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

1. barrycarter

As a picky detail, that's a definite integral. http://fwd4.me/0lMo

2. anonymous

oh you're right I meant improper

3. anonymous

i would start by using u substitution u = ln(x) du = 1/x dx $\rightarrow \int\limits_{e}^{\infty}\frac{1}{u^{p}} du$

4. anonymous

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

5. anonymous

oh wait nevermind it's the infinity part sorry

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

apply general power rule $\int\limits\limits_{?}^{?} u^{-p} = \frac{u^{-p+1}}{-p+1}$

11. anonymous

Evaluated from e to t: (with lnx substituted back in for u) $\rightarrow \frac{\ln(t)^{-p+1}}{-p+1} - \frac{1}{-p+1}$

12. anonymous

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

13. anonymous

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

14. anonymous

I think it's divergent when p<1

15. anonymous

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

16. anonymous

if p = then it would also be divergent I think

17. anonymous

p=1

18. anonymous

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

19. anonymous

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

20. anonymous

you mean divergent i think, then yes if limit exists and is real number, it is convergent

21. anonymous

oh right sorry haha backwards again

22. anonymous

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

23. anonymous

no problem :)