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Is this integral convergent?

  1. Jan 25, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    For what values of a and b is the integral in the attachment convergent?


    2. Relevant equations



    3. The attempt at a solution
    By the comparison test, as well as the fact that arctanx/x diverges, I believe a-b<1. Is that correct?
     

    Attached Files:

  2. jcsd
  3. Jan 25, 2013 #2

    Mark44

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    Here's the integral, in LaTeX:
    $$ \int_1^{\infty} \frac{x^a~arctan(x)~dx}{2 + x^b}$$
     
  4. Jan 25, 2013 #3

    LCKurtz

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    No, that is not correct. For example a = 0 and b = 1. You might try making use of the fact that arctan(x) is bounded.
     
  5. Jan 25, 2013 #4
    I meant to write, a-b<-1.
     
  6. Jan 25, 2013 #5

    LCKurtz

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    I get that too, assuming a and b are nonnegative. Is that given? You didn't show your argument though.
     
  7. Jan 25, 2013 #6
    It is not given that a and b are nonnegative.
     
  8. Jan 25, 2013 #7

    LCKurtz

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    So you have more to do...
     
  9. Jan 25, 2013 #8
    My argument was that it had to be greater than or equal to (pi/4)/(x^p), where p is greater than 1, and less than or equal to (pi/2)/(x^p), as then, by the comparison test, the original integral would converge.
     
  10. Jan 25, 2013 #9

    LCKurtz

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    Sorry, but to continue this discussion, I need to see your work, showing assumptions on a and b, your equations/inequalities, and your conclusions or where you are stuck. Unfortunately, I have to leave for a few hours, but that will give you time to post the details.
     
  11. Jan 25, 2013 #10
    Wouldn't the inequality be as I stated above just before you were leaving, namely:
    (π/4)/xb-a<arctanx/xb-a<(π/2)/xb-a
    Hence, for the integral to converge then by the comparison test b-a>1?
    Why is that incomplete/incorrect?
     
  12. Jan 25, 2013 #11

    LCKurtz

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    Let's be clear about something. You did not state that inequality anywhere before the post to which I am responding. And you still haven't shown any work which can be analyzed to see what you are thinking. And the reason that is incomplete/incorrect is because it is incomplete.

    For example, if a=-2 and b = -3 you don't have b-a>1 yet the integral converges. So until you post more complete arguments as I have suggested in post #9 instead of just answers, we don't have anything to discuss.
     
  13. Jan 25, 2013 #12
    First of all, I DID post that just before you were leaving. You should check carefully and read posts well before you accuse anyone of anything or make any demands/claims.
    Second, there is no need to be rude and use such tone. If you don't wish to help in a civil manner, do me a big favour and don't!
     
  14. Jan 25, 2013 #13

    micromass

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    I don't see it either...

    He was not being rude. He is merely stating the rules of this forum. That is: you must show your work and your thinking process. This is meant to make our help more effective. If you show all your work, then it's much more easy for us to see where you're going wrong and how to help you.
     
  15. Jan 25, 2013 #14
    What about this post then, posted at 21:24: "My argument was that it had to be greater than or equal to (pi/4)/(x^p), where p is greater than 1, and less than or equal to (pi/2)/(x^p), as then, by the comparison test, the original integral would converge."
    ??
    It was not typed using LaTex, but it was there all along and was precisely what I wrote again for him to read upon his return.
     
  16. Jan 25, 2013 #15
    And if you do not consider that good enough an answer, it is not because I am holding back something, but because that is all I know and all I am capable of deriving myself at this stage. He could have tried being more efficient and helpful, instead of making ridiculous assertions and demands.
     
  17. Jan 25, 2013 #16

    micromass

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    LCKurtz has been very efficient and helpful in this thread. He always is efficient and helpful. I don't really appreciate it when you attack a member who is trying to help you. Please don't do this again.

    What we want of you is a clean argument.
    For example

    Write out the actual inequality and prove it.

    Show this explicitely.
     
  18. Jan 25, 2013 #17
    First of all, as I am experiencing difficulties with my keyboard you would have to excuse my not using LaTex at this point. It is certainly not my choice.
    Regarding my argument:
    I realised that arctanx/(x^p) does not converge unless p were greater than 1. Hence, I was trying to express the original expression as arctanx/(x^p) and demand that p is greater than 1. In order to make sure the original expression converges, I needed to find something convergent which is greater (following the rationale of the comparison test). So I noted that arctanx was bounded from above by pi/2, and so I wrote that the original expression was < (pi/2)(x^a)/[2+x^b] < (pi/2)/(x^(b-a)). I then demanded b-a to be greater than 1.
    I realise that this answer might be/is incomplete, but that is all I can muster on my own at this point.
    Now if you care to help, I'd appreciate it.
     
  19. Jan 26, 2013 #18

    vela

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    Well why don't you see why your argument doesn't work for the example LCKurtz gave back in post #11?
     
  20. Jan 26, 2013 #19
    I do see that, but am not sure how to correct it, which is why my last post asked for assistance.
     
  21. Jan 26, 2013 #20

    vela

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    Examine each step of your argument to see where and why it breaks down. LCKurtz has already given you a hint, noting that your basic argument works when a and b are non-negative. Find which step doesn't work if you drop that assumption. That might give you ideas of how to fix up your proof.
     
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