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[tex] \int^{∞}_{-∞}\frac{1}{x^{2}}dx [/tex]

Also, does the function being even help us at all?

BiP

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- Thread starter Bipolarity
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- #1

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[tex] \int^{∞}_{-∞}\frac{1}{x^{2}}dx [/tex]

Also, does the function being even help us at all?

BiP

- #2

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We have to split it up into two integrals (can you see why? What happens at x=0?), giving us...

[tex]\int_{a}^{b} \frac{1}{x^2} + \int_{c}^{d} \frac{1}{x^2}[/tex]

Now integrate and take a, b, c, and d to the appropriate limits and see what happens.

- #3

jedishrfu

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seems it would be undefined at zero and hence not be evaluable.

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The function being integrated is not even continuos on the domain of integration...

- #5

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The function being integrated is not even continuos on the domain of integration...

So is it computable or not? Can't we consider it an improper integral of type II?

BiP

- #6

chiro

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So is it computable or not? Can't we consider it an improper integral of type II?

BiP

If you doing a Riemann integration, you need this to be Riemann-Integrable let alone continuous (which is only one condition of being Riemann-Integrable).

- #7

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If you doing a Riemann integration, you need this to be Riemann-Integrable let alone continuous (which is only one condition of being Riemann-Integrable).

According to http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx

the integrals can be computed even if there is a discontinuity in the integrand provided that each of the partitioned integrals converges.

BiP

- #8

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According to http://tutorial.math.lamar.edu/Classes/CalcII/ImproperIntegrals.aspx

the integrals can be computed even if there is a discontinuity in the integrand provided that each of the partitioned integrals converges.

BiP

Do they converge? Have you tried computing the limits?

- #9

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Can you explain why a function has to be continuous on the domain of integraion?The function being integrated is not even continuos on the domain of integration...

- #10

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This is a boundless integral. What did you think of when you read that? What do you think of when you're dealing with infinity in calculus??

Anyway, here's what your first step should look like to see if this improper integral diverges or converges.

[tex]{\int_{-\infty}^{+\infty}\frac{1}{x^{2}} dx= \lim_{a\to+\infty} \lim_{b\to-\infty}\left. \frac{-1}{x}\right|_{b}^{a}}[/tex].

I think you can solve it from there.

OOPSSS I messed up, I forgot to think about x=0 (was too focus on making my latex look cool :/). Anyway, the questions I asked should help develop some more intuition on improper integrals.

Anyway, here's what your first step should look like to see if this improper integral diverges or converges.

[tex]{\int_{-\infty}^{+\infty}\frac{1}{x^{2}} dx= \lim_{a\to+\infty} \lim_{b\to-\infty}\left. \frac{-1}{x}\right|_{b}^{a}}[/tex].

I think you can solve it from there.

OOPSSS I messed up, I forgot to think about x=0 (was too focus on making my latex look cool :/). Anyway, the questions I asked should help develop some more intuition on improper integrals.

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- #12

jgens

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I don't know if dextercioby was referring to the more general spaces like L^2 and measures/integrals on those spaces as opposed to the much less general spaces.

A function does not have to be continuous on its domain to be Riemann integrable.

- #13

chiro

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A function does not have to be continuous on its domain to be Riemann integrable.

http://en.wikipedia.org/wiki/Riemann_integral#Integrability

A function on a compact interval [a,b] is Riemann integrable if and only if it is bounded and continuous almost everywhere (the set of its points of discontinuity has measure zero, in the sense of Lebesgue measure). This is known as the Lebesgue integrability condition or Lebesgue's criterion for Riemann integrability or the Riemann—Lebesgue theorem.[2] Note that this should not be confused with the notion of the Lebesgue integral of a function existing; the result is due to Lebesgue, and uses the notion of measure zero, but does not refer to or use Lebesgue measure more generally, or the Lebesgue integral.

Unless you are talking some really exotic functions, the almost everywhere criteria will not come into play and the normal idea of continuity will be still required.

- #14

jgens

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Unless you are talking some really exotic functions, the almost everywhere criteria will not come into play and the normal idea of continuity will be still required.

