Improper integral with a root

Problem

Prove that $\displaystyle\int_0^1\frac{dx}{\sqrt{x}} = 2$ (improper integral).

Solution

The function $\dfrac{1}{\sqrt x}$ has a vertical asymptote at $x=0$: the integral is improper at the lower limit. We treat it as a limit: $\int_0^1\frac{dx}{\sqrt{x}} = \lim_{\varepsilon\to 0^+}\int_\varepsilon^1 x^{-1/2}\,dx = \lim_{\varepsilon\to 0^+}\bigl[2\sqrt{x}\bigr]_\varepsilon^1 = \lim_{\varepsilon\to 0^+}\bigl(2 - 2\sqrt{\varepsilon}\bigr) = \boxed{2}.$ Since $2\sqrt{\varepsilon}\to 0$, the integral converges to the value $2$ (consistent with the criterion: $\int_0^1 x^{-p}dx$ converges for $p<1$, and here $p=\tfrac12$).

Links

Topics: Integral
Concepts: Improper integral · Limit
Skills: Limiting-case analysis · Prove · Integrate
Exercise types: Integral computation