• Support PF! Buy your school textbooks, materials and every day products Here!

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

  • Thread starter irycio
  • Start date
  • #1
97
1

Homework Statement



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



Homework Equations



Well... :P

The Attempt at a Solution



Since integrating is linear, we can write [tex] \int_{-1}^1 \frac{1}{x} dx = \int_{-1}^0 \frac{dx}{x} + \int_0^1 \frac{dx}{x} [/tex]. Since first integral is [tex]-\infty[/tex] and the 2nd is [tex]\infty [/tex], their sum is not known.


But I don't like this prove. Anyone can come up with a better one?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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
[itex]\displaytype\lim_{\epsilon\to 0}\int_{-1}^{-\epsilon}\frac{1}{x}dx+ \lim_{\delta\to 0}\int_{\delta}^1\frac{1}{x}dx[/itex]

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",
[tex]\lim_{\epsilon\to 0}\left(\int_{-1}^{-\epsilon} \frac{1}{x}dx+ \int_\epsilon^1 \frac{1}{x}dx\right)[/tex]
which is, in fact, 0.)
 
  • #3
97
1
Well, since anti-derivative of [tex] \frac{1}{x} [/tex] is Log|x|, we eventually get [tex] lim_{\epsilon -> 0} Log|-\epsilon| - lim_{\delta -> 0} Log|delta| [/tex], which again is [tex] \infty - \infty [/tex]. 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?
 
Last edited:
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
955
Well, since anti-derivative of [tex] \frac{1}{x} [/tex] is Log|x|, we eventually get [tex] lim_{\epsilon -> 0} Log|-\epsilon| - lim_{\delta -> 0} Log|\delta| [/tex], which again is [tex] \infty - \infty [/tex]. 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
[tex]\lim_{\epsilon\to 0}\left( ln|\epsilon|- ln|\epsilon|\right)= \lim_{\epsilon\to 0} 0= 0[/tex]

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".
 
  • #5
97
1
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?
 

Related Threads on Proof that int_(-1)^1 1/x != 0

Replies
13
Views
1K
Replies
7
Views
1K
Replies
8
Views
2K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
1
Views
750
Replies
6
Views
1K
  • Last Post
Replies
1
Views
4K
Top