1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Evaluate or show that is divergent

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data
    evaluate or show that is divergent.

    2. Relevant equations
    [itex]\displaystyle\int_0^3 {\frac{1}{x-1}dx}[/itex]


    3. The attempt at a solution

    [itex]\displaystyle\int_0^1 {\frac{1}{x-1}dx}[/itex]

    +


    [itex]\displaystyle\int_1^3 {\frac{1}{x-1}dx}[/itex]


    lim[itex]_{b->1^-} \displaystyle\int_0^b {\frac{1}{x-1}dx}[/itex]

    +

    lim[itex]_{a->1^+} \displaystyle\int_a^3 {\frac{1}{x-1}dx}[/itex]


    ∞ - ln|-1| + ln|2| - ∞ ->> would the two infinities cancel out? is this an indeterminate form?
     
  2. jcsd
  3. Mar 15, 2013 #2

    Zondrina

    User Avatar
    Homework Helper

    Hm, was the original integral all together and then you split it or was it two different integrals all together?
     
  4. Mar 15, 2013 #3
    it was all together but there was a discontinuity at 1 where i had split it up into two.
     
  5. Mar 15, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    ∞ - ∞ is undefined. No, you can't cancel them out. If the integral diverges on [0,1) then it diverges on [0,3].
     
  6. Mar 15, 2013 #5

    Zondrina

    User Avatar
    Homework Helper

    Okay so you originally had ##\int_{0}^{3} \frac{1}{x-1} dx##

    You can make this a bit easier by comparison :

    ##\int_{0}^{3} \frac{1}{x-1} dx ≥ \int_{0}^{3} \frac{1}{x} dx = lim_{a→0^+} \int_{a}^{3} \frac{1}{x} dx = lim_{a→0^+} [ ln(x) ]_a^3 = ln(3) - lim_{a→0^+} ln(a) → -∞##
     
  7. Mar 15, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    1/(x-1) is NOT greater than 1/x. That's certainly not true at x=(1/2), for example. And there is nothing wrong with what whatlifeforme was doing. The only error was in the conclusions.
     
  8. Mar 15, 2013 #7

    Zondrina

    User Avatar
    Homework Helper

    Ohhh my mistake. I think I've been dealing with way too many series recently.
     
  9. Mar 15, 2013 #8
    what conclusions? do ∞ and -∞ not cancel out?
     
  10. Mar 15, 2013 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, that one. Isn't that what I said?
     
  11. Mar 16, 2013 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The whole point of writing "a" and "b" is to NOT get ∞! You should have
    [tex]ln|a-1|- ln|-1|+ ln|2|- ln|b- 1|= ln(2)+ ln|(a- 1)/(b- 1)|.

    No, you cannot cancel the two "infinities". With a and b independent, that last term is not 0.
     
  12. Mar 17, 2013 #11
    i'm still stuck here.
     
  13. Mar 17, 2013 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Why? You showed the integral from 0 to 1 diverges. That means the whole integral diverges. You were done then.
     
  14. Mar 17, 2013 #13
    are you sure that is a valid conclusion? so i would simply state the integral diverges on 0 to 1 thus it diverges on 0 to 3?
     
  15. Mar 17, 2013 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, I am sure. Infinities can't be 'cancelled'. If it diverges on 0 to 1 then nothing else can fix that.
     
  16. Mar 17, 2013 #15
    what if i did a direct comparison test.

    1/x-1 > 1/x

    [itex]\displaystyle\int_0^3 {\frac{1}{x} dx}[/itex]

    ln|3| - ln|0|= divergent right?

    smaller diverges so large diverges.
     
  17. Mar 17, 2013 #16

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you want to apply the comparison test you need both functions to be nonnegative on [0,3]. 1/(x-1) isn't. It's not even defined at x=1.
     
  18. Mar 17, 2013 #17
    update: looks like i did not need to subtract ∞ from ∞ after all.

    limit[itex]_{x->0-}[/itex] = -∞
    limit[itex]_{x->0+}[/itex] = ∞

    Thus, I would have:

    limit[itex]_{b->1-} ln|x-1| ^{b}_{0}[/itex] = ∞ - ln|-1|

    +

    limit[itex]_{a->1+} ln|x-1| ^{a}_{3} [/itex] = ln|2| + ∞

    thus, ∞ + ln|2| + ∞.

    does this look correct?
     
  19. Mar 17, 2013 #18

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Well, no! Your second limit should be from a to 3, not from 3 to a. Just like you said in your original post. The integral from 0 to 1 gives you an infinitely large negative and the integral from 1 to 3 gives an infinitely large positive. Why are you pushing this so hard and what exactly are you trying to accomplish. As I keep telling you, you had the answer to begin with.
     
  20. Mar 17, 2013 #19
    i'm just really stressed about getting this right. and taking the first part

    [itex]\displaystyle\int_0^1 {\frac{1}{x-1} dx}[/itex] and stating that it diverges, thus,

    [itex]\displaystyle\int_0^3 {\frac{1}{x-1} dx}[/itex] diverges.

    doesn't seem like a solid way of answering the question.
     
  21. Mar 17, 2013 #20
    also, ln|0| - ln|0| should equal 0. as they are both the same infinity.

    some infinities are larger than others, but in this case they should be the same.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Evaluate or show that is divergent
Loading...