anonymous
  • anonymous
Can anyone help me get started with this indefinite integral problem? http://db.tt/9NCNpBTd
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
barrycarter
  • barrycarter
As a picky detail, that's a definite integral. http://fwd4.me/0lMo
anonymous
  • anonymous
oh you're right I meant improper
dumbcow
  • dumbcow
i would start by using u substitution u = ln(x) du = 1/x dx \[\rightarrow \int\limits_{e}^{\infty}\frac{1}{u^{p}} du\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.