Answer
The integral converges and is equal to: $2$
Work Step by Step
First, we will compute the integral for the negative interval
$\int_{-\infty}^{\infty} e^{-|x|} \ dx= \int_{\infty}^0 e^x \ dx \\=\lim\limits_{R \to \infty} \int_{R}^0 e^x \ dx \\=\lim\limits_{R \to \infty} (1-e^{-R}) \\=1$
Next, we will compute the integral for the positive interval
$\int_{0}^{\infty} e^{-|x|} \ dx= \int_{0}^{\infty} e^{-x} \ dx \\=\lim\limits_{R \to \infty}\int_{0}^{\infty} e^{-x} \ dx \\=\lim\limits_{R \to \infty} (1-e^{-R}) \\=1$
Hence, the integral converges and yields the result: $\int_{-\infty}^{\infty} e^{-|x|} \ dx=1+1=2$
