Question

integral of 1/x from -1 to 1...
I guess the answer should be zero... But I get that it is divergent...

Answer 1

The integral of 1/x is ln(x)+C since d/dx[ln(x)] = 1/x

Answer 2

Yes. But now how do you solve the asymptote. This is an improper integral.

Answer 3

i think the fact that ln(-1) doesn't produce a real number means that the integral will be divergent

Answer 4

But, are you not calculating the signed area from -1 to 1. The figure is symetric. Equal amounts of negative and positive area? Final answer should not be 0?

Answer 5

I see what you mean, but based on the FTC (fundamental theorem of calculus), we know that

\[\Large \int_{a}^b f(x)dx = F(b)-F(a)\]

where

\[\Large \frac{d}{dx}[F(x)] = f(x)\]

Answer 6

no. function has to be continuous. Function is not continuous.

Answer 7

I think the problem here is that the right area under the curve is infinitely large

Answer 8

So is the area on the left side

Answer 9

So you're effectively saying \[\Large \infty - \infty = 0\] but the expression \[\Large \infty - \infty\] is indeterminant

Answer 10

Hmm. I don't think so. Because both infinity values are exactly the same.

Answer 11

The reason why \[\Large \infty - \infty\] is indeterminant is because you can't say the two infinities are equal (like you can with a variable), so you don't know the value of \[\Large \infty - \infty\]

Answer 12

they aren't, infinity grows forever and ever, but there are different "sizes" to infinity

Answer 13

If you don't mind me asking, which grade/university level are you? I'm a basic n00b: grade 12.

Answer 14

I'm a college student, but that doesn't automatically make me right all the time. I could be wrong of course. However, I do feel I'm correct here.

Answer 15

\(\int 1/x \;dx = ln|x| + C\)

The -1 is not significant. The ONLY consideration is around x = 0.

Answer 16

I do know that \[\Large \infty - \infty\] is an indeterminant form

Answer 17

A function does not have to be continuous. It only has to converge.

Answer 18

Tkhunny, to what does that condition apply to? FTC?
How would you go by to solve the problem?

Answer 19

not sure if this confirms, but might as well post it

http://www.wolframalpha.com/input/?i=integral(1%2Fx%2Cx%3D-1..1)

that calculator says the integral diverges

Answer 20

1) It's symmetrical. Ignore x < 0.
2) Standard Operating Procedure

\(lim_{a\rightarrow 0}\left(\int\limits_{a}^{1}\dfrac{1}{x}\;dx\right) = lim_{a \rightarrow 0} \left(-ln(a)\right)\) Which limit doesn't exist.

Answer 21

\(\displaystyle\lim_{x\to b} f(x)\) <== LaTeX Practice

Answer 22

Of course, the Cauchy Principal Value integral is zero (0), but that's a different idea for another day.

Answer 23

Hmm, okay, I am still on the same problem. tkhunny, what happens then?

Answer 24

As soon as you determine that the limit fails to exist, you are done.