Evaluating this integral for any integer $n>1$ : $\int \limits _0^ {\infty} \ln^n\frac{|1-x|}{|1+x|}dx$ How do I evaluate this integral for any integer $n>1$
$$\int \limits _0^ {\infty} \ln^n\frac{|1-x|}{|1+x|}\,dx$$ 
Note: for $n=1$ just we use integral by part and the integral will be divergent .My problem what about $n >1$
Thank you for any kind of help  
 A: Consider that the integral may be split up as follows;
$$\int_0^1 dx \log^n{\left (\frac{1-x}{1+x} \right )} + \int_1^{\infty} dx \log^n{\left (\frac{x-1}{x+1} \right )}$$
The first integral may be evaluated for any positive integer $n$ by subbing $e^{-u}=(1-x)/(1+x)$; the result is that the integral is equal to
$$2 (-1)^n \int_0^{\infty} du \frac{e^{-u}}{(1+e^{-u})^2} u^n$$
We may then expand the denominator in a Taylor series and reverse the order of integration and summation to get
$$2 (-1)^n \sum_{k=0}^{\infty} (-1)^k (k+1) \int_0^{\infty} du \, u^n \, e^{-(k+1) u} $$
Evaluating the integral, we get
$$2 (-1)^n n! \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^n} = 2 (-1)^n n! \left (1-\frac1{2^{n-1}} \right ) \zeta(n)$$
This holds even when $n=1$ (by taking a limit or evaluating the sum directly).
The second integral may be made to look like the first integral by subbing $x \mapsto 1/x$; the result is:
$$\int_0^1 dx \log^n{\left (\frac{1-x}{1+x} \right )} \frac1{x^2} $$
Using the same substitution, we get
$$2 (-1)^n \int_0^{\infty} du \frac{e^{-u}}{(1-e^{-u})^2} u^n  $$
Doing the same thing as in the first integral, we get that the second integral is
$$2 (-1)^n n! \sum_{k=0}^{\infty} \frac{1}{(k+1)^n} = 2 (-1)^n n!  \zeta(n)$$
Note that $n \gt 1$ here necessarily.  Thus, we may finally write down the result:

$$ \int_0^{\infty} dx \log^n{\left |\frac{x-1}{x+1} \right |} = 4 (-1)^n n! \left (1-\frac1{2^{n}} \right ) \zeta(n)$$

ADDENDUM
When $n=2 m$, the expression simplifies to
$$\int_0^{\infty} dx \log^{2 m}{\left |\frac{x-1}{x+1} \right |} = 2 (-1)^{m + 1} (2^{2 m} - 1) B_{2 m} \pi^{2 m}$$
where $B_{2 m}$ is a Bernoulli number.
