# Cauchy Principal Value Integral of Simple Rational Functions

How do I find the Cauchy Principal Value of the following integral (or how do I show if the value does not exist at all?):

$\int\limits_{-\infty}^\infty\frac{\text{dx}}{x-2}$

The integrand possesses a discontinuity at $x=2$. The limits of integration are also not finite. I am familiar with finite limits only.

And, what will happen if the integrand is

$\frac{1}{(x-2)(x-3)}$

or

$\frac{1}{(x-2)^2(x-3)}$

?

## 2 Answers

For a given function, its Cauchy Principal Value is defined by: $$p.v. = \lim _{ x\rightarrow \infty }{ \int _{ -x }^{ x }{ f\left( t \right)dt } }$$ So you have to solve the integral between that limits and then solve the limit. You would have to be careful with $x=2$ if you are looking for the area under the function, which can't be the case because that function is not convergent.

• Believe here, the Cauchy principal value at 2 has to be taken before the indefinite integral is taken. Since the limit as $x\rightarrow2$ sums there at zero, the rest of the integral can now be taken. Cauchy principal values are not just taken as the limit goes to infinity. – hkr May 3 '16 at 15:34

I tried something like that.

$\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\int\limits_{-R}^{2-\epsilon}\frac{\text{dx}}{x-2}+\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\int\limits_{2+\epsilon}^{R}\frac{\text{dx}}{x-2}$

$=\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\left.\ln(x-2)\right\rvert_{-R}^{2-\epsilon}+\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\left.\ln(x-2)\right\rvert_{2+\epsilon}^{R}$

$=\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\ln\frac{2-\epsilon-2}{-R-2}+\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\frac{R-2}{2+\epsilon-2}$

$=\lim\limits_{\substack{R\rightarrow\infty\\\epsilon\rightarrow 0}}\ln\left(\frac{-\epsilon}{-R-2}\times\frac{R-2}{\epsilon}\right)$

$=\lim\limits_{R\rightarrow\infty}\ln\left(\frac{R-2}{R+2}\right)$

$=\lim\limits_{R\rightarrow\infty}\ln\left(\frac{1-2/R}{1+2/R}\right)$

$=\ln(1)=0$

Is it correct?