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