[proofplan] The proof proceeds in three stages. First, the [Plemelj Formulae](/theorems/556) convert the singular integral equation into a sum-type boundary condition $C^+(x) + C^-(x) = -ig(x)$ for the Cauchy transform $C$ of $f$. Second, an auxiliary function $\varphi(z) = \sqrt{z^2 - 1}$ with anti-symmetric boundary values $\varphi^+ = -\varphi^-$ converts the sum-type condition into a jump-type problem for the product $w = \varphi C$. Third, the [Solution Of The Jump Problem](/theorems/557) produces $w$ explicitly, and the Plemelj jump formula recovers $f$ with an arbitrary constant from the kernel of the Hilbert transform on $[-1,1]$. [/proofplan] [step:Reduce the integral equation to a sum-type boundary condition via the Plemelj formulae] The integral equation $\frac{1}{\pi}\operatorname{p.v.}\!\!\int_{-1}^{1}\frac{f(t)}{t-x}\,dt = g(x)$ reads $2iH(x) = g(x)$ in the notation of the Cauchy--Hilbert transform framework, where $H$ is the Hilbert transform of $f$ on $[-1,1]$. The sum formula from the [Plemelj Formulae](/theorems/556) gives $C^+(x) + C^-(x) = 2H(x)$, where $C(z) = \frac{1}{2\pi i}\int_{-1}^{1}\frac{f(t)}{t-z}\,dt$. Substituting $H(x) = -ig(x)/2$: \begin{align*} C^+(x) + C^-(x) = -ig(x), \qquad -1 < x < 1. \end{align*} [/step] [step:Introduce the auxiliary function $\varphi(z) = \sqrt{z^2 - 1}$ and convert to a jump problem] Define $\varphi(z) = \sqrt{z^2 - 1}$ with the branch cut along $[-1,1]$, chosen so that $\varphi(z) \sim +z$ as $z \to \infty$. To compute the boundary values, introduce local polar coordinates: $r_1 = |z - 1|$, $\theta_1 = \arg(z - 1)$, $r_2 = |z + 1|$, $\theta_2 = \arg(z + 1)$ with $-\pi \leq \theta_j \leq \pi$. Then $\varphi(z) = (r_1 r_2)^{1/2} e^{i(\theta_1 + \theta_2)/2}$. On the upper side of the cut ($-1 < x < 1$, approaching from above): $\theta_1 = \pi$, $\theta_2 = 0$, giving: \begin{align*} \varphi^+(x) = \sqrt{(1-x)(1+x)} \cdot e^{i\pi/2} = i\sqrt{1-x^2}. \end{align*} On the lower side: $\theta_1 = -\pi$, $\theta_2 = 0$, giving $\varphi^-(x) = -i\sqrt{1-x^2}$. Therefore $\varphi^+(x) = -\varphi^-(x)$. Define $w(z) := \varphi(z) C(z)$. The boundary values of $w$ satisfy: \begin{align*} w^+(x) - w^-(x) &= \varphi^+(x) C^+(x) - \varphi^-(x) C^-(x) \\ &= \varphi^+(x) C^+(x) + \varphi^+(x) C^-(x) \\ &= \varphi^+(x)(C^+(x) + C^-(x)) \\ &= i\sqrt{1-x^2} \cdot (-ig(x)) = \sqrt{1-x^2}\,g(x), \end{align*} where the second line uses $\varphi^-(x) = -\varphi^+(x)$. [/step] [step:Solve the jump problem for $w$ and recover the inversion formula for $f$] By the [Solution Of The Jump Problem](/theorems/557) (and its generalisation allowing $O(|z|^n)$ growth), the function $w$ satisfying $w^+(x) - w^-(x) = \sqrt{1-x^2}\,g(x)$ and $w(z) = O(1)$ as $z \to \infty$ is: \begin{align*} w(z) = \frac{1}{2\pi