Integrating $f(x) = \frac{1}{x}$ over $[-1,1]$

Question

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?

@CheerfulParsnip, I understand that, but in my opinion this integral should be equal to $0$, because it is logically appealing — Markoff Chainz, Aug 12 '19 at 15:21
To answer your second question: Yes, see 'Cauchy Principal Value Integral' — projectilemotion, Aug 12 '19 at 15:22
Yikes! "Logically appealing" doesn't count for anything in mathematics. There are axioms and theorems which must be adhered to! — JohnColtraneisJC, Aug 12 '19 at 15:23
@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. — hmakholm left over Monica, Aug 12 '19 at 15:25
@JohnColtraneisJC math should strengthen your intuition, not other way around — Markoff Chainz, Aug 12 '19 at 15:25
I have a strong sense of both...Hence the reason I already know why this is an improper integral. — JohnColtraneisJC, Aug 12 '19 at 15:26
@JohnColtraneisJC do you have a sense that area to the left of $0$ equals to the area to the right of $0$ but with opposite sign? Then integral must be equal to $0$ — Markoff Chainz, Aug 12 '19 at 15:28
I've covered Fourier Series before and its a pretty standard concept used over and over with them. — JohnColtraneisJC, Aug 12 '19 at 15:32
Just a side thought here: math is a discipline that ought to serve us in other disciplines in order to understand what's around us and as to how to make our lifes better and perhaps if I may say that, to make it more comfortable. With that all in mind, if (I want to stress "if") there is some practical application to the integral in question here, would 0 as an answer be desired, or undefined? — imranfat, Aug 12 '19 at 16:28
Related: https://math.stackexchange.com/questions/2600482/symmetric-integral-of-1-x, https://math.stackexchange.com/questions/2829348/what-is-with-this-integral-int-11-frac-1x-dx, https://math.stackexchange.com/questions/3063452/how-can-i-prove-that-int-11-frac1x-dx-0, and surely many other older questions too. — Hans Lundmark, Aug 12 '19 at 16:52
@Hans Lundmark None of the (good) references you give say that there is a hidden part in this iceberg, i.e. interpretation within distribution theory. This what I have attempted to do in my answer. — Jean Marie, Aug 12 '19 at 17:31
@MichaelRozenberg I didn't quite understand what point you were trying to make to me, but I'm glad to see the question got a fruitful answer with the Cauchy Principal Value. — Cheerful Parsnip, Aug 12 '19 at 19:59

score 4 · Accepted Answer · edited Jun 12 '20 at 10:38

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.

Integrating $f(x) = \frac{1}{x}$ over $[-1,1]$

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