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

  • Thread starter suspenc3
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  • #1
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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]
 
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Answers and Replies

  • #2
StatusX
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Try the substitution u=sqrt(x).
 
  • #3
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so....2du=1/root(x)dx.. it becomes just ln(u^2)?
 
  • #4
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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?
 
  • #5
StatusX
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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.
 
  • #6
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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?
 
  • #7
StatusX
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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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