Integral\[\int_{-\infty}^{\infty}\frac{dx}x\]

- TuringTest

Integral\[\int_{-\infty}^{\infty}\frac{dx}x\]

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- anonymous

0

- TuringTest

incorrect

- anonymous

This integral does not converge

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## More answers

- anonymous

indeterminate?

- lgbasallote

int dx/x = lnx
ln( inf) - ln (-inf)
inf - inf...l'hospital!!!

- TuringTest

correct, but why is it not zero?
can we make it zero?

- anonymous

ln(-infinity) isn't defined

- TuringTest

yes, but that's not the main problem

- TuringTest

that can't be fixed
there is a way to fix this integral to make it give zero, as it intuitively should

- TuringTest

perhaps "fix" is the wrong word
but there is something that can be done to get the logical answer of 0

- anonymous

why isn't this zero? the integral is area under the curve of (1/x) from -infty to infty, it must be zero?

- TuringTest

but we have a singularity at x=0

- Zarkon

it is not...the integral does not converge for several reasons

- TuringTest

Yes Zarkon, you taught me about this, I'm trying to see how others take it

- Zarkon

ic

- anonymous

what about the logical, intuitive way you were talking about? it was a trap?

- TuringTest

the 'intuition' I would have (before Zarkon enlightened me) is that because 1/x is odd, this is zero
but the singularity, amongst other reasons, prevents this trick

- anonymous

okay, this might be a little stupid or a lot stupid to ask but why is \[\int _{\infty} ^{\infty} \frac1x = 0\]

- TuringTest

\[\int_{-a}^{a}f(x)dx=0\]if f(x) is odd, so one may come to the conclusion that\[\int_{-\infty}^{\infty}\frac{dx}x=0\]as I did in an earlier problem

- anonymous

I think this must be the case if f(x) is even, maybe
\[\int_{-a}^{a} f'(x) dx = [f(x)]_{-a}^{a} = f(a) - f(-a)= 0\]

- anonymous

i shouldn't have answered zero as the answer, something is wrong with my head... now when i think zero doesn't even make sense

- TuringTest

\[\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx\]if f(x) is even

- TuringTest

and yeah, I realized that zero makes no sense too after a couple minutes as well
hence I find this whole dilemma interesting

- anonymous

oh yeah, thanks...

- TuringTest

do you want me to tell you how physicists get away with saying that this integral is 0 ?
or do you want to investigate it yourself?

- anonymous

i don't have a clue how physicists do it, it'd be better you tell me

- TuringTest

as Zarkon said, the integral really doesn't converge (I'm going to say just because of the singularity at x=0) so there is a trick to avoid this called the Cauchy principle value (CPV) that sort of circumvents the point x=0 evenly in both directions from x=0
i.e. they split the integral and make what would be the middle term x=0
into x=-a and x=a respectively
http://en.wikipedia.org/wiki/Cauchy_principal_value

- TuringTest

I think this is a really good thing to know about improper integrals, and I just learned about it, so I wanted to bring it up here :)

- TuringTest

physicists sometimes use the CPV without stating it, so that is why I mentioned them

- anonymous

thanks, really nice of you to do so... i think some threads on openstudy must be made resource threads or wiki threads maybe (like this one)

- TuringTest

lol, well it's got the wiki link on it!
but thanks, fun discussion :)

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