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

[coki2000]

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.

 

[Cyosis]

This integral doesn't have a primitive in terms of elementary functions.

Edit: didn't notice it was a definite integral.

 

[Hurkyl]

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...

 

[coki2000]

If i integrete this integral by parts

\int_{0}^{1}\frac{ln(x+1)}{x}dx

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

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

This time integral of ln(x+1)/x from 0 to 1 is not integratable but answer of this is (pi^2)/6.

 

[Cyosis]

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).

 

[guerom00]

The message #4 is in contradiction with the title of the thread…
Which integral is it ?

 

[Gib Z]

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.