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Improper Integrals

  1. Feb 21, 2008 #1
    1. The problem statement, all variables and given/known data
    Determine whether the improper integral converges and, if so, evaluate it.

    the limits of integration are 1 to 2

    the integrand is
    (dx/(xlnx))




    2. Relevant equations



    3. The attempt at a solution

    so first I found the indefinite integral which was ln(ln(x)) + c

    now usually these problems have infinite for one of the limits of integration in which case that limit becomes a variable such as a or b, with this one I did notice that the the ln(ln(1)) is going to be taking the ln(0) which you cannot do I'm not sure what to do from here though or what limit I am supposed to be evaluating to see if it converges or not.
     
  2. jcsd
  3. Feb 21, 2008 #2

    NateTG

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    Have you dealt with things like:
    [tex]\int_{0}^{1} \frac{1}{x^2} dx[/tex]
     
  4. Feb 21, 2008 #3
    oh if the exponent is >= 1 then the integral is infinite right?
     
    Last edited: Feb 21, 2008
  5. Feb 21, 2008 #4
    so if f(x)>=g(x)>=0

    if the integral of a to infinite of f(x) converges then the integral of a to infinite of g(x) conveges and
    if the integral of a to infinite of g(x) diverges then the integral of a to infinite of f(x) diverges

    how would I go about selecting a function that is less then the one I was given though.
     
  6. Feb 21, 2008 #5

    NateTG

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    You should be able to work it out with:
    [tex]\lim_{a \rightarrow 1^+} \int_a^2 \frac{1}{x \ln x} dx [/tex]
     
  7. Feb 21, 2008 #6
    ok I figured that limit is positive infinite correct? which would mean it diverges ?
     
  8. Feb 21, 2008 #7

    NateTG

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    I get that it diverges also.
     
  9. Feb 21, 2008 #8
    what is the integral of 1/xlnx dx
     
  10. Feb 21, 2008 #9

    NateTG

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    [tex]\ln(\ln x)+C[/tex]
    (Just like you have.)
     
  11. Feb 21, 2008 #10
    integ dx/xlnx, take the sub lnx=t, dx/x=dt, from here we get

    integ dt/t=lnt
    but since he is taking a definite integral he need not go back to the original variable, so that means
    lim (e-->0) integ(from 1+e to 2) of dx/xlnx=lim(e-->0) ln(t) on the interval 1+e to 2, hence

    lim(e-->0) [ln2-ln(1+e)]=ln2
    so it converges!
     
  12. Feb 21, 2008 #11

    I see what you did with the substitution, but you got the limits wrong.

    [tex]\int^2_1\frac{1}{xlnx}dx[/tex] [tex]u=lnx => du = \frac{dx}_{x}[/tex]
    Now you have to plug in you original limits to find out what the limits in u are.
    [tex]ln(1) = 0 , ln(2) = ln(2)[/tex]
    in the end you end up with;
    [tex]ln(ln2) - ln(0) [/tex]
    [tex] ln(x) -> -\infty as x -> 0[/tex]
    So it diverges.
     
    Last edited: Feb 21, 2008
  13. Feb 21, 2008 #12
    where are you getting e from?
     
  14. Feb 21, 2008 #13
    Oh yeah, sorry my bad. I totally missed this part. I can't believe i forgot to change the limits of integration after i made that substitution!!!!!
     
  15. Feb 21, 2008 #14
    I just let e =epsylon, sorry i should have defined that!
     
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