Question

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

Answer 1

As a picky detail, that's a definite integral.

http://fwd4.me/0lMo

Answer 2

oh you're right I meant improper

Answer 3

i would start by using u substitution

u = ln(x)
du = 1/x dx

\[\rightarrow \int\limits_{e}^{\infty}\frac{1}{u^{p}} du\]

Answer 4

ok I've got that. Where is the part where it is improper though?

Answer 5

oh wait nevermind it's the infinity part sorry

Answer 6

then I think I would set t > e and take the integral from e to t

Answer 7

you could do that...then later take the limit as t->infinity

Answer 8

yeah, but how can I take the general integral of that without knowing what p is?

Answer 9

I think I might need to do cases for when p is -1 and when it isn't maybe

Answer 10

apply general power rule

\[\int\limits\limits_{?}^{?} u^{-p} = \frac{u^{-p+1}}{-p+1}\]

Answer 11

Evaluated from e to t: (with lnx substituted back in for u)

\[\rightarrow \frac{\ln(t)^{-p+1}}{-p+1} - \frac{1}{-p+1}\]

Answer 12

ok now all I need to do is see where the lim t-> inf is equal ro inf and when it's not

Answer 13

right...take lim ln(t) and lim 1/ln(t)

Answer 14

I think it's divergent when p<1

Answer 15

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

Answer 16

if p = then it would also be divergent I think

Answer 17

p=1

Answer 18

yes, 1/0 - 1/0 --> infinity

Answer 19

so as long as the limit doesn't exist or goes to infinity then it's convergent then right?

Answer 20

you mean divergent i think, then yes

if limit exists and is real number, it is convergent

Answer 21

oh right sorry haha backwards again

Answer 22

I understand now, thank you very much for your help!

Answer 23

no problem :)