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Math Help - Help with convergent/divergent integral

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    Help with convergent/divergent integral

    Is this integral convergent or divergent?

    <br />
\int_0^\infty\frac{1}{\sqrt{x^2}}\,dx
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  2. #2
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    Quote Originally Posted by jconfer View Post
    Is this integral convergent or divergent?

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


    <br />
\int_0^\infty\frac{1}{\sqrt{x^2}}\,dx =\int_{0}^{\infty}\frac{1}{x}dx<br />
=\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.
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