Integration of a divergent integral using the Cauchy Principal Value I already posted this question (Integral with singularity on the real axis with complex integration) but I think I need to clarify better what want.
I want to know how do I evaluate the following integral using the Cauchy Principal Value
$\int_{-\infty}^{\infty}\dfrac{1}{x^2-1}$.
I could start saying that the integrand should vanishes when we have large |x| and proceed to evaluate through a contour that passes right above or right below the singularities, but  what contour should I take when circling each singularity? Should the answers be different when circling the singularities with different contours?
I didn't find any explanation of this in the materials/books of complex variable I have, so a detailed answer would be appreciated. Thanks in advance!
 A: Here is a solution using the Fundamental Theorem of Calculus:
\begin{align}
I&=\text{p.v. }\int_{-\infty}^{+\infty} \frac{dx}{x^2-1}\\ \\
&=\lim_{b\to\infty} \text{arctanh } x\Big|_{-b}^b \\ \\
&=\lim_{b\to\infty} 0 \\\\
&=0
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{a, b \in\mathbb{R}_{>1}}$:

\begin{align}
\mrm{P.V.}\int_{-a}^{b}{\dd x \over x^{2} - 1} & =
{1 \over 2}\,\mrm{P.V.}\int_{-a}^{b}{\dd x \over x - 1} -
{1 \over 2}\,\mrm{P.V.}\int_{-a}^{b}{\dd x \over x + 1}
\\[5mm] & =
{1 \over 2}\,\mrm{P.V.}\int_{-a - 1}^{b - 1}{\dd x \over x} -
{1 \over 2}\,\mrm{P.V.}\int_{-a + 1}^{b + 1}{\dd x \over x}
\\[5mm] & =
{1 \over 2}\,\mrm{P.V.}\int_{-a - 1}^{a + 1}{\dd x \over x} +
{1 \over 2}\int_{a + 1}^{b - 1}{\dd x \over x} -
{1 \over 2}\,\mrm{P.V.}\int_{-a + 1}^{a - 1}{\dd x \over x} -
{1 \over 2}\int_{a - 1}^{b + 1}{\dd x \over x}
\end{align}

Note that
  $\rule{0pt}{1cm}\ds{\mrm{P.V.}\int_{-c}^{c}{\dd x \over x} =
\lim_{\epsilon \to 0^{+}}\pars{%
\int_{-c}^{-\epsilon}{\dd x \over x} + \int_{\epsilon}^{c}{\dd x \over x}} =
\lim_{\epsilon \to 0^{+}}\bracks{%
\ln\pars{\verts{\epsilon \over c}} + \ln\pars{\verts{c \over \epsilon}}} = 0}$.

Then,
\begin{align}
&\mrm{P.V.}\int_{-a}^{b}{\dd x \over x^{2} - 1} =
{1 \over 2}\,\ln\pars{b - 1 \over a + 1} -
{1 \over 2}\,\ln\pars{b + 1 \over a - 1} =
{1 \over 2}\ln\pars{{b - 1 \over b + 1}\,{a - 1 \over a + 1}}
\\[5mm] &\
\implies
\lim_{a \to \infty}\lim_{b \to \infty}\mrm{P.V.}\int_{-a}^{b}
{\dd x \over x^{2} - 1} =
\lim_{b \to \infty}\lim_{a \to \infty}\mrm{P.V.}\int_{-a}^{b}
{\dd x \over x^{2} - 1} = 0
\\[5mm] &\
\implies
\bbx{\mrm{P.V.}\int_{-\infty}^{\infty}{\dd x \over x^{2} - 1} = 0}
\end{align}
A: Choose the contour $C$ as the large semicircle on the upoer half plane, with indent at singularities such that the poles are not included into the semicircle.
Decomposing the integral into 4 parts:
$$\oint_C=\int_{arc}+\int^R_{-R}+\int_\text{left small semicircle}+\int_\text{right small semicircle}$$.
By Cauchy’s theorem $$\oint_C=0$$
The arc integral obviously vanishes as $R$ approaches infinity.
The third integral is equal to 
$$\frac12\oint_{|z-(-1)|=r}\frac1{z^2-1}dz=\frac12\text{Res}_{z=-1} \frac1{z^2-1}=-\pi i/4$$
Similarly, the fourth integral equals $\pi i/4$.
Surprisingly, the required integral equals $0$!
p.s. please tell me where you do not understand so I can elaborate accordingly.
ADDED:
We can loosely see that this integral has the principal value of zero by:
$$\int^\infty_{-\infty}\frac1{z^2-1}dz=\frac12\left(\int^\infty_{-\infty}\frac1{z-1}dz+\int^\infty_{-\infty}\frac1{z+1}dz\right)=\frac12(0+0)=0$$
with the well-known result
$$\text{P}\int^\infty_{-\infty}\frac1{z-a}dz=0$$ for real $a$.
