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anonymous

  • 4 years ago

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

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  1. barrycarter
    • 4 years ago
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    As a picky detail, that's a definite integral. http://fwd4.me/0lMo

  2. anonymous
    • 4 years ago
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    oh you're right I meant improper

  3. dumbcow
    • 4 years ago
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    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
    • 4 years ago
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    ok I've got that. Where is the part where it is improper though?

  5. anonymous
    • 4 years ago
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    oh wait nevermind it's the infinity part sorry

  6. anonymous
    • 4 years ago
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    then I think I would set t > e and take the integral from e to t

  7. dumbcow
    • 4 years ago
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    you could do that...then later take the limit as t->infinity

  8. anonymous
    • 4 years ago
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    yeah, but how can I take the general integral of that without knowing what p is?

  9. anonymous
    • 4 years ago
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    I think I might need to do cases for when p is -1 and when it isn't maybe

  10. dumbcow
    • 4 years ago
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    apply general power rule \[\int\limits\limits_{?}^{?} u^{-p} = \frac{u^{-p+1}}{-p+1}\]

  11. dumbcow
    • 4 years ago
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    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
    • 4 years ago
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    ok now all I need to do is see where the lim t-> inf is equal ro inf and when it's not

  13. dumbcow
    • 4 years ago
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    right...take lim ln(t) and lim 1/ln(t)

  14. anonymous
    • 4 years ago
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    I think it's divergent when p<1

  15. dumbcow
    • 4 years ago
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    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
    • 4 years ago
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    if p = then it would also be divergent I think

  17. anonymous
    • 4 years ago
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    p=1

  18. dumbcow
    • 4 years ago
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    yes, 1/0 - 1/0 --> infinity

  19. anonymous
    • 4 years ago
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    so as long as the limit doesn't exist or goes to infinity then it's convergent then right?

  20. dumbcow
    • 4 years ago
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    you mean divergent i think, then yes if limit exists and is real number, it is convergent

  21. anonymous
    • 4 years ago
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    oh right sorry haha backwards again

  22. anonymous
    • 4 years ago
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    I understand now, thank you very much for your help!

  23. dumbcow
    • 4 years ago
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    no problem :)

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