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quick convergence/divergence Q watch

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    Does
    \int^\infty_1 \frac{(ln(x))^s}{x} dx

    diverge for any real value of s where there are 3 cases with s>0 s=0 and s<0?

    thanks for any help/advice.
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    It definitely diverges for s>1. After all, \sum ^{\infty}_1 \frac{1}{n} diverges, so something bigger than is eventually larger than 1/n on each interval is going to diverge as well, no? You could formalize this idea with the integral test. So you need to consider other cases than s>1.
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    oh yeah, cheers
    also, if you integrate using the substitution u = ln(x)
    you get

    [\frac{ln(x)^{s+1}}{s+1}]^\infty_1

    and when s<-1 you get the second half of the evaluation divided by zero, does this mean the whole function diverges or is it undefined?

    thanks a lot
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    (Original post by aerot85)
    and when s<-1 you get the second half of the evaluation divided by zero, does this mean the whole function diverges or is it undefined?
    When s<-1 you get (1/(s+1))*(ln(x))^a where a is negative so? And when s=-1 it is undefined.


    Spoiler:
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    It is undefined
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    yeah i thought that was the case, the question asked us to tell us for which ranges of s the integral converges, but as far as i could work out they all diverge. Thanks for your help.
 
 
 
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