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  "body": "\n## Statement\n\nLet $n \\ge 0$ be an integer and let $P_n$ be the [[Legendre polynomial|QFYGJVA996]] of degree $n$.\nFor every $x \\in \\C$,\n$$P_n(x) = \\frac{1}{\\pi} \\int_0^\\pi \\left( x + \\sqrt{x^2 - 1} \\, \\cos \\varphi \\right)^n \\, d\\varphi,$$\nwhere $\\sqrt{x^2 - 1}$ is either square root of $x^2 - 1$.\n\n## Remarks\n\nThe choice of square root does not matter: the substitution $\\varphi \\mapsto \\pi - \\varphi$ changes the sign of $\\cos \\varphi$.\nProof idea: multiply by $t^n$ and sum over $n$ for small $t$.\nThe integral $\\int_0^\\pi \\frac{d\\varphi}{a - b \\cos \\varphi} = \\frac{\\pi}{\\sqrt{a^2 - b^2}}$ with $a = 1 - tx$ and $b = t \\sqrt{x^2 - 1}$ gives $(1 - 2xt + t^2)^{-1/2}$.\nThis is the generating function of the Legendre polynomials, so the coefficients agree.\nCheck for $n = 2$: the integral equals $x^2 + \\frac{x^2 - 1}{2} = \\frac{3x^2 - 1}{2} = P_2(x)$, since the mean of $\\cos \\varphi$ is $0$ and the mean of $\\cos^2 \\varphi$ is $\\frac12$.\nFor $-1 \\le x \\le 1$ the integrand is $\\left( x + i \\sqrt{1 - x^2} \\cos \\varphi \\right)^n$, whose modulus is at most $\\left( x^2 + (1 - x^2) \\right)^{n/2} = 1$.\nHence $\\abs{P_n(x)} \\le 1$ on $[-1, 1]$.\n\n## Explanations\n\n### L1 human\n\nThe formula writes each Legendre polynomial as an average of $n$-th powers.\nThe quantity being raised to the power moves along a segment as the angle $\\varphi$ goes from $0$ to $\\pi$.\nIt turns a polynomial into an integral that is easy to bound, which is useful for estimates.\n\n### L2 human\n\nLaplace's integral is one of the standard integral representations of $P_n$, next to Rodrigues' formula and the generating function.\nIts main use is in estimates: it gives the bound $\\abs{P_n(x)} \\le 1$ on $[-1, 1]$ in one line.\nFor $x \u003e 1$ and the positive root, the integrand is largest at $\\varphi = 0$, where it equals $\\left( x + \\sqrt{x^2 - 1} \\right)^n$.\nSo $P_n(x) \\le \\left( x + \\sqrt{x^2 - 1} \\right)^n$, and [[Laplace's method|9G7K8DBT9Y]] gives the precise growth as $n \\to \\infty$.\nThe same idea, with Gegenbauer or Jacobi weights, gives integral representations for those families.\n",
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