i}\int_{-1}^{1}\frac{\sqrt{1-t^2}\,g(t)}{t-z}\,dt + A_1, \end{align*} where $A_1 \in \mathbb{C}$ is a constant. The growth constraint $w(z) = O(1)$ follows from $\varphi(z) = O(z)$ and $C(z) = O(1/z)$ (the latter by the [Analyticity Of The Cauchy Transform](/theorems/555)). The Cauchy-type integral is $O(1/z)$, so only a constant polynomial is permitted. To recover $f(x)$, use the jump formula $f(x) = C^+(x) - C^-(x)$ from the [Plemelj Formulae](/theorems/556). Since $w^\pm(x) = \varphi^\pm(x) C^\pm(x) = \pm i\sqrt{1-x^2}\,C^\pm(x)$, we have $C^\pm(x) = w^\pm(x)/(\pm i\sqrt{1-x^2})$, and: \begin{align*} f(x) = C^+(x) - C^-(x) = \frac{w^+(x)}{i\sqrt{1-x^2}} + \frac{w^-(x)}{i\sqrt{1-x^2}} = \frac{1}{i\sqrt{1-x^2}}(w^+(x) + w^-(x)). \end{align*} The sum formula from the [Plemelj Formulae](/theorems/556) applied to $w$ gives $w^+(x) + w^-(x) = 2H_w(x)$, where $H_w$ is the Hilbert transform of the jump data $\sqrt{1-t^2}\,g(t)$. The constant $A_1$ contributes equally to $w^+$ and $w^-$ (it has no jump), adding $2A_1/(i\sqrt{1-x^2})$ to $f(x)$. Writing $A = 2A_1/i$ and computing: \begin{align*} f(x) &= \frac{1}{i\sqrt{1-x^2}} \cdot \frac{1}{\pi i}\operatorname{p.v.}\!\!\int_{-1}^{1}\frac{\sqrt{1-t^2}\,g(t)}{t-x}\,dt + \frac{A}{\sqrt{1-x^2}} \\ &= \frac{-1}{\pi\sqrt{1-x^2}}\operatorname{p.v.}\!\!\int_{-1}^{1}\frac{\sqrt{1-t^2}\,g(t)}{t-x}\,dt + \frac{A}{\sqrt{1-x^2}}. \end{align*} The constant $A$ is arbitrary because $A/\sqrt{1-x^2}$ lies in the kernel of the Hilbert transform on $[-1,1]$: one can verify directly that $\operatorname{p.v.}\!\!\int_{-1}^{1}\frac{1}{(t-x)\sqrt{1-t^2}}\,dt = 0$ for $-1 < x < 1$. [guided] Why does the constant term $A_1$ produce the homogeneous solution $A/\sqrt{1-x^2}$? The constant $A_1$ adds to $w(z)$ without affecting the jump $w^+ - w^-$ (a constant has zero jump). But it does affect $f$: the recovery formula involves $w^+ + w^-$, and the constant contributes $2A_1$ to this sum. Dividing by $i\sqrt{1-x^2}$ and writing $A = 2A_1/i$ gives the homogeneous term. Why is $A/\sqrt{1-x^2}$ in the kernel of the Hilbert transform? The principal value integral $\operatorname{p.v.}\!\!\int_{-1}^{1}\frac{dt}{(t-x)\sqrt{1-t^2}}$ can be evaluated by contour integration or by the substitution $t = \cos\theta$, which transforms it to $\operatorname{p.v.}\!\!\int_0^\pi \frac{-\sin\theta\,d\theta}{(\cos\theta - x)\sin\theta} = -\int_0^\pi \frac{d\theta}{\cos\theta - x}$. This integral equals zero by the residue calculus (the integrand, viewed as a function of $e^{i\theta}$, has a pole inside the unit circle whose residue contributes exactly $\pi$ to the integral from $0$ to $\pi$ and $-\pi$ from $\pi$ to $2\pi$, but the integral from $0$ to $\pi$ alone vanishes by the symmetry of the Chebyshev weight). [/guided] [/step]