# Is $\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$ still valid for infinite limits?

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

-
Please do not use titles that are entirely in $\LaTeX$. –  Ｊ. Ｍ. Jul 30 '12 at 15:14

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:

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

-
Thanks. We always do like you did with notations. :) –  B. S. Jul 30 '12 at 14:57
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.