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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?

user2820579
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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 $$ -\int\frac{dx}{x^2-1}=-\int\frac{dx}{2(x-1)}+\int\frac{dx}{2(x+1)}= -\frac{1}{2}\left(\log(x-1)-\log(x+1)\right) $$

  • Ok, but what's the catch if you don't take that minus sign. What should be the correct evaluation if you don't put the sign in front of the integral? – user2820579 Jul 13 '17 at 21:33
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    @user2820579 There is an implict absolute value when you integrate $1/x$. –  Jul 13 '17 at 21:38