Can this integral be computed?

Bipolarity

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.

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

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

BiP

Number Nine

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...
[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.

jedishrfu

the general solution is: integral(x^-2) = -1 / x

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

dextercioby

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

Bipolarity

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

BiP

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).

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

Number Nine

Do they converge? Have you tried computing the limits?

oay

Can you explain why a function has to be continuous on the domain of integraion?

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.

romsofia

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.

jgens

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

chiro

http://en.wikipedia.org/wiki/Riemann...#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.

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.

chiro

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.

HallsofIvy

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".

chiro

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.