[proofplan] We derive the Plemelj formulae by deforming the integration contour: excise a small interval of length $2\varepsilon$ around $z_0$ and replace it with a semicircular arc of radius $\varepsilon$. On the left side the semicircle sweeps an angle of $+\pi$ and on the right side $-\pi$, producing the half-residue terms $\pm \frac{1}{2}f(z_0)$ via Taylor expansion. The integral over the excised contour converges to the Hilbert transform $H(z_0)$. The sum and jump formulae are immediate algebraic consequences. [/proofplan] [step:Compute the boundary value $C^+(z_0)$ from the left of $\gamma$] Deform the contour to $\gamma_\varepsilon + c_\varepsilon^+$, where $\gamma_\varepsilon$ is $\gamma$ with a section of length $2\varepsilon$ about $z_0$ removed, and $c_\varepsilon^+$ is a semicircle of centre $z_0$ and radius $\varepsilon$ lying to the left of $\gamma$. Since the integrand $f(\xi)/(\xi - z)$ is analytic between the original and deformed contours (for $z$ on the left side of $\gamma$ near $z_0$), Cauchy's integral theorem justifies the deformation: \begin{align*} C^+(z_0) = \lim_{\varepsilon \to 0} \frac{1}{2\pi i}\int_{\gamma_\varepsilon + c_\varepsilon^+} \frac{f(\xi)}{\xi - z_0}\,d\xi. \end{align*} On $c_\varepsilon^+$, parametrise $\xi = z_0 + \varepsilon e^{i\theta}$ for $\theta_0 \leq \theta \leq \theta_0 + \pi$ (where $\theta_0$ depends on the local orientation of $\gamma$ at $z_0$). Then $d\xi = i\varepsilon e^{i\theta}\,d\theta$ and $\xi - z_0 = \varepsilon e^{i\theta}$, so: \begin{align*} \int_{c_\varepsilon^+} \frac{f(\xi)}{\xi - z_0}\,d\xi = \int_{\theta_0}^{\theta_0 + \pi} \frac{f(z_0 + \varepsilon e^{i\theta})}{\varepsilon e^{i\theta}} \cdot i\varepsilon e^{i\theta}\,d\theta = i\int_{\theta_0}^{\theta_0 + \pi} f(z_0 + \varepsilon e^{i\theta})\,d\theta. \end{align*} Since $f$ is continuous at $z_0$, $f(z_0 + \varepsilon e^{i\theta}) = f(z_0) + O(\varepsilon)$ uniformly in $\theta$. Integrating over the interval of length $\pi$: \begin{align*} \lim_{\varepsilon \to 0} \int_{c_\varepsilon^+} \frac{f(\xi)}{\xi - z_0}\,d\xi = i\pi f(z_0). \end{align*} The integral over $\gamma_\varepsilon$ converges by definition to the Cauchy principal value: \begin{align*} \lim_{\varepsilon \to 0} \frac{1}{2\pi i}\int_{\gamma_\varepsilon} \frac{f(\xi)}{\xi - z_0}\,d\xi = H(z_0). \end{align*} Combining the two contributions: \begin{align*} C^+(z_0) = H(z_0) + \frac{1}{2\pi i} \cdot i\pi f(z_0) = H(z_0) + \frac{1}{2}f(z_0). \end{align*} [guided] The contour deformation replaces the original contour $\gamma$ (along which $z_0$ is a singular point of the integrand) with a detour that avoids $z_0$. Why is this valid? The function $f(\xi)/(\xi - z)$ is analytic in $\xi$ on the region between $\gamma$ and $\gamma_\varepsilon + c_\varepsilon^+$, provided $z$ is on the left side of $\gamma$ and close enough to $z_0$. Cauchy's integral theorem then guarantees that the integral over the original contour equals the integral over the deformed contour. The semicircle computation is the key step. On $c_\varepsilon^+$, the $\varepsilon$ in the denominator $\xi - z_0 = \varepsilon e^{i\theta}$ cancels the $\varepsilon$ in $d\xi = i\varepsilon e^{i\theta}\,d\theta$, leaving a finite integral $i\int f(z_0 + \varepsilon e^{i\theta})\,d\theta$. As $\varepsilon \to 0$, continuity of $f$ at $z_0$ gives $f(z_0 + \varepsilon e^{i\theta}) \to f(z_0)$ uniformly, so the integral converges to $i\pi f(z_0)$ (the angular extent is $\pi$, not $2\pi$, because we take a semicircle, not a full circle). Division by $2\pi i$ produces the factor $\frac{1}{2}f(z_0)$. [/guided] [/step] [step:Compute the boundary value $C^-(z_0)$ from the right of $\gamma$] The argument is analogous to the left boundary value, with the semicircle $c_\varepsilon^-$ now lying on the right side of $\gamma$. The key difference is the orientation: the semicircle sweeps from $\theta_0$ to $\theta_0 - \pi$ (clockwise), so the angular integral runs over an interval of length $\pi$ in the negative direction. The same parametrisation and continuity argument give: \begin{align*} \lim_{\varepsilon \to 0} \int_{c_\varepsilon^-} \frac{f(\xi)}{\xi - z_0}\,d\xi = -i\pi f(z_0), \end{align*} where the sign change arises because the semicircle is traversed clockwise, reversing the sign of the angular integral. The excised contour integral still converges to $2\pi i \cdot H(z_0)$. Combining: \begin{align*} C^-(z_0) = H(z_0) + \frac{1}{2\pi i} \cdot (-i\pi f(z_0)) = H(z_0) - \frac{1}{2}f(z_0). \end{align*} [/step] [step:Derive the sum and jump formulae by adding and subtracting] Adding the results of the two previous steps: \begin{align*} C^+(z_0) + C^-(z_0) = \left(H(z_0) + \tfrac{1}{2}f(z_0)\right) + \left(H(z_0) - \tfrac{1}{2}f(z_0)\right) = 2H(z_0). \end{align*} Subtracting: \begin{align*} C^+(z_0) - C^-(z_0) = \left(H(z_0) + \tfrac{1}{2}f(z_0)\right) - \left(H(z_0) - \tfrac{1}{2}f(z_0)\right) = f(z_0). \end{align*} [/step]