Integrating $f(x) = \frac{1}{x}$ over $[-1,1]$ I am reading a book where it says that improper integral $$\int_{-1}^{1}\frac{1}{x}\,dx$$ is undefined because $$\lim_{b \to 0^-}\int_{-1}^{b}\frac{1}{x}\,dx + \lim_{b \to 0^+}\int_{b}^{1}\frac{1}{x}\,dx$$ are unbounded.
I wonder, is it just a deficiency of definition of improper integral or is it universally accepted among mathematicians that $\int_{-1}^{1}\frac{1}{x}\,dx$ is undefined?
Function is odd, so in my opinion it is intuitively clear that this integral should be equal to $0$. Are there other definitions of integral that assign value of $0$ to this expression?
 A: In fact, behind your question, there is a very interesting mathematical "character" which is
$$PV \left( \frac{1}{x} \right).$$
We could avoid it, remaining with classical analysis tools as in this question whose interest is to introduce the concept of (Cauchy) Principal Value (abbreviated as "PV").
The rigorous way to attack this issue is to define it as a "distribution" in the framework of... "distribution theory", through its action on a generic "test function" $\varphi$ :
$$PV \left( \frac{1}{x} \right)(\varphi) := \lim_{\varepsilon\to 0} \int_{|x|>\varepsilon} \left( \frac{1}{x} \varphi(x) \right) dx 
$$
Please note

*

*the "shrinking hole" $(-\varepsilon,\varepsilon)$


*the fact that integral bounds are $(-\infty,\infty)$ (not limited to $[-1,1]$). See (https://en.wikipedia.org/wiki/Cauchy_principal_value).
There are different ways to handle "$PV \left( \frac{1}{x} \right)$" :

*

*as the limit when $\varepsilon \to 0$ of odd functions defined by :
$$f_{\varepsilon}(x):=\frac{x}{\varepsilon^2+x^2},$$
an astute way to overcome singularity $x=0$ !
In particular $\int_{[-a,a]}f_{\varepsilon}(x)dx=0$, whatever $a>0$...

*

*as the derivative of the even function $\log|x|$ (this one having the two "passports" : "ordinary function" and "regular distribution") (Derivative of ln|x| is the principal value of 1/x. Distribution Theory.).


*through its Fourier Transform,
$$\widehat{PV \left( \frac{1}{x} \right)} = -i\pi\,\text{sign} (\xi)$$
(Fourier transform of the distribution PV $\left( \frac{1}{x} \right)$)...
Remark : Distribution PV $\left( \frac{1}{x}\right)$ behaves, apart from properties due to singularity $0$, as ordinary function $1/x$. Thus we can await a differentiation formula generalizing $(1/x)'=-1/x^2$. Here it is :
$$\left(PV \left( \frac{1}{x} \right)\right)' = -FP \left( \frac{1}{x^2} \right)$$
(Derivative of principal value distribution $1/x$ is equal to finite part distribution $-1/x^2$?) where FP is for "Finite Part", which is another distribution. The  concept of "finite part", introduced by Hadamard in classical analysis around 1900, is  different from the "Principal Value" concept. See for that (https://www.ntu.edu.sg/home/mwtang/hypersie.pdf).
Nevertheless, one must be very cautious for some operations such as this one :
$$\begin{cases}(\delta \times x) \times PV(\frac1x)&=&0& \text{whereas}\\\delta \times (x \times PV(\frac1x))&=&\delta&\end{cases}$$
due to the fact that the product of distributions isn't associative as recalled here.
