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  "body": "\n## Definition\n\nLet $f$ and $g$ be [[holomorphic functions|0NAHXR1FRH]] on a domain $\\Omega \\subseteq \\C$, let $C$ be a [[contour|06ZGPYTWK8]] in $\\Omega$, and let\n$$I(\\lambda) = \\int_C g(z)\\, e^{\\lambda f(z)}\\,dz, \\qquad \\lambda \u003e 0.$$\nA simple [[saddle point|0JW6MDHAZD]] of $f$ is a point $z_0$ with $f'(z_0) = 0$ and $f''(z_0) \\ne 0$.\nA *steepest descent path* through $z_0$ is a path through $z_0$ on which $\\operatorname{Im} f$ is constant.\nAlong it, $\\operatorname{Re} f$ must decrease strictly as one moves away from $z_0$.\nThe *method of steepest descent* finds the behaviour of $I(\\lambda)$ as $\\lambda \\to \\infty$ in three steps:\n\n1. by Cauchy's theorem, replace $C$ by a contour with the same ends, or ending in the same valleys of $\\operatorname{Re} f$ at infinity.\n   The new contour consists of steepest descent paths through saddle points of $f$ and of pieces where $\\operatorname{Re} f$ is smaller;\n2. on each steepest descent path, where $e^{\\lambda f}$ has constant phase, apply [[Laplace's method|9G7K8DBT9Y]];\n3. keep the contributions of the saddle points with the largest value of $\\operatorname{Re} f$.\n\n## Remarks\n\nLet $z_0$ be a simple saddle with $g(z_0) \\ne 0$, and let $\\theta = \\frac12 \\big(\\pi - \\arg f''(z_0)\\big)$.\nThe angle $\\theta$, defined modulo $\\pi$, is the direction in which the path crosses $z_0$, taken with the orientation of the contour.\nThe contribution of $z_0$ is\n$$g(z_0)\\, e^{\\lambda f(z_0)}\\, e^{i\\theta} \\sqrt{\\frac{2\\pi}{\\lambda \\abs{f''(z_0)}}}\\,\\big(1 + o(1)\\big).$$\nIndeed $f(z_0 + r e^{i\\theta}) \\approx f(z_0) + \\frac12 f''(z_0) r^2 e^{2i\\theta}$, and $f''(z_0) e^{2i\\theta}$ is real and negative.\n\nExample: $I(\\lambda) = \\int_{\\R} e^{i\\lambda x^2}\\,dx$, with $f(z) = i z^2$ and $g = 1$.\nThe only saddle is $z_0 = 0$ with $f''(0) = 2i$, so $\\theta = \\pi/4$.\nOn the line $z = r e^{i\\pi/4}$ one has $f(z) = -r^2$, which is real and decreasing in $\\abs{r}$, so this line is the steepest descent path.\nRotating the contour to it gives exactly $I(\\lambda) = e^{i\\pi/4} \\sqrt{\\pi/\\lambda}$.\n\nNon-example: for $f(z) = z^2$ the real axis passes through the saddle $0$, but $\\operatorname{Re} f = x^2$ increases away from $0$ along it.\nSo it is a path of steepest ascent; the steepest descent path is the imaginary axis.\n",
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