Integration over a singular point yields finite value?

Question

Suppose we have $$I=\int_{-2}^{1}{\frac{1}{x^2}dx}$$

Evaluating the integral directly without considering the discontinuity, $$I=-\left[\frac{1}{x}\right]_{-2}^1=-\frac{3}{2}$$

The actual integral doesn't converge, but what does $-\frac{3}{2}$ represent? What is its significance?

Is it safe to say that it is the area from $-2$ to $0^-$ and $0^+$ to $1$?

Answer 1

The value $-3/2$ is the Hadamard finite part of the integral in question, which is a way to assign a meaningful finite value to an otherwise divergent quantity via meromorphic continuation. Denoting $\mathcal H$ as the finite part integral we write $$ \begin{align} \mathcal H\int_{-2}^1\frac{\mathrm dx}{x^2} &=\lim_{\epsilon\to 0^+}\left(\left(\int_{-2}^{-\epsilon}+\int_\epsilon^1\right)\frac{\mathrm dx}{x^2}-\frac{2}{\epsilon}\right)\\ &=\lim_{\epsilon\to 0^+}\left(-x^{-1}\Big|_{-2}^{-\epsilon}-x^{-1}\Big|_\epsilon^1-\frac{2}{\epsilon}\right)\\ &=\lim_{\epsilon\to 0^+}\left(\frac{1}{\epsilon}-\frac{1}{2}-1+\frac{1}{\epsilon}-\frac{2}{\epsilon}\right)\\ &=-\frac{3}{2}. \end{align} $$ This result can be derived directly from the antiderivative of $1/x^2$ because in this particular example no limit needs to be evaluated (since all terms involving $\epsilon$ simply cancel out).

Answer 2

In most cases, you can't integrate over a discontinuous interval. In cases like this, you can try splitting it up into two improper integrals, however this still yields $-\frac{3}{2}$, and each integral has a term that divides by zero. Hence, this integral is divergent, and the $-\frac{3}{2}$ has no significance. Additionally, because $\frac{1}{x^2}>0$ for all $x\in \mathbb{R},$ the integral would be positive if it was defined. This integral is divergent.

This BPRP video goes into more detail: https://www.youtube.com/watch?v=bEWIdRu4ctM

Answer 3

It is certainly not the area from $-2$ to $0^-$ and $0^+$ to $1$. That area is infinite (and shouldn't be negative).

Instead it's the area of the complement, i.e. from $-\infty$ to $-2$ and $1$ to $\infty,$ but with a reversed sign, or as the area from $1$ to $-2$ going via a point at infinity (wrapping around from $+\infty$ to $-\infty$).

Assume $-\infty < a < 0 < b < \infty$. Then the area under $\frac{1}{x^2}$ from $-\infty$ to $a$ plus from $b$ to $\infty$ is $$ \int_{-\infty}^{a} \frac{1}{x^2} dx + \int_{b}^{\infty} \frac{1}{x^2} dx = \left[ -\frac{1}{x} \right]_{-\infty}^{a} + \left[ -\frac{1}{x} \right]_{b}^{\infty} = -\frac{1}{a} + \frac{1}{b} \\ = - \left[ -\frac{1}{x} \right]_a^b = - \left[ \int \frac{1}{x^2} dx \right]_a^b = \left[ \int \frac{1}{x^2} dx \right]_b^a $$