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Math Help - convergence of an integral

  1. #1
    Super Member Random Variable's Avatar
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    convergence of an integral

    Does  \int_{0}^{\infty} \frac{\ln^{n} x}{1+x^{2}} \ dx converge for all  n \in \mathbb{N} ?

    If it does, it's not too hard to show that it equals 0 when n is odd and 2n! \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{n+1}} when n is even.


    Is x^{2} always eventually going to grow much faster than \ln^{n} x no matter how large  n is ?
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  2. #2
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    Opalg's Avatar
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    Re: convergence of an integral

    Quote Originally Posted by Random Variable View Post
    Does  \int_{0}^{\infty} \frac{\ln^{n} x}{1+x^{2}} \ dx converge for all  n \in \mathbb{N} ?

    If it does, it's not too hard to show that it equals 0 when n is odd and 2n! \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{n+1}} when n is even.


    Is x^{2} always eventually going to grow much faster than \ln^{n} x no matter how large  n is ?
    As x\to\infty, \ln x increases more slowly than any positive power of x. Thus \ln x \ll x^{1/(2n)} and therefore \frac{\ln^nx}{x^2} < \frac1{x^{3/2}}, which is enough to ensure convergence of your integral at ∞.

    There is also a singularity at 0. But near 0 the integral looks like \int_0\ln^nx\,dx, and that also converges.
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