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Math Help - About a improper integral

  1. #1
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    About a improper integral

    I would like to ask about this integral:
    integral from 2 to infinity [ ln x /(exp(x) +1) ]

    I think it diverges, but I don't know how to show it diveges.

    first, I thought [ ln x /(exp(x) +1) ]can't be integrate,
    so I considered if it can be proved by comparing it with simpler integral.
    I had try to find a smaller integral and then prove it diveges,
    hence it could conclude that integral from 2 to infinity [ ln x /(exp(x) +1) ] also diveeges.
    However, I could't think out such these smaller integral.

    So how can I proved it, or it actually converges?
    Please help me.
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  2. #2
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    Re: About a improper integral

    Quote Originally Posted by wong999 View Post
    I would like to ask about this integral:
    integral from 2 to infinity [ ln x /(exp(x) +1) ]

    I think it diverges, but I don't know how to show it diveges.

    first, I thought [ ln x /(exp(x) +1) ]can't be integrate,
    so I considered if it can be proved by comparing it with simpler integral.
    I had try to find a smaller integral and then prove it diveges,
    hence it could conclude that integral from 2 to infinity [ ln x /(exp(x) +1) ] also diveeges.
    However, I could't think out such these smaller integral.

    So how can I proved it, or it actually converges?
    Please help me.
    note for x > 2 ...

    \frac{\ln{x}}{e^x + 1} < \frac{\ln{x}}{e^x} < \frac{x}{e^x}

    and \int_2^\infty xe^{-x} \, dx = \frac{3}{e^2} converges
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  3. #3
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    Re: About a improper integral

    Quote Originally Posted by skeeter View Post
    note for x > 2 ...

    \frac{\ln{x}}{e^x + 1} < \frac{\ln{x}}{e^x} < \frac{x}{e^x}

    and \int_2^\infty xe^{-x} \, dx = \frac{3}{e^2} converges
    huh...how stupid I am
    thank you for helping
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