Can anyone help me get started with this indefinite integral problem? http://db.tt/9NCNpBTd

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Can anyone help me get started with this indefinite integral problem? http://db.tt/9NCNpBTd

Mathematics
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As a picky detail, that's a definite integral. http://fwd4.me/0lMo
oh you're right I meant improper
i would start by using u substitution u = ln(x) du = 1/x dx \[\rightarrow \int\limits_{e}^{\infty}\frac{1}{u^{p}} du\]

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ok I've got that. Where is the part where it is improper though?
oh wait nevermind it's the infinity part sorry
then I think I would set t > e and take the integral from e to t
you could do that...then later take the limit as t->infinity
yeah, but how can I take the general integral of that without knowing what p is?
I think I might need to do cases for when p is -1 and when it isn't maybe
apply general power rule \[\int\limits\limits_{?}^{?} u^{-p} = \frac{u^{-p+1}}{-p+1}\]
Evaluated from e to t: (with lnx substituted back in for u) \[\rightarrow \frac{\ln(t)^{-p+1}}{-p+1} - \frac{1}{-p+1}\]
ok now all I need to do is see where the lim t-> inf is equal ro inf and when it's not
right...take lim ln(t) and lim 1/ln(t)
I think it's divergent when p<1
good..because then the ln(t) stays on top which goes to infinity if p>1, ln(t) goes to denominator which goes to 0
if p = then it would also be divergent I think
p=1
yes, 1/0 - 1/0 --> infinity
so as long as the limit doesn't exist or goes to infinity then it's convergent then right?
you mean divergent i think, then yes if limit exists and is real number, it is convergent
oh right sorry haha backwards again
I understand now, thank you very much for your help!
no problem :)

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