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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?

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The fun part is you don't need to solve the integral in order to confirm it. It would be true for any function that has its variable under modulus. – SF. Jan 21 '13 at 10:10
Maybe if you sketched the integrand before trying calculus, a plan of attack might suggest itself. – Dilip Sarwate Jan 21 '13 at 13:13

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: $$f(-x) = f(x)$$ $$\int_{-\infty}^{\infty} f(x)dx =2\int_{0}^{\infty} f(x)dx$$

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Please write your integrals properly, the notation $\int f(x)$ is undefined and should be replaced by $\int f(x)dx$ or $\int f(t)dt$ or whichever you like. – Did Jan 21 '13 at 8:30

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:

enter image description here

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The punchline is of course that for even functions $f(x)$ we have $$\int_{-\infty}^{\infty} f(x) dx = \lim_{a \to \infty} \int_{-a}^a f(x) dx = \lim_{a \to \infty} 2 \cdot \int_0^a f(x) dx = 2 \int_0^{\infty} f(x) dx.$$ – JavaMan Jan 21 '13 at 8:43
@JavaMan: Indeed, your comment makes the answer potent. Thanks. – Babak S. Jan 21 '13 at 8:46
Nice combination: hints, graph, and comment! +1 – amWhy Feb 11 '13 at 0:06

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