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Math Help - Improper Integral

  1. #1
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    Improper Integral

    here's the problem:
    \int_{7}^{\infty} \frac{\ln(x)}{x} dx

    I used substitution, letting u = ln(x) which gave me du=\frac{1}{x}dx and the resulting:
     \lim_{b\rightarrow \infty}\int_{ln(7)}^{ln(b)}{u} du

    This gave me  \lim_{b\rightarrow \infty}\frac{ln(b)^2}{2} - \frac{ln(7)^2}{2}

    Since infinity ends up in a numerator, I would think the answer is that the solution DNE, and it diverges, but webwork won't accept that. Where did I go wrong? Thanks MHF!!
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  2. #2
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    Re: Improper Integral

    Quote Originally Posted by jjtjp View Post
    here's the problem:
    \int_{7}^{\infty} \frac{\ln(x)}{x} dx
    I used substitution, letting u = ln(x) which gave me du=\frac{1}{x}dx and the resulting:
     \lim_{b\rightarrow \infty}\int_{ln(7)}^{ln(b)}{u} du
    This gave me  \lim_{b\rightarrow \infty}\frac{ln(b)^2}{2} - \frac{ln(7)^2}{2}
    Since infinity ends up in a numerator, I would think the answer is that the solution DNE, and it diverges, but webwork won't accept that. Where did I go wrong? Thanks MHF!!
    I think that your problem is webwork.

    That may well be the very worst online course delivery system for mathematics there is.
    Thanks from jjtjp
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  3. #3
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    Re: Improper Integral

    You are correct, the integral diverges.
    Thanks from jjtjp
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  4. #4
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    Re: Improper Integral

    Is my math correct, though? Because when I graph \frac{ln(x)}{x} it definitely approaches zero making me think I messed up somewhere.
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  5. #5
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    Re: Improper Integral

    okay, nvm. It was a webwork issue. I do hate how if you are wrong in webwork, it just tells you you're wrong, no explanation or anything. Thanks for verifying guys. I was thinking I was crazy after about 15 min.
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  6. #6
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    Re: Improper Integral

    Quote Originally Posted by jjtjp View Post
    Is my math correct, though? Because when I graph \frac{ln(x)}{x} it definitely approaches zero making me think I messed up somewhere.
    An integral is really an infinite sum, and just because terms in a series tend to 0 is no guarantee that the series will still converge.
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