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Find the integral of ln(x)/sqrt(x) as x is between 0 and 1

  1. Jul 13, 2006 #1
    Hi Im trying to figure out the following..Ive tried substitutions but cant find any that work:

    [tex]\int_0^1 \frac{lnx}{ \sqrt x}dx[/tex]

    HA @ 0 so
    [tex]\int_0^1 \frac{lnx}{ \sqrt x} dx[/tex]
    =
    [tex]\lim_{t\rightarrow 0^+} \int_t^1 \frac{lnx}{\sqrt x}dx[/tex]
     
    Last edited: Jul 13, 2006
  2. jcsd
  3. Jul 13, 2006 #2

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    Try the substitution u=sqrt(x).
     
  4. Jul 13, 2006 #3
    so....2du=1/root(x)dx.. it becomes just ln(u^2)?
     
  5. Jul 13, 2006 #4
    if this is right I end up with

    [tex]\lim_{t\rightarrow 0^+} 2 - \frac{s\sqrt t}{t}[/tex]

    so would this be divergent?
     
  6. Jul 13, 2006 #5

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    I don't know what s is there, but yes, this is divergent. Another way to find this is to note that this function is less than ln(x), whose integral diverges to negative infinity on this interval.
     
  7. Jul 13, 2006 #6
    ok, the s should be a two...so basically if the integral ever has [tex] \frac{a}{0}[/tex]...then it is called divergent?..or whenever an answer cannot be found..it is divergent?
     
  8. Jul 13, 2006 #7

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    This integral is divergent because the limit you mention in your first post is infinity. "if the integral ever has a/0" isn't precise enough an expression for me to say anything meaningful about. In general, for improper integrals like this one, you have to take a limit as you approach vertical asymptopes or as [itex]x \rightarrow \pm \infty[/tex]. If the result after taking all these limits is infinity, the integral is said to diverge. Whether you can find the answer or not is your problem, not the integral's.
     
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