Integral of ln(x)/(x+1) from 1 ot 0

  • Thread starter coki2000
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  • #1
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Hello,
what is it's solvation.I tried to solve it by parts but i found undefinable term(from ln(0)).Please give me an advice.Thanks.
 

Answers and Replies

  • #2
Cyosis
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This integral doesn't have a primitive in terms of elementary functions.

Edit: didn't notice it was a definite integral.
 
  • #3
Hurkyl
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Hello,
what is it's solvation.I tried to solve it by parts but i found undefinable term(from ln(0)).Please give me an advice.Thanks.
That should be a clue that you've forgotten about improper integrals....
 
  • #4
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If i integrete this integral by parts

[tex]\int_{0}^{1}\frac{ln(x+1)}{x}dx[/tex]

ln(x+1)=u ,1/(x+1)dx=du and 1/xdx=dv, lnx=v

[tex]\int_{0}^{1}\frac{ln(x+1)}{x}dx=ln(x+1)lnx\mid_{0}^{1}-\int_{0}^{1} \frac{lnx}{x+1}dx[/tex]

This time integral of ln(x+1)/x from 0 to 1 is not integratable but answer of this is (pi^2)/6.
 
  • #5
Cyosis
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You can integrate it just fine, however there does not exist a primitive in terms of elementary functions. You can evaluate this integral by using the power series of ln(x+1).
 
  • #6
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The message #4 is in contradiction with the title of the thread…
Which integral is it ?
 
  • #7
Gib Z
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The message #4 is in contradiction with the title of the thread…
Which integral is it ?
No it isn't. The post showed that each integral yields the other. And as Cyosis mentioned, using the Taylor Series for ln(1+x) centered at x=0, we can get a series representation of the integral, and it so happens in this case it simplifies to a simple combination of well known constants.
 

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