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.

Maybe this question is a stupid one but, let me ask it here just to be sure. :) We know that under continuity of function $f$ on an interval $[a,b]$ wherein $a<b$: $$\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$$ Now,

Does this equality remain valid if we replace $a$ and $b$ with $-\infty$ and $+\infty$ respectively?

I just want to be sure that, if the definition of definite integral can be extended for infinity (as for upper and lower limits). Thanks

share|improve this question
    
Please do not use titles that are entirely in $\LaTeX$. –  J. M. Jul 30 '12 at 15:14
add comment

2 Answers

up vote 3 down vote accepted

Yes, but that is a definition of what it means to integrate 'backwards' along a (finite) interval. But a consequence of this definition and the definition of integration over an infinite interval is that the result does hold.

Pardon my disgusting abuse of notation:

$$\displaystyle \int_{-\infty}^{\infty} = \lim_{a \to -\infty} \lim_{b \to \infty} \int_a^b = \lim_{a \to -\infty} \lim_{b \to \infty} -\int_b^a = -\lim_{a \to -\infty} \lim_{b \to \infty} \int_b^a = -\int_{\infty}^{-\infty}$$

share|improve this answer
    
Thanks. We always do like you did with notations. :) –  B. S. Jul 30 '12 at 14:57
add comment

Even when $a$ and $b$ are finite, this formula is a pure convention. Since the improper integral is just a limit of proper integrals, you can extend this convention, provided that $f$ is integrable in the generalized sense.

share|improve this answer
    
Thank you very much. –  B. S. Jul 30 '12 at 14:55
add comment

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.