Proof that int_(-1)^1 1/x = 0

[irycio]

Homework Statement

I'd like to prove the inexistence of $$\int_{-1}^1 \frac{1}{x} dx$$, or at least that it's not 0.

Homework Equations

Well... :P

The Attempt at a Solution

Since integrating is linear, we can write $$\int_{-1}^1 \frac{1}{x} dx = \int_{-1}^0 \frac{dx}{x} + \int_0^1 \frac{dx}{x}$$. Since first integral is $$-\infty$$ and the 2nd is $$\infty$$, their sum is not known.But I don't like this prove. Anyone can come up with a better one?

 

[HallsofIvy]

Especially since you did not prove that each of those two integrals is infinite!
Nor is "infinity- infinity" (in the limit sense) non-existent. Simply saying it "is not known" does not mean it does not exist.

The basic definition of such an improper integral is
$\displaytype\lim_{\epsilon\to 0}\int_{-1}^{-\epsilon}\frac{1}{x}dx+ \lim_{\delta\to 0}\int_{\delta}^1\frac{1}{x}dx$

You can take the anti-derivative of 1/x, put it into each of those and see what happens to the limits.

(This is NOT the same as the "Cauchy Principle value",
$$\lim_{\epsilon\to 0}\left(\int_{-1}^{-\epsilon} \frac{1}{x}dx+ \int_\epsilon^1 \frac{1}{x}dx\right)$$
which is, in fact, 0.)

 

[irycio]

Well, since anti-derivative of $$\frac{1}{x}$$ is Log|x|, we eventually get $$lim_{\epsilon -> 0} Log|-\epsilon| - lim_{\delta -> 0} Log|delta|$$, which again is $$\infty - \infty$$. Basically, I can't see much difference between those 2 limits you wrote (one with epsilon and delta and one for Cauchy's principle value).E:Oh, I see now. The latter one, for Cauchy, is 0 as it is a limit of a difference of, say, a-a, whereas the first one is a difference of left-side limit and right-side limit. Or not?

 

[HallsofIvy]

Well, since anti-derivative of $$\frac{1}{x}$$ is Log|x|, we eventually get $$lim_{\epsilon -> 0} Log|-\epsilon| - lim_{\delta -> 0} Log|\delta|$$, which again is $$\infty - \infty$$. Basically, I can't see much difference between those 2 limits you wrote (one with epsilon and delta and one for Cauchy's principle value).


E:Oh, I see now. The latter one, for Cauchy, is 0 as it is a limit of a difference of, say, a-a, whereas the first one is a difference of left-side limit and right-side limit. Or not?

Yes, that is true. For the Cauchy principal value, you have
$$\lim_{\epsilon\to 0}\left( ln|\epsilon|- ln|\epsilon|\right)= \lim_{\epsilon\to 0} 0= 0$$

But for the pther, since the two one-sided limits do not exist, their difference does not exist. (Which is NOT the same as saying "infinity - infinity does not exist".

 

[irycio]

Whoa, whoa, whoa, wait ;). The left-sided limit definitely doesn't exist, since in Reals there is no logarithm of negative number. But why would the right-sided limit not exist? I mean, infinity is a kind of a limit, isn't it?