# adding a "negative area" and a positive area when the two are infinite

If we take the integral of, say, $\sin x$ from $x=0$ to $x= 2\pi$, we get $0$, which is adding the area between the curve and the $x$-axis between $0$ to $\pi$, plus the so-called "negative area" from $\pi$ to $2\pi$. So far, so good. But what we deal with another function with this type of symmetry, but where the two areas are each infinite (one positive, one negative)?

For example, to take $\int_{-1}^1\!\frac{dx}{x}$, we first break the integral into two integrals at the discontinuity $x=0$, but from there we can look at it in two ways, with very different results:

1. For each integral we get infinity; i.e., we say the limit for each one doesn't exist, so the sum doesn't exist. Ergo, the integral diverges.
2. the limit of $A$ plus the limit of $B$ equals the limit of $(A+B)$ (if they both go to the same limit, from the same direction), so taking the integrals we get $(\lim_{x\to 0^+}\ln|x|) - \lim_{x\to 0^+}(\ln|x|)$, which is $\lim_{x\to 0^+} (\ln|x|-\ln|x|)= \lim_{x\to 0^+}(|x|-|x|) = \lim_{x\to 0^+} (0) = 0$.

Which one of these two viewpoints is correct, and why is the other one incorrect? Thanks in advance.

The standard definition of an improper integral forces you to break up the integral into separate pieces for each "bad point," and to take the limits separately, then add them together. For example, for $\int_{-1}^1\!\frac{dx}{x}$ we have \begin{align*} \int_{-1}^1\!\frac{dx}{x} &= \left[\lim_{a\to 0^-}\int_{-1}^a\!\frac{dx}{x}\right] + \left[\lim_{b\to 0^+}\int_b^1\!\frac{dx}{x}\right]\\ &= \left[\lim_{a\to 0^-}\log\left|a\right|\right] + \left[\lim_{b\to 0^+}-\log\left|b\right|\right], \end{align*} which as you've noted does not exist. This is the "correct" way (by standard conventions) to evaluate the integral.
However, others have noticed that in certain cases it seems like the symmetry should imply that the integral is defined with a certain value (in the case above, you get $0$), and when you take the symmetry into account and take the limits "symmetrically," you wind up with what is known as the "Cauchy principal value" of the integral (see also here). Keep in mind that this method of assigning values to improper integrals is not standard.
Note that $\int_{-1}^1 \frac1xdx$ by definition of improper integrals is $$\lim_{h\to0^-,k\to0^+}\int_{-1}^h\frac1xdx+\int_k^1\frac1xdx$$ which is an undefined limit as the value depends on how $(h,k)$ moves towards $(0,0)$.