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Math Help - Integration goes wrong? ( ln(x)/x )

  1. #1
    Member Pranas's Avatar
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    Integration goes wrong? ( ln(x)/x )

    \displaystyle \[I = \int\limits_0^{ + \infty } {\frac{{\ln x}}{x}} dx = \mathop {\lim }\limits_{\gamma  \to  + \infty } \left( {\int\limits_{1/\gamma }^\gamma  {\frac{{\ln x}}{x}dx} } \right)\]


    Because \displaystyle \[\int {\frac{{\ln x}}{x}dx = \int {\ln x \cdot d\left( {\ln x} \right)} }  = \frac{{{{\left( {\ln x} \right)}^2}}}{2}\]


    \displaystyle \[I = \frac{1}{2}\mathop {\lim }\limits_{\gamma  \to  + \infty } \left( {{{\left( {\ln \gamma } \right)}^2} - {{\left( {\ln \frac{1}{\gamma }} \right)}^2}} \right) = \frac{1}{2}\mathop {\lim }\limits_{\gamma  \to  + \infty } \left( {{{\left( {\ln \gamma } \right)}^2} - {{\left( { - \ln \gamma } \right)}^2}} \right)\]

    So it seems like zero, bus this feels kind of wrong to me. Besides, computer software assumes that integral diverges.
    What should I do?
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  2. #2
    A Plied Mathematician
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    Couple of things:

    1. \infty-\infty is undefined. It's like dividing by zero: not allowed.

    2. Your integral most definitely diverges on the positive end. Examining the integral

    \displaystyle\int_{e}^{\infty}\frac{\ln(x)}{x}\,dx  \ge \int_{e}^{\infty}\frac{1}{x}\,dx,

    you see that it diverges. I'd have to think a bit more about what happens at the lower limit.
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  3. #3
    Member Pranas's Avatar
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    Quote Originally Posted by Ackbeet View Post
    Couple of things:

    1. \infty-\infty is undefined. It's like dividing by zero: not allowed.

    2. Your integral most definitely diverges on the positive end. Examining the integral

    \displaystyle\int_{e}^{\infty}\frac{\ln(x)}{x}\,dx  \ge \int_{e}^{\infty}\frac{1}{x}\,dx,

    you see that it diverges. I'd have to think a bit more about what happens at the lower limit.
    Somewhat of a similar behaviour:

    \displaystyle \[\int\limits_{ - \pi /2}^{\pi /2} {\tan x}  \cdot dx = 0\]

    I find no other way but to accept this by definition even though there's like \infty-\infty involved.

    I'm afraid situation here might be relatively similar
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  4. #4
    A Plied Mathematician
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    Quote Originally Posted by Pranas View Post
    Somewhat of a similar behaviour:

    \displaystyle \[\int\limits_{ - \pi /2}^{\pi /2} {\tan x}  \cdot dx = 0\]

    I find no other way but to accept this by definition even though there's like \infty-\infty involved.

    I'm afraid situation here might be relatively similar
    That integral does not exist, either. If you want to mess around with Cauchy Principal Values, you might get some meaningful answer, probably zero as you've stated. However, the function simply is not Riemann integrable.
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  5. #5
    Forum Admin topsquark's Avatar
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    I'm inclined to go along with along with WolframAlpha. \displaystyle \lim_{\gamma \to \infty} \left ( ln^2(\gamma) - ln^2(\gamma) \right ) = \lim_{\gamma \to \infty} 0 = 0

    -Dan
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  6. #6
    A Plied Mathematician
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    Quote Originally Posted by topsquark View Post
    I'm inclined to go along with along with WolframAlpha. \displaystyle \lim_{\gamma \to \infty} \left ( ln^2(\gamma) - ln^2(\gamma) \right ) = \lim_{\gamma \to \infty} 0 = 0

    -Dan
    I would claim that's a Cauchy Principal Value of the integral. Normally, if I'm not mistaken, you have to evaluate the limits first, and then subtract.
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