The point is that continuity is quite a bit stronger than Riemann integrability. That the integral in question fails to converge is not due to a lack of continuity (since the integrand is continuous almost everywhere), but rather due to its unboundedness.

- #15

chiro

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The point is that continuity is quite a bit stronger than Riemann integrability. That the integral in question fails to converge is not due to a lack of continuity (since the integrand is continuous almost everywhere), but rather due to its unboundedness.

You pointed out something about continuity and I replied with that statement. This reply was a general reply about Riemann Integrability, not a reply specific to the statement of the OP.

- #16

HallsofIvy

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A function does not have to be continuous on its domain to be Riemann integrable.

No, it doesn't. The not at all "exotic" function f(x)= 1 for [itex]0\le x< 1[/itex], f(x) [itex]1\le x\le 2[/itex] is integrable from 0 to 2 even though it is not continuous at x= 0. The "almost everywhere" certainly does come into play- the function is discontinuous at a single point which has measure 0 and so is continuous "almost everywhere".http://en.wikipedia.org/wiki/Riemann_integral#Integrability

Unless you are talking some really exotic functions, the almost everywhere criteria will not come into play and the normal idea of continuity will be still required.

- #17

chiro

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No, it doesn't. The not at all "exotic" function f(x)= 1 for [itex]0\le x< 1[/itex], f(x) [itex]1\le x\le 2[/itex] is integrable from 0 to 2 even though it is not continuous at x= 0. The "almost everywhere" certainly does come into play- the function is discontinuous at a single point which has measure 0 and so is continuous "almost everywhere".

This is a really trivial example (having the end-points not included or having a discontinuity at an endpoint is something we should all know how to handle). Same reason why Integral over all these regions [a,b], (a,b], [a,b), (a,b) are equal if the measure is the standard infinitesimal one and the function has the expected properties.

This has more to do with the consideration and clarification of how different intervals can be considered to be equal so that you can focus on the mapping that has all the right properties. It's like the whole 0.9999 (repeated forever) = 1 scenario, and we just use the trick that since the intervals are the same, we can use this to remove these "discontinuities" since it doesn't change the value of the final integral.

The really tough examples include discontinuities everywhere in any interval or having functions like f(x) = 1 if x irrational, 0 otherwise.

- #18

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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.

- #19

HallsofIvy

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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.This is a really trivial example (having the end-points not included or having a discontinuity at an endpoint is something we should all know how to handle). Same reason why Integral over all these regions [a,b], (a,b], [a,b), (a,b) are equal if the measure is the standard infinitesimal one and the function has the expected properties.

This has more to do with the consideration and clarification of how different intervals can be considered to be equal so that you can focus on the mapping that has all the right properties. It's like the whole 0.9999 (repeated forever) = 1 scenario, and we just use the trick that since the intervals are the same, we can use this to remove these "discontinuities" since it doesn't change the value of the final integral.

The really tough examples include discontinuities everywhere in any interval or having functions like f(x) = 1 if x irrational, 0 otherwise.

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This is a boundless integral. What did you think of when you read that? What do you think of when you're dealing with infinity in calculus??

Anyway, here's what your first step should look like to see if this improper integral diverges or converges.

[tex]{\int_{-\infty}^{+\infty}\frac{1}{x^{2}} dx= \lim_{a\to+\infty} \lim_{b\to-\infty}\left. \frac{-1}{x}\right|_{b}^{a}}[/tex].

I think you can solve it from there.

OOPSSS I messed up, I forgot to think about x=0 (was too focus on making my latex look cool :/). Anyway, the questions I asked should help develop some more intuition on improper integrals.

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!

- #21

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except for that one useful answer by romsofia:

I recall pointing him in the exact same direction in the very first response:tongue2:

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I recall pointing him in the exact same direction in the very first response:tongue2:

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

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

You are forgiven. I pardon you.

*removes from death list*

- #24

chiro

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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.

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.

- #25

pwsnafu

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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.

No, a continuous function is Riemann integrable over

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