## Calculus (3rd Edition)

The integral converges and is equal to: $2$
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: $\int_{-\infty}^{\infty} e^{-|x|} \ dx=1+1=2$