Integral of rational functions over the asymptote,with bounds

Question

If one were to consider the integral from -1 to 1 of 1/x, why would the areas not cancel out? From my limited Caclulus 1 knowledge, the pure geometric area should cancel to be come 0, and when evaluated mathematically, the natural logarithm of the absolute value of the limit of some arbitrary variable should be able to be subtracted, as the absolute value of the limit of that variable as it approaches -1 and 1, respectively should cancel to become 0. Can someone please explain why this would not be accepted as an answer or logic in most cases?

Answer 1

HINT

Intuitively, what you are saying is something like this. Define $D_n = \{1/n < |x| < n, x \in \mathbb{R}\}$ for all $n \in \mathbb{Z}^+$ and note that $\int_{D_n} dx/x = 0$ by symmetry. Similarly, note that $D_n \to \mathbb{R}$ as $n \to \infty$, and so you would like to define $$ \int_{\mathbb{R}} \frac{dx}{x} = \int_{\lim_n D_n} \frac{dx}{x} = \lim_{n \to \infty} \int_{D_n} \frac{dx}{x} = 0. $$

There are a couple of problems that you have formally exchanging the integral with the limit. Do you know under what conditions can you do that? Does your case actually fall into one of those where you would be allowed to do that?

Not only do you have possible issues at $x \to \pm \infty$ to resolve, but you would ideally like to write $$ \int_{-\infty}^\infty f(x)dx = \int_{-\infty}^0 f(x)dx + \int_0^\infty f(x) dx, $$ and integrability around $x=0$ becomes a real problem as well...