# 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?

• $\infty-\infty$ is what is known as an indeterminate form. – Cheerful Parsnip Aug 12 at 15:19
• @Cheerful Parsnip See please better the question. – Michael Rozenberg Aug 12 at 15:20
• @CheerfulParsnip, I understand that, but in my opinion this integral should be equal to $0$, because it is logically appealing – Markoff Chainz Aug 12 at 15:21
• To answer your second question: Yes, see 'Cauchy Principal Value Integral' – projectilemotion Aug 12 at 15:22
• @JohnColtraneisJC: "Logically appealing" counts for a lot. It's one of the things that inform which definitions and axioms we bother to spend effort studying. – Henning Makholm Aug 12 at 15:25

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").

But the best way to attack rigorously this issue is to define it as a "distribution" in the framework of... "distribution theory", through its action on a generic "test function" $$\varphi$$ :

$$PV \left( \frac{1}{x} \right)(\varphi) := \lim_{\varepsilon\to 0} \int_{|x|>\varepsilon} \left( \frac{1}{x} \varphi(x) \right) dx$$

• 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 "logically appealing" ways to handle the mathematical object "$$PV \left( \frac{1}{x} \right)$$" :

• 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,

$$\widehat{PV \left( \frac{1}{x} \right)} = -i\pi\,\text{sign} (\xi)$$

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", a concept introduced by Hadamard in classical analysis which is different from the "Principal Value" concept. See for that (https://www.ntu.edu.sg/home/mwtang/hypersie.pdf).