How come $\int_{-\infty}^\infty \! e^{-|t|} \, \mathrm{d} t = 2\int_{0}^\infty \! e^{-|t|} \, \mathrm{d} t$?

Question

In a text book problem, $\int_{-\infty}^\infty \! e^{-|t|} \, \mathrm{d} t$ is said to equal $2\int_{0}^\infty \! e^{-|t|} \, \mathrm{d} t$. I cannot understand how this conclusion is reached.

I may be close to understanding it, but it seems weird and overly simplistic, so it feels like I've made a mistake. It may be because

$$\int_{-n}^n \! x^2 \, \mathrm{d} t = \frac{n^3}{3} - (-\frac{n^3}{3}) = 2\frac{n^3}{3} = 2\int_{0}^n \! x^2 \, \mathrm{d} t$$

but it feels weird and overly simple. Is that really it, or am I missing something crucial?

Answer 1

Much much simpler: $$\int_{-\infty}^{\infty} f(x)dx = \int_{0}^{\infty} f(x)dx + \int_{-\infty}^{0} f(x)dx = \int_{0}^{\infty} f(x)dx + \int_{0}^{\infty} f(-x)dx $$ Since in your case: $$f(-x) = f(x)$$ $$\int_{-\infty}^{\infty} f(x)dx =2\int_{0}^{\infty} f(x)dx$$

Answer 2

Not that the integrand, $\exp(-|t|)$ is an even contious function on $\mathbb R$. There is a fact that for an even function $f(x)$ we have; $$\int_{-a}^{a}f(x)dx=2\int_0^af(x)dx,a>0$$

Here is the plot of integrand: