MHB How can I find the Cauchy principal value of this integral?

[Another1]

How can I find Cauchy principal value. of this integral

\[ n(x) = \int_{a}^{b} \frac{d \omega}{\omega ' ^2 - x^2} \] Where $ a<x<b $

I case $a = 0, b = 3, x = 1$ We get

\[ n(1) = \int_{0}^{3} \frac{d \omega}{\omega ' ^2 - 1^2} = −0.3465735902799727 \] The result shown is the Cauchy principal value.

You can check this answer with https://www.integral-calculator.com/

 

[HOI]

First, you are integrating with respect to "$\omega$" but the integrand has $\omega'$ rather than $\omega$! What is the relationship between $\omega$ and $\omega'$? If that is just a typo, or otherwise irrelevant, the denominator factors as $(\omega- x)(\omega+ x)$ and you can use "partial fractions" to separate them.

 

[HOI]

$\int_a^b \frac{d\omega}{\omega^2- x^2}=\frac{1}{2x} \int_a^b \frac{d\omega}{\omega- x}- \frac{1}{2x} \int_a^b\frac{d\omega}{\omega+ x}$

$= \frac{1}{2x}(ln(b- x)- ln(a- x)- ln(b+ x)+ ln(a+ x))= \frac{1}{2x}ln\left(\frac{(b- x)(a+ x)}{(b+ x)(a- x)}\right)$.

That IS the "Cauchy Principal value" simply because I didn't worry about "limits" if $x= a$ and $x= b$. Any "difficulties" at a or b will cancel.