# Thread: Trouble with logarithmic Integral

1. ## Trouble with logarithmic Integral

Hello!!

I am trying to understand how to calculate the following integral:

$\displaystyle \int_{0}^{1} \frac{lnx}{x+1}$
I already know that the answer is $\displaystyle -\frac{\pi}{12}$
but why?
Can anyone help me on this?

2. Originally Posted by zenctheo
Hello!!

I am trying to understand how to calculate the following integral:

$\displaystyle \int_{0}^{1} \frac{lnx}{x+1}$.
$\displaystyle \int_0^1 {\frac{{\ln x}} {{1 + x}}\,dx} = \sum\limits_{k = 0}^\infty {( - 1)^k \left\{ {\int_0^1 {x^k \ln x\,dx} } \right\}} = - \sum\limits_{k = 1}^\infty {\frac{{( - 1)^{k + 1} }} {{k^2 }}} .$

And this is fairly well-known, so the rest follows.

3. Originally Posted by Krizalid
$\displaystyle \int_0^1 {\frac{{\ln x}} {{1 + x}}\,dx} = \sum\limits_{k = 0}^\infty {( - 1)^k \left\{ {\int_0^1 {x^k \ln x\,dx} } \right\}} = - \sum\limits_{k = 1}^\infty {\frac{{( - 1)^{k + 1} }} {{k^2 }}} .$

And this is fairly well-known, so the rest follows.
Thanks for the reply. But I still don't get it:

From where do you derive this substitution:

$\displaystyle \int_0^1 {\frac{{\ln x}} {{1 + x}}\,dx} \sum\limits_{k = 0}^\infty {( - 1)^k \left\{ {\int_0^1 {x^k \ln x\,dx} } \right\}}$.

If it's possible can you direct me to read an article or a webpage in order to understand the concept behind this.

Sorry for the inconvenience.

4. It's okay.

You know from the geometric series $\displaystyle \frac{1} {{1 - x}} = \sum\limits_{k = 0}^\infty {x^k } ,\,\left| x \right| < 1,$ from there you can find the sum for $\displaystyle \frac1{1+x}.$

After the interchange between the sum & the integral, yields the integral which can be solved by substitution $\displaystyle u=-\ln x,$ you'll end up with the Gamma Function. Finally you'll get the last sum.