Does $ \int x^{-2} \, \mathrm{d}{x} $ have a singularity?

Question

How do you integrate $ \dfrac{1}{x^{2}} $ from $ 0 $ to, say, $ a $? Can you get a principal value? What is the divergence: $ + \infty $ or $ - \infty $?

Answer 1

Let $$I_\epsilon =\int_{\epsilon}^a \frac{dx}{x^2}.$$ We say that the improper integral $$\int_0^a \frac{dx}{x^2}$$ converges (exists) if $\lim_{\epsilon\to 0^+}I_\epsilon$ exists. In this case, the limit does not exist. The integral also does not exist in the PV sense.

There is a reasonable case for saying that the integral is $+\infty$, since $I_\epsilon$ blows up as $\epsilon$ approaches $0$ from the right.

Answer 2

We would evaluate the improper integral as follows:

$$ \int_0^a \frac 1{x^2}\,dx = \lim_{\epsilon \to 0} \int_\epsilon^a \frac 1{x^2}\,dx = \lim_{\epsilon \to 0}\left[\frac{1}{\epsilon} - \frac{1}{a} \right] = +\infty $$