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Math Help - integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

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    integral from 1 to infiinity of ln x / (x^2) equals 1. How?! (IMPROPER INTEGRATION)

    I integrated the function to become

    ln x / (x^2)

    u = lnx
    du = 1/x
    dv = 1/x^2
    v = -1/x

    UV - int VDU

    lnx * -1/x - - int of 1/x^2 becomes -1/x

    so

    - lnx/x - lnx (or close to this)

    So where the hell does this mystical one come from? It looks as if it'd be divergant from this!!! Please help!!!!!!!
    Last edited by skinsdomination09; December 17th 2012 at 04:52 PM.
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    Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?! (IMPROPER INTEGRATIO

    Quote Originally Posted by skinsdomination09 View Post
    I integrated the function to become

    ln x / (x^2)

    u = lnx
    du = 1/x
    dv = 1/x^2
    v = -1/x

    UV - int VDU

    lnx * -1/x - - int of 1/x^2 becomes -1/x

    so

    - lnx/x - lnx (or close to this)

    So where the hell does this mystical one come from? It looks as if it'd be divergant from this!!! Please help!!!!!!!

    Look at this webpage.
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    Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

    \int_1^\infty \frac{\ln{x}}{x^2} \, dx

    u = \ln{x} ... du = \frac{1}{x} \, dx

    dv = \frac{1}{x^2} dx ... v = -\frac{1}{x}

    \lim_{b \to \infty} \int_1^b \frac{\ln{x}}{x^2} \, dx =  \lim_{b \to \infty} \left[-\frac{\ln{x}}{x}\right]_1^b - \int_1^b -\frac{1}{x^2} \, dx

    \lim_{b \to \infty} \left[-\frac{\ln{x}}{x}\right]_1^b - \left[\frac{1}{x} \right]_1^b

    \lim_{b \to \infty} \left[-\frac{\ln{b}}{b}\right] - \left[\frac{1}{b} - 1 \right]_1^b = 1
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    Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?! (IMPROPER INTEGRATIO

    Quote Originally Posted by Plato View Post
    delete
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    Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

    Quote Originally Posted by skeeter View Post
    \int_1^\infty \frac{\ln{x}}{x^2} \, dx

    u = \ln{x} ... du = \frac{1}{x} \, dx

    dv = \frac{1}{x^2} dx ... v = -\frac{1}{x}

    \lim_{b \to \infty} \int_1^b \frac{\ln{x}}{x^2} \, dx =  \lim_{b \to \infty} \left[-\frac{\ln{x}}{x}\right]_1^b - \int_1^b -\frac{1}{x^2} \, dx

    \lim_{b \to \infty} \left[-\frac{\ln{x}}{x}\right]_1^b - \left[\frac{1}{x} \right]_1^b

    \lim_{b \to \infty} \left[-\frac{\ln{b}}{b}\right] - \left[\frac{1}{b} - 1 \right]_1^b = 1

    Makes more sense, Thanks But wouldn't the ln b / b (since it's approaching infinity) be indetirminate so it would be divergant. My original mistake was that I integrated the second half twice by accident, but I still don't get how you can just "ignore" ln b /b since ln(infinity) doesn't equal 0 or anything.

    Oh... is it bc the top is growing slower than the bottom so its just a really small number over a big one and thus goes to zero? Simple Alg. + early Calc skills holding me back
    Last edited by skinsdomination09; December 17th 2012 at 05:17 PM.
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    Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

    L'Hopital ...

    \displaystyle \lim_{b \to \infty} \frac{\ln{b}}{b} = \lim_{b \to \infty} \frac{1}{b} = 0
    Thanks from skinsdomination09
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    Re: integral from 1 to infiinity of ln x / (x^2) equals 1. How?!!?!?! (PLEASE HELP)!!

    Quote Originally Posted by skeeter View Post
    L'Hopital ...

    \displaystyle \lim_{b \to \infty} \frac{\ln{b}}{b} = \lim_{b \to \infty} \frac{1}{b} = 0
    Wow. I'm an idiot lol. Thanks
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