TITLE: Why is $\int_{-1}^{1} \frac{1}x \mathrm{d}x$ divergent?
QUESTION [8 upvotes]: Isn't
$$\int_{-1}^{1} \frac{1}x \mathrm{d}x=\lim_{\epsilon\to 0^{+}} \int_{-1}^{-\epsilon} \frac{1}x \mathrm{d}x+\int_{-\epsilon}^{\epsilon} \frac{1}x \mathrm{d}x+\int_{\epsilon}^{1} \frac{1}x \mathrm{d}x=0 ?$$

REPLY [20 votes]: First, $\frac 1 x$ isn't defined on $[-1,1]$ (because of what happens in $0$). You could get around this by considering it defined on $[-1,0) \cup (0,1]$, but then you've got another problem, much more serious: the function is unbounded, and the concept of "Riemann integral" is defined only for bounded functions (and bounded intervals).
Finally, you might try to use the concept of "improper integral of the second kind". This doesn't work, either: $\int \limits _{-1} ^1 \frac 1 x \ \Bbb d x = \int \limits _{-1} ^0 \frac 1 x \ \Bbb d x + \int \limits _0 ^1 \frac 1 x \ \Bbb d x = \infty - \infty$ which is indeterminate.
What you are trying to do is to give a meaning to that integral using the concept of "principal value", in the framework of distribution theory. But this is clearly not the same as saying that your function is integrable with integral $0$.