I am trying to demonstrate that the principal value via partial fractions of (exercise VII.6.1 from Gamelin's Complex Analysis)
$$ \operatorname{P.V.}\int_0^\infty\frac{dx}{1-x^2}=\lim_{\epsilon\rightarrow0}\,\left[\int_0^{1-\epsilon} \frac{dx}{1-x^2} + \int_{1+\epsilon}^\infty \frac{dx}{1-x^2}\right]=0, $$ My problem is that the second integral involves the limit
$$ \lim_{x\rightarrow\infty}\operatorname{log}\frac{1+x}{1-x} = i\pi, $$
and therefore the integral is not zero! Am I misunderstanding something?
You're better off rewriting your second integral as $$ \int_{1+\epsilon}^\infty \frac{dx}{1-x^2}=-\int_{1+\epsilon}^\infty\frac{dx}{x^2-1}. $$ This way the denominator is always positive and you can write $$ -\int\frac{dx}{x^2-1}=-\int\frac{dx}{2(x-1)}+\int\frac{dx}{2(x+1)}= -\frac{1}{2}\left(\log(x-1)-\log(x+1)\right) $$