Why is it necessary to take$\int \frac{1}{x}~ dx$, and split it into two pieces, left and right side? Lets say its bounded by $2$ and $-2$?

Question

The answer to this question is posted, but i don't understand what it means that as $b\to 0^+$ that this goes to infinity, and why is the integral split up in two?

Answer 1

Last term miss $\lim_{b \to 0+}$ - when you put it there you'll have answer: $$\lim_{b \to 0^+}\left( \ln v|_{b}^{2} \right) = \lim_{b \to 0^+}\left( \ln 2 - \ln b \right) = \ln 2 + \infty= +\infty $$

Answer 2

A function is Rieman-integrable on a compact set $A$ if $$ \int_{A}f < \infty $$

Formally Riemann integral exists only on compact sets, so it's not well-defined if either the function, as in your case, or the interval (e.g. $x \to \infty$) is unbounded. Therefore, to extend Riemann integral to either case, the limit is replaced by some value, e.g. $\varepsilon>0$, as in your case. Such integral is known as an improper integral. $$ \lim_{\varepsilon \to 0^{+}} \int_{\varepsilon}^{b}\frac{dx}{x} = \lim_{\varepsilon \to 0^{+}} (F(b) -F(\varepsilon)) $$ Again, in your case this limit doesn't exist (or equal to $\infty$), so the function is not Riemann-integrable.

EDIT concerning your interval of $[-2,2]$: note it contains a discontinuity at $0$. By construction, Riemann integral is a limit of upper and lower sums, and the value in each interval lies between upper ($\sup$) and lower ($\inf$) values of the function $(x_i <t <x_{i+1})$: $$ \inf f(x^{\ast}_t)(x_{i+1} - x_i)\\ \sup f(x^{\ast}_t)(x_{i+1} - x_i) $$ Obviously $ -\infty < \inf f(x^{\ast}_t) < \sup f(x^{\ast}_t) < \infty$, and if this condition is violated, Riemann integral doesn't exist in this interval.