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Prove that integral is convergent

  1. Jun 16, 2011 #1
    1. The problem statement, all variables and given/known data

    i need to know the integral of x^alpha times (lnx)^ beta from 0 to 0.5
    the question is if alpha greater than -1 prove that integral convergent
    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 16, 2011 #2

    micromass

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    Re: integral

    So what did you try?
     
  4. Jun 16, 2011 #3
    Re: integral improper

    I don't think you'll be able to actually do that integral in closed form, but you can identify when it exists. You'll need to find a bound for log(x) as x approaches 0.
     
  5. Jun 16, 2011 #4
    Re: integral improper

    i dont want to know what is the answer , i just want to prove how does it converge at zero
     
  6. Jun 16, 2011 #5
    Re: integral improper

    i tried all methods
     
  7. Jun 16, 2011 #6

    micromass

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    Re: integral improper

    Try a substitution first to get rid of the logarithm.
     
  8. Jun 16, 2011 #7
    Re: integral

    i tried the neighbourhood at zero since the problem is at zero , but i dont think lnx has a neighbourhood at zero , so i tried change by varibale x= e^t but still didnt work
    then i tried integration by parts , and that was a disaster im stuck
    and last i tried the theory which says that if you mulitply by x ^m where m is less than 1 and you reach the limit zero you will have a convergent integral but that only works with real numbers not variables so what do you think i should do? please make sure
     
  9. Jun 16, 2011 #8
    Re: integral improper

    i tried
     
  10. Jun 16, 2011 #9

    micromass

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    Re: integral improper

    So, what did you get?
     
  11. Jun 16, 2011 #10
    Re: integral improper

    i told you in the other thread its the same question they told me to write where i told you
     
  12. Jun 16, 2011 #11

    micromass

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    Re: integral

    What did you get after the substitution x=et?
     
  13. Jun 16, 2011 #12
    Re: integral

    man it becomes from -infinte to zero et ^a times x ^beta but still nothing i cnt go anywhere with that
     
  14. Jun 16, 2011 #13

    micromass

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    Re: integral

    So you get

    [tex]\int_{-\infty}^0{e^{\alpha+1}t^\beta dt}[/tex]

    right?

    Now, apply partial integration to reduce the value of beta (i.e. to make beta negative)
     
  15. Jun 16, 2011 #14
    Re: integral

    and what do i get please can you solve it i was trying all day please
     
  16. Jun 16, 2011 #15

    micromass

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    Re: integral

    I'm sorry, that would be against the rules. I'm afraid you will have to solve it.

    Alternatively, do you know the following test:

    If [itex]\int_{-\infty}^0{g}[/itex] converges and [itex]\lim_{x\rightarrow-\infty}{\frac{f}{g}}=0[/itex], then [itex]\int_{-\infty}^0{f}[/itex] converges as well.

    Applying this with an appropriate g could also give you the solution.
     
  17. Jun 17, 2011 #16
    Re: integral

    To summarize everything so far (there have been a few errors), the original integral is

    [tex]\int_0^{0.5} x^\alpha (\ln{x})^\beta dx [/tex]

    Micromass suggested the substitution x = et, which leads to

    [tex]\int_{-\infty}^{-\ln{2}} e^{t(\alpha + 1)} t^{\beta} dt [/tex]

    Note that the upper limit is -ln(2) = ln(0.5), not 0.

    From here, try parts with u = tβ. The following may help:

    [tex]\int_a^b u dv = vu \; \big|_a^b - \int_a^b v du [/tex]

    Specifically, for what values of [itex]\alpha[/itex] and [itex]\beta[/itex] will [itex]vu \; \big|_a^b[/itex] converge?
     
    Last edited: Jun 17, 2011
  18. Jun 18, 2011 #17

    hunt_mat

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    Re: integral

    Why not try looking at certain values of [itex]\beta[/itex], say for example [itex]\beta =1[/itex], then your integral becomes:
    [tex]
    \int_{0}^{\frac{1}{2}}x^{\alpha}\ln xdx=\int_{-\infty}^{-\ln 2}ye^{(\alpha +1)y}dy
    [/tex]
    upon using the substitution [itex]y=\ln x[/itex]. Now note that the interval of integration is over a negative region which means that [itex]\alpha +1>0[/itex] and so [itex]\alpha >-1[/itex], which is what we were trying to show in the first place. Can you now show this for general [itex]\beta[/itex]?
     
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