## Can this integral be computed?

Is it possible to compute the precise value of this integral? If not, can we at least show whether it converges or diverges? It seems that the integrand is undefined on both limits of integration.

$$\int^{∞}_{-∞}\frac{1}{x^{2}}dx$$

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

BiP
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Do you know how to handle improper integrals? We have to split it up into two integrals (can you see why? What happens at x=0?), giving us... $$\int_{a}^{b} \frac{1}{x^2} + \int_{c}^{d} \frac{1}{x^2}$$ Now integrate and take a, b, c, and d to the appropriate limits and see what happens.
 the general solution is: integral(x^-2) = -1 / x seems it would be undefined at zero and hence not be evaluable.

Blog Entries: 9
Recognitions:
Homework Help

## Can this integral be computed?

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

 Quote by dextercioby 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

 Quote by Bipolarity 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).

 Quote by chiro 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/Class...Integrals.aspx
the integrals can be computed even if there is a discontinuity in the integrand provided that each of the partitioned integrals converges.

BiP

 Quote by Bipolarity According to http://tutorial.math.lamar.edu/Class...Integrals.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?

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

 Quote by oay Can you explain why a function has to be continuous on the domain of integraion?
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.
 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. $${\int_{-\infty}^{+\infty}\frac{1}{x^{2}} dx= \lim_{a\to+\infty} \lim_{b\to-\infty}\left. \frac{-1}{x}\right|_{b}^{a}}$$. 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.

 Quote by chiro 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.

 Quote by jgens A function does not have to be continuous on its domain to be Riemann integrable.
http://en.wikipedia.org/wiki/Riemann...#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.

 Quote by chiro 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.

 Quote by jgens 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.

Recognitions:
Gold Member
No, it doesn't. The not at all "exotic" function f(x)= 1 for $0\le x< 1$, f(x) $1\le x\le 2$ 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".
 Quote by HallsofIvy No, it doesn't. The not at all "exotic" function f(x)= 1 for $0\le x< 1$, f(x) $1\le x\le 2$ 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".