Reinterpreting improper integrals that require Cauchy principal value to be defined

Question

This question concerns the Cauchy principal value. Consider the improper integral $$\int_{-∞}^{∞}\frac{1+x}{1+x^2}dx$$ which is divergent, and then its Cauchy principal value $$\lim_{u \to ∞} \int_{-u}^{u}\frac{1+x}{1+x^2}dx=\pi$$ As I have yet to study Lebesgue integration, but I was wondering whether it resolves problems of this nature - or if any other interpretation of the integral does. Just to clarify, by resolving I mean that the value of the integral does not have to be assigned to integrals that would otherwise be undefined.

Answer 1

by resolving I mean that the value of the integral does not have to be assigned to integrals that would otherwise be undefined.

Well, you do not have to assign values to divergent integrals. Mathematicians use the Cauchy principal value when they want to use an integral which unfortunately happens to diverge. For example, consider the Poisson formula that gives the values of a harmonic function $u$ in the upper half-plane $H=\{(x,y):x\in\mathbb R,y>0\}$ in terms of its boundary values $u^*$: $$u(x,y)=\frac{y}{\pi}\int_{-\infty}^\infty \frac{u^*(x-t)}{t^2+y^2}\,dt \tag1$$ When a function with boundary values $u^*(x)=1+x$ comes along, we have two ways to react:

1. "Formula (1) does not apply because the integral diverges"
2. "Formula (1) can be used if the integral is understood as Cauchy principal value"

I believe the second approach is more constructive™ and therefore is in the interest of the mathematical community. It yields $$\lim_{A\to+\infty}\frac{y}{\pi}\int_{-A}^A \frac{1+x-t}{t^2+y^2}\,dt =\lim_{A\to+\infty} \frac{2}{\pi}\tan^{-1}(A/y)(1+x) = 1+x \tag2$$ which, while not very exciting, is a correct formula for a harmonic function with the given boundary values.

---

Does the Lebesgue theory of integration help us work with formulas such as (1)? Short answer: not at all. The Lebesgue theory is agnostic of the order structure of the domain of integration: it takes it as a whole, not as a sequence of expanding intervals. The concept of Cauchy principal value is alien to it. The Lebesgue integral has many wonderful features (most importantly, powerful convergence theorems and the completeness of $L^p$ spaces) but it is not a flexible tool for dealing with divergent improper integrals. For example, the (conditionally) convergent integral $\int_0^\infty \frac{\sin x}{x}\,dx$ is deemed divergent by the Lebesgue theory.

Of course, one can take the Lebesgue integral over $[-A,A]$ and then let $A\to\infty$ with the same result as above. But the resulting notion of $\int_{-\infty}^\infty$ is closer in spirit to the Riemann integral; it depends on the order structure of the line and is not as stable under convergence of integrands as integrals of $L^1$ functions.

Answer 2

I'm not sure sure whether Lebesgue integration helps to solve that integral, though I think it doesn't. Perhaps a little complex analysis?

$$f(z)=\frac{1+z}{1+z^2}=\frac{1+z}{(z+i)(z-i)}$$

and we can take the contour determined by

$$C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;|z|=R\;,\;\text{Im}(z)\ge 0\}\;,\;\;R>>0$$

Within the domain bounded by the above curve we've only one simple pole, so

$$Res_{z=i}(f)=\lim_{z\to i}(z-i)f(z)\stackrel{\text{l'Hospital}}=\frac{1+i}{2i}$$

so by Cauchy's Residue Theorem

$$2\pi i\cdot\frac{1+i}{2i}=\oint\limits_{C_R}f(z)dz+\int_{-R}^R\frac{1+x}{1+x^2}dx$$

and passing to the limit when $\,R\to\infty\,$ and taking the real parts we get

$$\pi=\int_{-\infty}^\infty\frac{x+1}{x^2+1}dx$$

getting btw that the imaginary part has also the same value...

It should be noted that the imaginary part is due entirely to the loop integral; the imaginary part of the integral along the real axis is zero, as expected.