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Math Help - Evaluating improper integral

  1. #1
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    Evaluating improper integral

    \int_e^{\infty} \frac{1}{x(\ln x)^3} dx

    First i made a u-substitution by making u = ln x.
    After which i got -\frac{1}{2(\ln x)^2}
    Then i went back to compute the improper integral again , but couldn't solve it.

    Did i made any wrong move?
    Thanks

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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by xcluded View Post
    \int_e^{\infty} \frac{1}{x(\ln x)^3} dx

    First i made a u-substitution by making u = ln x.
    After which i got -\frac{1}{2(\ln x)^2}
    Then i went back to compute the improper integral again , but couldn't solve it.

    Did i made any wrong move?
    Thanks

    That's the correct substitution, but you set it back up incorrectly. Once the substitution is made, \ln x should not appear in the new integral.

    It turns out that we should get \int_e^{\infty}\frac{\,dx}{x\left(\ln x\right)^3}\xrightarrow{u=\ln x}{}\int_1^{\infty}\frac{\,du}{u^3}

    Does this make sense?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by xcluded View Post
    \int_e^{\infty} \frac{1}{x(\ln x)^3} dx

    First i made a u-substitution by making u = ln x.
    After which i got -\frac{1}{2(\ln x)^2}
    Then i went back to compute the improper integral again , but couldn't solve it.

    Did i made any wrong move?
    Thanks

    Since \frac{d}{dx}\left[ -\frac{1}{2 (\ln(x))^2}\right]=\frac{1}{x(\ln(x))^3} by the fundamental theorem of calculus:

    \int_e^{\infty} \frac{1}{x(\ln x)^3}\; dx= \lim_{\zeta \to \infty}\left[-\frac{1}{2(\ln(x))^2}\right]_e^{\zeta}= <br /> <br />
 \frac{1}{2}-\lim_{\zeta \to \infty} \frac{1}{2(\ln(\zeta))^2}=\frac{1}{2}

    CB
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    That's the correct substitution, but you set it back up incorrectly. Once the substitution is made, \ln x should not appear in the new integral.

    It turns out that we should get \int_e^{\infty}\frac{\,dx}{x\left(\ln x\right)^3}\xrightarrow{u=\ln x}{}\int_1^{\infty}\frac{\,du}{u^3}

    Does this make sense?
    Hmm why does the lower limit becomes 1 ?
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  5. #5
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    Quote Originally Posted by xcluded View Post
    Hmm why does the lower limit becomes 1 ?
    You substitute u = ln(x) with the lower limit of x being e. So the lower limit of u is ln(e)=1
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  6. #6
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    Thanks everyone !

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