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  "body": "\n## Example\n\nLet $E_1(x) = \\int_1^\\infty e^{-x\\mu}\\, \\mu^{-1}\\, d\\mu$ for $x \u003e 0$ be the [[exponential integral|EV6ZDQ0NDY]], and more generally $E_n(x) = \\int_1^\\infty e^{-x\\mu}\\, \\mu^{-n}\\, d\\mu$.\nThe *Milne problem* asks for a nonzero function $J$ on $[0, \\infty)$ of at most linear growth with\n$$J(\\tau) = \\frac{1}{2} \\int_0^\\infty E_1(\\abs{\\tau - t})\\, J(t)\\, dt, \\quad \\tau \\ge 0.$$\nThis is a homogeneous [[Wiener–Hopf equation|QYY4NMDY5H]] with kernel $k(x) = \\frac{1}{2} E_1(\\abs{x})$.\nIn astrophysics $\\tau$ is the optical depth in a plane-parallel grey atmosphere in radiative equilibrium, and $J$ is the mean intensity.\n\nCheck of the main features.\nThe kernel has total mass $1$: $\\int_{\\R} \\frac{1}{2} E_1(\\abs{x})\\, dx = \\int_1^\\infty \\mu^{-2}\\, d\\mu = 1$.\nSo on the whole line every linear function is a solution, and the cut at $\\tau = 0$ makes the problem nontrivial.\nUsing $E_n' = -E_{n-1}$ for $n \\ge 2$, one computes for $\\tau \\ge 0$\n$$\\frac{1}{2} \\int_0^\\infty E_1(\\abs{\\tau - t})\\, dt = 1 - \\frac{1}{2} E_2(\\tau), \\qquad \\frac{1}{2} \\int_0^\\infty E_1(\\abs{\\tau - t})\\, t\\, dt = \\tau + \\frac{1}{2} E_3(\\tau).$$\nHence the Eddington guess $J = \\tau + \\frac{2}{3}$ gives a right side that differs from $J$ by $\\frac{1}{2} E_3(\\tau) - \\frac{1}{3} E_2(\\tau)$.\nThis error tends to $0$ as $\\tau \\to \\infty$, but at $\\tau = 0$ it is $\\frac{1}{4} - \\frac{1}{3} = -\\frac{1}{12}$, since $E_2(0) = 1$ and $E_3(0) = \\frac{1}{2}$.\n\n## Remarks\n\nThe exact solution, up to a constant factor, is $J(\\tau) = \\tau + q(\\tau)$, where $q$ is the Hopf function.\nIt is found by the [[Wiener–Hopf method|H4NXEY6QTZ]].\nThe function $q$ increases from $q(0) = 1/\\sqrt{3}$ to a limit $q(\\infty) \\approx 0.7104$.\nThe Fourier symbol of the kernel is $\\frac{\\arctan \\xi}{\\xi}$, which equals $1$ at $\\xi = 0$, so $1 - \\hat k$ vanishes at $0$; this zero is why the solution grows linearly.\n\n## Explanations\n\n### L1 human\n\nLight travels through the outer layers of a star, is absorbed and is re-emitted in all directions.\nIn equilibrium, the light at each depth is an average of the light at all other depths, weighted by how much of it survives the trip.\nDeep inside, the intensity grows linearly with depth, but near the surface there is a correction because no light comes in from outside.\nThe Milne problem asks for this correction, and its exact answer was one of the first successes of the Wiener–Hopf method.\n",
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