Why $\int^1_{-1}\frac{1}{\sqrt{x}}dx$ experiences no singularity at $x=0$?

Question

Why $\dfrac{1}{\sqrt{x}}$ is not singular at $x=0$?

My book says, in general, $\displaystyle\int^a_{-a}\dfrac{1}{x^n}dx$ converges for $n>1$, exists as a Cauchy principal value for $n=1$, and experiences no singularity for $n<1$.

Answer 1

Although the integrand is singular, it is an improper integral that converges. We interpret $\int_0^a\frac 1{\sqrt x}dx=\lim_{b \to 0}\int_b^a\frac 1{\sqrt x}dx$. We can then do the integral $$\left.\lim_{b \to 0}\int_b^a\frac 1{\sqrt x}dx=\lim_{b \to 0}2\sqrt x\right|_b^a=2\sqrt a$$ without the singularity of the integrand causing a problem at $0$. Intuitively $\frac 1{\sqrt x}$ doesn't go off to infinity fast enough to cause a problem.

Answer 2

That formulation is kind of problematic - exactly how are you defining $\frac1{x^n}$ for negative $x$ and non-integer $n$?

What we have here is that the infinite singularity in $f(x)=x^{-n}$ at $x=0$ is absolutely integrable for $n<1$; that is, $\int_0^a x^{-n}\,dx$ converges to some finite value. As such, if we have a reasonable definition of $x^{-n}$ on both sides, we can just integrate across the singularity (or on an interval with one endpoint there) and have the integral converge, either as an improper Riemann integral or as a Lebesgue integral. Yes, $\frac1{\sqrt{x}}$ is singular there - but the singularity isn't strong enough to break the integral.