Convergence of Definite Improper Integrals of the Form $1/x$ Given a simple integral of the form: $$ \int ^1 _{-1} \frac{1}{x} \, dx =\lim _{a\rightarrow 0} \int ^1 _a \frac{1}{x} \, dx + \int ^a _{-1} \frac{1}{x} \, dx$$
Is it possible to say that this integral converges? I was told explicitly by my professor that this sort of improper definite integral can be said to converge in the sense of "Cauchy" but I can't find anything to back up his claim. I had posted another question about this where I was told if you can't assign a finite value to one of these integrals, it can't be said to exist. So for this simple example, we have $$\lim _{a\rightarrow 0} \int ^1 _a \frac{1}{x} \, dx = \lim _{a \rightarrow 0} [\ln (1)- \ln(a)] = -\infty$$
So you end up with $\infty - \infty$ for the entire interval which is undefined.  He never clarified what he really meant by "Cauchy", so I'm left to guess he means the integral test for convergence, which doesn't seem to apply here since we're asking wethere this definite integral is defined in the first place. 

$$
\int_{-1}^{-a} \frac{dx} x + \int_a^1 \frac{dx}x = 0 \to 0.
$$
However:
\begin{align}
& \int_{-1}^{-2a} \frac{dx} x + \int_a^1 \frac{dx}x = \Big(\log|{-2a}| - \log |{-1}|\Big) + \Big( \log 1 - \log a \Big) \\[10pt]
= {} & \log(2a) - \log a = \log \frac{2a} a = \log 2 \approx 0.693\ldots 
\end{align}
As $a\downarrow 0$, the sets $(-1,-a)\cup(a,1)$ and $(-1,-2a)\cup(a,1)$ both approach $(-1,1)$, but the way in which the bounds approach $0$ alters the value of the integral.  The first one is singled out as the "principal value", in a sense that the conventional language attributes to Cauchy.  What the actual history is, and hence how much credit or blame for this Cauchy deserves, is a different question.
(A bit more precisely: $\displaystyle\bigcup_{a>0} (-1,-a)\cup(a,1)$ and $\displaystyle\bigcup_{a>0} (-1,-a)\cup(a,1)$ are both equal to $(-1,0)\cup(0,1)$, which includes all of $[1,1]$ except a set whose measure is $0$.)
