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.
[Math] Reinterpreting improper integrals that require Cauchy principal value to be defined
improper-integralsintegrationphilosophy
Best Answer
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.