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

This Question is Closed

barrycarter
 4 years ago
Best ResponseYou've already chosen the best response.0As a picky detail, that's a definite integral. http://fwd4.me/0lMo

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh you're right I meant improper

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i would start by using u substitution u = ln(x) du = 1/x dx \[\rightarrow \int\limits_{e}^{\infty}\frac{1}{u^{p}} du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok I've got that. Where is the part where it is improper though?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait nevermind it's the infinity part sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then I think I would set t > e and take the integral from e to t

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you could do that...then later take the limit as t>infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, but how can I take the general integral of that without knowing what p is?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I might need to do cases for when p is 1 and when it isn't maybe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0apply general power rule \[\int\limits\limits_{?}^{?} u^{p} = \frac{u^{p+1}}{p+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Evaluated from e to t: (with lnx substituted back in for u) \[\rightarrow \frac{\ln(t)^{p+1}}{p+1}  \frac{1}{p+1}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok now all I need to do is see where the lim t> inf is equal ro inf and when it's not

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right...take lim ln(t) and lim 1/ln(t)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think it's divergent when p<1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0good..because then the ln(t) stays on top which goes to infinity if p>1, ln(t) goes to denominator which goes to 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if p = then it would also be divergent I think

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes, 1/0  1/0 > infinity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so as long as the limit doesn't exist or goes to infinity then it's convergent then right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you mean divergent i think, then yes if limit exists and is real number, it is convergent

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh right sorry haha backwards again

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand now, thank you very much for your help!
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