Can this integral be computed?

oay

I think Ivy meant to show a trivial function which was not continuous at x=1 (which is what I was inferring earlier), but wrote it incorrectly.

Either way, continuity is not a necessity for integrability.

Is it just me, or does Ivy very often not check her posts before pressing the "Send" button"? I've seen a lot of itex problems recently.

If you're reading this, Ivy, please accept this as constructive! You are obviously very mathematically intelligent.

HallsofIvy

I have no idea what you are talking about. The whole point was your assertion earlier that integrable functions have to be continuous and then, later, that except for "exotic" functions, integrable functions have to be continous. This example shows that, unless you are going to call piecewise defined functions "exotic", neither of those is true. A function is Riemann integrable if its set of all discontinuities has measure 0 and the discontinuities are no worse than "jump" discontinuities. Any function with a finite (or even countable) number of jump discontinuities is integrable.

Unkraut

Interesting. A rather simple question and two pages of theoretically dubious digressions about mathematical theory which don't address the question, except for that one useful answer by romsofia:

Yeah, just to expand on this (taking care of x=0) you just try to calculate [tex]\int_{-\infty}^{0}\frac{1}{x^{2}} dx+\int_{0}^{+\infty}\frac{1}{x^{2}} dx=\left. \frac{-1}{x}\right|_{-\infty}^{0}+\left. \frac{-1}{x}\right|_{0}^{\infty}[/tex].

Regarding the statement in the OP that "it seems that the integrand is undefined on both limits of integration", that's wrong. The integrand is undefined at [tex]x=0[/tex] but the limits for x=±∞ are equal to zero, so they don't pose a problem. The problem is x=0 where the integrand is undefined. So you can't just integrate over any interval that contains 0 in the naive way that romsofia suggested (although he/she noticed his/her mistake; I'm just trying to clarify).
"Also, does the function being even help us at all?"
That just shows that [tex]\int_{-\infty}^{0}\frac{1}{x^{2}} dx=\int_{0}^{+\infty}\frac{1}{x^{2}} dx[/tex] (symmetry).

I'm always annoyed by smartass people adding more confusion to questions than there originally was. (If you're not a smartass I'm not addressing you. No offense meant.) I hope I'm not making the same mistake.

About Riemann integrability: Yeah, as others have noticed, chiro was talking ********. A continuous function is Riemann integrable. Continuity is a sufficient condition for Riemann integrability. [tex]\frac{1}{x^2}[/tex] is continuous on [tex]\mathbb{R}-{\{0\}}[/tex], so it's Riemann integrable on every compact interval which does not cross x=0. So you can integrate over such intervals in the "naive" way, (just substracting the indefinite integral at the boundary points). Blah!

Number Nine

I recall pointing him in the exact same direction in the very first response

Unkraut

Oh, my bad. I overlooked that. Sorry for my smug reply. May the force be with you.

Number Nine

You are forgiven. I pardon you.
*removes from death list*

chiro

I didn't say continuity wasn't a sufficient condition!

This whole idea of almost everywhere is just a trick that we use where we can equate an interval like [a,b] to (a,b] to get the same result. You just partition up the integral so that if you this jumps, you use this trick to get rid of them while still maintaining the value of the integral.

These aren't the exotic integrals I was talking about: the exotic ones are the ones where you can't just partition them up easily like you can with a function like y = 1 from -1 to 1 and then y = 2 outside the region where you want to integrate from x = -5 to x = 5.

A function like the above that HallsOfIvy calls "exotic" is not what I would call exotic: Brownian motion is exotic, not a nice function with a jump discontinuity or two.

pwsnafu

No, a continuous function is Riemann integrable over a closed and bounded subinterval of ℝ, the statement fails when discussing integrals over all of ℝ. Consider ##\int_{-\infty}^\infty x \, dx##. This is continuous over ℝ, integrable on every bounded subset of ℝ, but not over all of ℝ. It does however have a principal value, namely zero.