Cauchy's formula on function at the pole not inside

Question

How can one define the integral when the pole is on the integration contour?

$$\int_{|w|=1}\dfrac{1}{i-w}\, dw$$

Answer 1

The integral of interest $I=\oint_{|w|=1}\frac{1}{i-w}\,dw$ does not exist since the pole of the integrand intersects the integration contour.

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However, the Cauchy Principal value of the integral does exist and is given by

$$\begin{align} \text{PV}\left(\oint_{|w|=1}\frac{1}{i-w}\,dw \right)&\equiv \lim_{\varepsilon\to 0^+}\int_\pi^{2\pi}\frac{i\varepsilon e^{i\theta}}{i-(i+\varepsilon e^{i\theta})}\,d\theta\\\\ &=-i\pi \end{align}$$

We defined the Cauchy Principal Value to be the limit of integration over a deformed unit circle as the arc length of the deformation approaches $0$. Here, the deformation is an "infinitesimal semi-circular dent" at $w=i$, parameterized by $w_{\text{dent}}=i+\varepsilon e^{i\theta}$, $\pi\le \theta\le 2\pi$.

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Alternatively and equivalently, we can define the Cauchy Principal Values as the limit of the simple arithmetic average of the integrals over circular contours of raddii $1+\varepsilon$ and $1-\varepsilon$ as $\varepsilon\to 0$. Proceeding, we find that

$$\begin{align} \text{PV}\left(\oint_{|w|=1}\frac{1}{i-w}\,dw \right)&\equiv \frac12\lim_{\varepsilon\to 0^+}\left(\underbrace{\oint_{|w|=1+\varepsilon}\frac{1}{i-w}\,dw}_{=-2\pi i}+\underbrace{\oint_{|w|=1-\varepsilon}\frac{1}{i-w}\,dw}_{=0}\right)\\\\ &=-i\pi \end{align}$$

as expected!