Help in this improper integral

Question

$\int_0^\infty\ln(\frac{1+x}{1-x}) \frac{dx}{x}$ can this integral be done by power series If yes I hope a full solution I dont need to use by parts

Answer 1

For $x>1$, $(1+x)/(1-x)<0$ so its logarithm would have an imaginary part. If that is $\pi i$, the integral has imaginary part $\pi i\int_1^\infty dx/x$ which diverges.

Maybe you wanted to evaluate $$\int_1^\infty\ln\left|\frac{1+x}{1-x}\right|\frac{dx}x$$ instead? You can break this into integrals over $(0,1)$ and $(1,\infty)$. In the latter, use the substitution $y=1/x$ to get $$\int_1^\infty\ln\left|\frac{1+x}{1-x}\right|\frac{dx}x =\int_0^1\ln\left|\frac{y+1}{y-1}\right|\frac{dy}y$$ which is the same as the integral over $(0,1)$. The integral over $(0,1)$ is $$\int_0^1\left(2\sum_{k=0}^\infty\frac{x^{2k+1}}{2k+1}\right)\frac{dx}x$$ etc.