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Thread: Convergent/Divergent Integral

  1. #1
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    Convergent/Divergent Integral

    Evaluate the following integral if it is convergent.

    $\displaystyle
    \int_1^{\infty}\frac{ln^2x}{x^3}dx$
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  2. #2
    Super Member General's Avatar
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    Quote Originally Posted by zer0e View Post
    Evaluate the following integral if it is convergent.

    $\displaystyle
    \int_1^{\infty}\frac{ln^2x}{x^3}dx$
    $\displaystyle \color{blue}\int_1^{\infty}\frac{ln^2x}{x^3}dx$
    =$\displaystyle \color{blue}\lim_{t\to\infty} \int_1^{t}\frac{ln^2x}{x^3}dx$.

    To solve the integral, use integration by parts with $\displaystyle \color{blue}u=ln^2(x)$ and $\displaystyle \color{blue}dv=\frac{1}{x^3}dx$
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  3. #3
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    Of course we evaluate as long as having its convergence:

    For all $\displaystyle x\ge1$ is $\displaystyle \ln x<\sqrt x,$ thus $\displaystyle \ln^2x<x$ hence $\displaystyle \frac{\ln^2x}{x^3}<\frac1{x^2},$ and the integral converges since $\displaystyle \int_{x\ge1}\frac{dx}{x^2}<\infty.$
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