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Thread: Help with convergent/divergent integral

  1. #1
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    Help with convergent/divergent integral

    Is this integral convergent or divergent?

    $\displaystyle
    \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?

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


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

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

    so

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


    You should be able to finish from here

    good luck.
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