Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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

share|improve this answer
4  
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

share|improve this answer
2  
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. –  B. S. Jan 21 '13 at 8:46
1  
Nice combination: hints, graph, and comment! +1 –  amWhy Feb 11 '13 at 0:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.