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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.

- Thread starter coki2000
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- #1

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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.

- #2

Cyosis

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

- #3

Hurkyl

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That should be a clue that you've forgotten about improper integrals....

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.

- #4

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[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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- #6

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The message #4 is in contradiction with the title of the thread…

Which integral is it ?

Which integral is it ?

- #7

Gib Z

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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.The message #4 is in contradiction with the title of the thread…

Which integral is it ?