Integral of $\frac{1}{\sqrt{|x|}}$

Question

Why is $$\int_{-1}^1 \frac{dx}{\sqrt{|x|}} < \infty?$$

The area seems infinite since there is an asymptote at $0$. Can someone explain? Thanks

Answer 1

Hint. One may observe that $$ \begin{align} \int_{-1}^1 \frac{1}{\sqrt{|x|}}\:dx&=2\int_0^1 \frac{1}{\sqrt{|x|}}\:dx \\\\&=2\int_0^1 \frac{1}{\sqrt{x}}\:dx \\\\&=4\left[\sqrt{x}\right]_0^1 \\\\&=4. \end{align} $$

Answer 2

The integrand is even so sufficient to show on the interval $[0,1]$. We have $$ \int \frac{dx}{\sqrt x} = \int x^{-1/2} dx = \frac{x^{1/2}}{1/2} = 2\sqrt{x}, $$ which easily evaluates both at $0$ and and $1$...

Answer 3

Let's restrict attention to the interval $[0, 1]$. (The situation in $[-1, 0]$ is symmetric.) If we try to compute $$ \lim_{t \rightarrow 0+} \int_{t}^{1} {dx \over \sqrt{x}}, $$ we see that, for each $t$ the antiderivative turns out to be $$ (-1) \sqrt{x}\; \big|^{1}_{t} = (-1) (1 - \sqrt{t}), $$ which has a right-sided limit at $t = 0$. The existence of the limit tells that the integral converges.

Answer 4

Perhaps this reasoning will be convincing: A change of variables merely re-arranges the area under a curve, it does not change the area. So let's try $$ x = u^2 $$ and look at $\int_0^1 \frac1{\sqrt{x}}\, dx$. By coincidence, when $x=0$ $u=0$ and when $x=1$, $u=1$.

If $x = u^2$ then $dx = 2u\, du$ so the integral becomes $$ \int_{u=0}^1 \frac{2u\, du}{u} = \int_{u=0}^1 2 du = 2 $$ OUr change of variables has transformed an ugly chimney into a nice box.