What's wrong with$$\int_{-1}^1\frac{dx}{x}=\ln|1|-\ln|-1|=0$$?

Question

anonymous · Answer

Yes, but I'm told that this is incorrect.

callisto · Answer

I guess you have to 'separate' the integral into 2 parts (from -1->0 and 0->1) and use lim to find the area?!
It seems i've encountered some problem like this and you have to use limit to do it..

anonymous · Answer

Gotcha.

anonymous · Answer

Because you're integrating $\ln x$ in a region underfined for the reals.

callisto · Answer

the term is something like 'improper integral'.. I'm not sure and i couldn't find that post.. sorry :(

callisto · Answer

Hmm the post is this http://openstudy.com/study#/updates/4f6ac4a6e4b014cf77c7e6e1 That's something similar to your problem

myininaya · Answer

Yes @callisto 's is right 
You have to break this up

myininaya · Answer

If both of the parts converge then find the sum of what both converge to 
and that is your answer

myininaya · Answer

If one part diverges then the whole thing diverges

myininaya · Answer

$$\lim_{a \rightarrow 0^-} \int\limits_{-1}^{a} \frac{1}{x} dx+\lim_{b \rightarrow 0^+} \int\limits_{b}^{1}  \frac{1}{x} dx$$

myininaya · Answer

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