx$$
Since the first integral includes the value $0$, causing discrepancies with $e^{-|x|}$.
This is more of a notation issue and I would like some insight. Thank you.
 A: Actually, there is no need to take limits here. The function has no singularities where it blows up to $\infty$ for any real value of $x$.
Since $e^{-|x|}$ is an even function, we have
$$
\int_{-\infty}^{\infty}e^{-|x|}=2\int_0^{\infty}e^{-|x|}=2\int_0^{\infty}e^{-x}=2.
$$
A: Since your integrand is bounded everywhere (including the origin) the question whether to consider $c=0$ is not so much the point, there is no limit process needed here. The question is how to include $\infty$ as a bound. 
This is done ba defining $\int_a^\infty f(x) dx := \lim_{r\rightarrow \infty} \int_a^r f(x) dx$
And if you split an integral from $-\infty$ to $ \infty$ the you have to use the same point for splitting it, of course.
(For the so called Lebesgue integral all this is not an issue, only for the Riemann integral)
A: There is no need to be cautious with $x$ as $x \to 0$. If you want to use limits on this problem, I'd do it for the limits at infinity. You can also use the fact that the area under your curve is symmetric about the $y$ axis, leaving you to evaluate $$\lim_{N \to \infty} 2\int_{0}^{\infty} e^{-|x|}\text{d}x $$ Restricting your interval of integration to $[0,\infty)$ will also let you drop the absolute value out of the exponent.
A: Your approach is needlessly complicated.  You can proceed as follows:
\begin{align}
\int_{-\infty}^\infty e^{-|x|}\,dx & = \int_{-\infty}^0  e^{-|x|} \, dx + \int_0^\infty e^{-|x|} \, dx \\[10pt]
& = \int_{-\infty}^0 e^{-(-x)} \, dx + \int_0^\infty e^{-x}\,dx \\[10pt]
& = \int_{-\infty}^0 e^x \, dx + \int_0^\infty e^{-x}\,dx
\end{align}
Then take antiderivatives, etc.
A: Notice, the given function $$f(x)=e^{-|x|}\implies f(-x)=e^{-|-x|}=e^{-|x|}=f(x)$$ hence $f(x)=e^{-|x|}$ is an even function
Now, using property of definite integral $\color{blue}{\int_{-a}^{a}f(x)dx=2\int_{0}^{\infty}f(x)dx}$, we get
$$\int_{-\infty}^{\infty}e^{-|x|}dx=2\int_{0}^{\infty}e^{-|x|}dx$$
Since, $|x|=x\ \forall \ \ 0\leq x$ hence 
$$2\int_{0}^{\infty}e^{-|x|}dx=2\int_{0}^{\infty}e^{-x}dx=2[-e^{-x}]_{0}^{\infty}$$ $$=2[-e^{-\infty}+e^{0}]=2[0+1]=2[1]=\color{red}{2}$$
