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Math Help - Direct Comparison Theorem

  1. #1
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    Direct Comparison Theorem

    Use the Comparison Theorem to determine whether the integral is convergent or divergent.



    So I'm thinking that I can compare that integrand to

    \int_1^\infty \frac{e^{-x}}{x}dx

    If that integral is divergent, then my original integral is also divergent. But, I get stuck on evaluating my substitution

    \frac{2 + e^{-x}}{x} \geq \frac{e^{-x}}{x} dx}

    \lim_{t \to \infty} \int_1^t \frac{e^{-x}}{x} dx

    let u = e^{-x}   du = -e^{-x}dx
    let dv = \frac{dx}{x}   v = ln(x)

    \lim_{t \to \infty} \int_1^t \frac{e^{-x}}{x} dx = \lim_{t \to \infty} e^{-x}\ln{x} + \int_1^t e^{-x}}\ln{x} dx

    And I'm not getting anywhere... Can anybody suggest something to get me on the right path? Thanks.
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  2. #2
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    Re: Direct Comparison Theorem

    note that \frac{2+e^{-x}}{x} > \frac{1}{x} for x > 1

    and, \int_1^\infty \frac{1}{x} \, dx diverges ...
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  3. #3
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    Re: Direct Comparison Theorem

    Thanks!
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