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  "body": "\n## Example\n\nLet $a$ and $b$ be real constants with $b \\ne 0$, and let $f$ be a given function on the interval $(-1, 1)$.\n*Carleman's equation* asks for a function $\\varphi$ on $(-1, 1)$ with\n$$a\\, \\varphi(x) + \\frac{b}{\\pi}\\, \\mathrm{PV}\\!\\int_{-1}^{1} \\frac{\\varphi(t)}{t - x}\\, dt = f(x), \\quad -1 \u003c x \u003c 1,$$\nwhere $\\mathrm{PV}$ is the [[Cauchy principal value|0SYR10HD4R]].\nIt is a [[singular integral equation|E30EA6ETZ4]] of the second kind with constant coefficients on an arc, and its singular part is the [[finite Hilbert transform|5JYBXH9BP6]].\n\nA check with an explicit solution uses two classical principal value integrals, valid for $-1 \u003c x \u003c 1$:\n$$\\frac{1}{\\pi}\\, \\mathrm{PV}\\!\\int_{-1}^{1} \\frac{\\sqrt{1 - t^2}}{t - x}\\, dt = -x, \\qquad \\frac{1}{\\pi}\\, \\mathrm{PV}\\!\\int_{-1}^{1} \\frac{dt}{\\sqrt{1 - t^2}\\,(t - x)} = 0.$$\nHence $\\varphi(t) = \\sqrt{1 - t^2}$ solves the equation with right side $f(x) = a\\sqrt{1 - x^2} - b x$.\nThe function $(1 - t^2)^{-1/2}$ solves it with right side $f(x) = a (1 - x^2)^{-1/2}$.\n\n## Remarks\n\nPut $\\Phi(z) = \\frac{1}{2\\pi i} \\int_{-1}^{1} \\frac{\\varphi(t)}{t - z}\\, dt$ for $z \\notin [-1, 1]$.\nBy the [[Sokhotski–Plemelj theorem|RNB6QX0GCA]], $\\Phi^+ - \\Phi^- = \\varphi$ and $\\Phi^+ + \\Phi^- = \\frac{1}{\\pi i}\\, \\mathrm{PV}\\!\\int_{-1}^{1} \\frac{\\varphi(t)}{t - x}\\, dt$ on the cut.\nThe equation becomes $(a + ib)\\, \\Phi^+ = (a - ib)\\, \\Phi^- + f$, a [[Riemann–Hilbert problem|RY8YQ9JZF9]] on the segment with constant coefficient $(a - ib)/(a + ib)$ of modulus $1$.\nSolving it gives the solutions in closed form, with powers of $1 - x$ and $1 + x$ at the endpoints whose exponents are fixed by the argument of $a + ib$.\nFor $a = 0$ the equation is the [[airfoil equation|W6Z6CWRE25]].\n\n## Explanations\n\n### L1 human\n\nThis is the simplest singular integral equation on an interval: the unknown appears once by itself and once inside a Cauchy principal value integral.\nThe integral blows up where $t = x$, so it is taken as a principal value.\nBecause the coefficients are constants, the equation can be solved exactly by turning it into a jump problem for holomorphic functions.\n",
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