Question

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

Answer 1

0

Answer 2

incorrect

Answer 3

This integral does not converge

Answer 4

indeterminate?

Answer 5

int dx/x = lnx

ln( inf) - ln (-inf)

inf - inf...l'hospital!!!

Answer 6

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

Answer 7

ln(-infinity) isn't defined

Answer 8

yes, but that's not the main problem

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

but we have a singularity at x=0

Answer 13

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

Answer 14

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

Answer 15

ic

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

oh yeah, thanks...

Answer 25

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?

Answer 26

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

Answer 27

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

Answer 28

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 :)

Answer 29

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

Answer 30

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)

Answer 31

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