# 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!

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.

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 $a$.

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}


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}