So, I've been wondering why PV integrals don't prove convergence of Integrals? For example, this integral $$ \int_{-1}^1 \frac{1}{x} \mathrm dx$$ converges to 0 using Cauchy Principle, and diverges by its definition.
What's the explanation for this?
Thanks in advance!
P.S - new to this site, I'm not sure if I used the math format correctly.
Edit: I've made it! Lol
On
PV integral and Lebesgue integrals are simply different things.
In particular, Lebesgue integral requires the integral to converge absolutely, meaning that $f$ is Lebesgue integrable if and only if $$ \int_X |f(x)|dx < \infty. $$ This is due to the fact that you define the integral only for positive functions, and you extend the definitions to signed functions in the following way $$ \int_X f(x)dx = \int_X f^+(x)dx - \int_X f^-(x)dx. $$ For example, $1/x$ is not Lebesgue integrable around zero because $$ \int_{-\delta}^{\delta} \frac{1}{|x|} = 2\int_0^\delta \frac{1}{x} = \infty, $$ i.e. is not absolutely integrable. To explain why the PV integral gives different results, consider a function $f\in L^1(X)$. Suppose that $f$ is defined and continuous everywhere but at a point $x_0$, where for example it goes to infinity. By dominated convergence theorem $$ \int_{\mathbb{R}} f(x)dx = \lim_{\delta\to 0} \int_{\mathbb{R}\setminus B_\delta(x_0)} f(x)dx. $$ Now, last guy CAN be well defined even for functions that are integrable but in a neighborhood of $x_0$, since we are integrating outside of it. Therefore an operator of the form $$ f\mapsto \lim_{\delta\to 0} \int_{\mathbb{R}\setminus B_\delta(x_0)} f(x)dx, $$ has a domain larger than $L^1(X)$. Moreover, this integrals turn out to be easier to compute than it looks like, by the use of tools coming from complex theory.
Yes, using PV this integral is equal to $0$.
We have singularity at $x=0$ so we will separate the Integral there: $$\int_{-1}^1 1/x\,dx\overbrace{=}^{\text{PV}}\lim_{a\to0^+}\int_{-1}^{-a}1/x\,dx+\int_a^11/x\,dx$$ Set $-u=x,\,-du=dx$ to get $$\lim_{a\to0^+}\int_{-1}^{-a}1/x\,dx+\int_a^11/x\,dx=\lim_{a\to0^+}\int_1^a -1/u\,-du+\int_a^1 1/x\,dx=\lim_{a\to0^+} -\int_a^1 1/u\,du+\int_a^1 1/x\,dx=0$$
Normal integral with singularity will be defined like the following:$$\int_{-1}^1 1/x\,dx=\lim_{a\to0^+,\,b\to0^-}\int_{-1}^{b}1/x\,dx+\int_a^11/x\,dx$$Now, if I do everything like in PV I will get $$\lim_{a\to0^+,\,b\to0^-} -\int_{-b}^1 1/u\,du+\int_a^1 1/x\,dx$$. This will result:$$\lim_{a\to0^+,\,b\to0^-}-(\ln(1)-\ln(-b))+(\ln(1)-\ln(a))=\lim_{a\to0^+,\,b\to0^-}\ln(-b)-\ln(a)=\lim_{a\to0^+,\,b\to0^-}\ln\left(-\frac ba\right)$$ Without knowing the relationship between $a,b$ we cannot avaluate this, for example, of $b=-2a$ we will get $\ln(2)$, with $b=-6a$ we will get $\ln(6)$, what PV tells us is to set $b=-a$.