# Thread: Help with convergent/divergent integral

1. ## Help with convergent/divergent integral

Is this integral convergent or divergent?

$
\int_0^\infty\frac{1}{\sqrt{x^2}}\,dx$

2. Originally Posted by jconfer
Is this integral convergent or divergent?

$
\int_0^\infty\frac{1}{\sqrt{x^2}}\,dx$
Note that $\sqrt{x^2}=|x|=\begin{cases}
-x \mbox{ if } x < 0 \\
x \mbox{ if } x \ge 0
\end{cases}
$

$
\int_0^\infty\frac{1}{\sqrt{x^2}}\,dx =\int_{0}^{\infty}\frac{1}{x}dx
=\int_{0}^{1}\frac{1}{x}dx+\int_{1}^{\infty}\frac{ 1}{x}dx$

for the first integral use the sub $u=\frac{1}{x} \mbox{ then }du=\frac{-1}{x^2}dx \iff-\frac{1}{u^2}du=dx$

so

$\int_{0}^{1}\frac{1}{x}dx=\int_{\infty}^{1}u \cdot \left( \frac{-1}{u^2}\right)du=\int_{1}^{\infty}\frac{1}{u}du$

You should be able to finish from here

good luck.