Evaluate $ \int_{-1}^1 \frac{1}{x}dx$. Question: Evaluate $\displaystyle \int_{-1}^1 \frac{1}{x}dx$.
I know this is a Type II improper integral. When I split I get $\displaystyle \int_{-1}^1 \frac{1}{x}dx=\int_{-1}^0 \frac{1}{x}dx+\int_0^{1} \frac{1}{x}dx$.
Then, $\displaystyle \int_{-1}^1 \frac{1}{x}dx=\lim_{b \rightarrow 0^-}\int_{-1}^b \frac{1}{x}dx+\lim_{a \rightarrow 0^+}\int_a^{1} \frac{1}{x}dx=\lim_{b \rightarrow 0^-}\ln|x|{\bigg|_{-1}^b}+\lim_{a \rightarrow 0^+}\ln|x|{\bigg|_a^1}$. Now when I evaluate this I get $-\infty+\ln(2)+\infty$. I know $\infty-\infty$ is not $0$. When I check a solution manual it says the answer is divergent, since $\displaystyle \lim_{b \rightarrow 0^-}\int_{-1}^b \frac{1}{x}dx$ is $-\infty$, and we do not have to even evaluate the other integral. But the other integral gives $\infty$.
I am confused. Because in limits, if both limits are divergent, then it does not necessarily mean the addition of them is also divergent. For example if $\displaystyle \lim_{x\rightarrow \infty}(x-x)$ is not divergent, even though $\lim_{x\rightarrow \infty}x=\infty$.
What is the difference in a regular limit calculation and an improper integral calculation? Could someone explain this, please?
Edit: The suggested post doesn’t help. I have already understood what that question is asking about. I know infinity-infinity is not zero. My question is about the difference between taking limit of a sum vs having a sum of limits already. Please don’t close this post. Could you please remove your votes for closing?
 A: You cannot just evalueate this integral, because it depends in how you split the integral.

*

*First split (symmetric):
$$\int_{-1}^{1} \frac{1}{x}dx=\lim_{\varepsilon \to 0} \left( \int_{-1}^{-\varepsilon}\frac{1}{x}dx+\int_{\varepsilon}^{1}\frac{1}{x}dx \right)=\lim_{\varepsilon \to 0} \left( \log(|-\varepsilon|)-\log(\varepsilon) \right)=0.$$

*Second split (non symmetric):
$$\int_{-1}^{1} \frac{1}{x}dx=\lim_{\varepsilon \to 0} \left( \int_{-1}^{-\varepsilon}\frac{1}{x}dx+\int_{c\varepsilon}^{1}\frac{1}{x}dx \right)=\lim_{\varepsilon \to 0} \left( \log(|-\varepsilon|)-\log(c\varepsilon) \right)=-\log(c).$$
What you can do is compute the "principal value of the integral".

Definition. Let $f$ continuous for $a \leq x<x_{0}$ and $x_{0}<x \leq b$ and absolutely convergent (in the sense that $\int_{a}^{b}|f|<\infty$). The principal value of $\int_{a}^{b}f$ is defined by
$$\text{PV}\int_{a}^{b}f(x)dx=\lim_{\varepsilon \to 0} \left( \int_{a}^{x_{0}-\varepsilon}f(x)dx+\int_{x_{0}+\varepsilon}^{b}f(x)dx \right).$$
Principal values always appear in harmonic analysis while you study singular integrals.
