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Does this integral converge or diverge?

  • Thread starter Dell
  • Start date
  • #1
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the given integral is:

[tex]\int[/tex][tex]\frac{dx}{ln(x)}[/tex] (from 0 to 1)


now i have a problem at both ends of the integral, because i have ln(0) is undefinde and ln(1) is 0 and 1/0 is undefined, so i can integrate from 0 to 0.5 and 0.5 to 1, but i tried to integrate this and just couldnt, i tried substitution
t=ln(x)
x=et
dt=dx/x

and then tried integration in parts but this just didnt work,

so then i thought maybe i need to find a similar function which i know more about or can easily find more about, but again i have problems at both limits, so i think i need 2 seperate functions, since i know that 1/ln(x) is negative in the limits 0-1, i will take the abs value of 1/ln(x) and look for positive functions...
i am looking for an integral that is either
*bigger than my original integral and converges
*smaller than my original integral and diverges

what i found was
[tex]\int[/tex]dx/[tex]\sqrt{x}[/tex] (from 0 to 0.4) which converges and is bigger than [tex]\int[/tex]dx/|ln(x)| (from 0 to 0.4 )

and my second integral is [tex]\int[/tex]1/x (from 0.6 to 1) which diverges and is smaller than [tex]\int[/tex]dx/|ln(x)| (from 0.6 to 1 )

therefore i know that the integral [tex]\int[/tex][tex]\frac{dx}{ln(x)}[/tex] (from 0 to 1) diverges near 1, and converges near 0, so all in all it diverges.

is this okay? is it alright that i didnt take into account the area between 0.4 and 0.6 since the function is constant there anyway??

can you see any better way to solve this??
 

Answers and Replies

  • #2
364
2
is this okay? is it alright that i didnt take into account the area between 0.4 and 0.6 since the function is constant there anyway??

can you see any better way to solve this??
I came to a similar conclusion using your first substitution t=ln|x|. I found this easier to visualize. The lower limit of zero transforms to negative infinity and it is clear that the integral of exp(t)/t converges at negative infinity. However the upper limit of 1 transforms to zero and it is clear that the integral of exp(t)/t diverges at zero, since the numerator goes to unity and the denominator is a divergent 1/t form.
 

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