Why is $\int_{-1}^{1} \frac{1}x \mathrm{d}x$ divergent? Isn't
$$\int_{-1}^{1} \frac{1}x \mathrm{d}x=\lim_{\epsilon\to 0^{+}} \int_{-1}^{-\epsilon} \frac{1}x \mathrm{d}x+\int_{-\epsilon}^{\epsilon} \frac{1}x \mathrm{d}x+\int_{\epsilon}^{1} \frac{1}x \mathrm{d}x=0 ?$$
 A: I presume your intention is that $\int_{-1}^{-\epsilon}\frac{1}{x}\,\mathrm{d}x = -\int_{\epsilon}^{1}\frac{1}{x}\,\mathrm{d}x$ by way of a simple substitution, and then the middle integral "should be zero since the integrand is odd".
Unfortunately, $\lim_{\epsilon\to0^{+}}\int_{-\epsilon}^{\epsilon}\frac{1}{x}\,\mathrm{d}x \neq 0$. The integrand is not defined at $0$, and $0$ is in the region of integration. This means that the integral is not defined either.
You can't really escape this problem. If you try to evaluate the integral as
\begin{align*}
\int_{-1}^{1}\frac{1}{x}\,\mathrm{d}x = \lim_{\epsilon\to 0^{-}}\int_{-1}^{\epsilon}\frac{1}{x}\,\mathrm{d}x + \lim_{\epsilon\to0^{+}}\int_{\epsilon}^{1}\frac{1}{x}\,\mathrm{d}x
\end{align*}
then you run into the problem that neither of your two new limits are convergent. For example,
$$\lim_{\epsilon\to0^{+}}\int_{\epsilon}^{1}\frac{1}{x}\,\mathrm{d}x = \lim_{\epsilon\to0^{+}}\big(\ln{1}-\ln{\epsilon}\big) = -\lim_{\epsilon\to0^{+}}\ln{\epsilon},$$
and this last clearly diverges to infinity.
A: 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 $0$.
