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  "body": "\n## Summary\n\nThe [[residue theorem|032HKYS329]] gives the values of many definite integrals of real functions.\nFor an integral $\\int_0^{2\\pi} R(\\cos \\theta, \\sin \\theta) \\, d\\theta$ with $R$ a rational function of two variables, the substitution $z = e^{i\\theta}$ changes it into a [[contour integral|00FWNP9TXY]] over the unit circle.\nThis substitution uses $\\cos \\theta = \\frac{1}{2}(z + z^{-1})$, $\\sin \\theta = \\frac{1}{2i}(z - z^{-1})$ and $d\\theta = \\frac{dz}{iz}$.\nFor an integral $\\int_{-\\infty}^{\\infty} f(x) \\, dx$, one closes the segment $[-R, R]$ with a semicircle of radius $R$ in the upper half-plane and lets $R \\to \\infty$.\nThis method gives the integral when the integral over the semicircle tends to $0$, for example by the [[estimation lemma|0DJMGPN8K9]] or by [[Jordan's lemma|0PNPG2PP9H]].\n\n## Remarks\n\nFor example, $\\frac{1}{1 + z^2}$ has one pole $i$ in the upper half-plane, with residue $\\frac{1}{2i}$, so $\\int_{-\\infty}^{\\infty} \\frac{dx}{1 + x^2} = 2\\pi i \\cdot \\frac{1}{2i} = \\pi$.\nWith Jordan's lemma, the same method gives $\\int_{-\\infty}^{\\infty} \\frac{e^{itx}}{x^2 + 1} \\, dx = \\pi e^{-|t|}$ for each real $t$.\nWhen the integrand has a pole on the real axis, the method usually gives a [[Cauchy principal value|0V3RDQV3ZX]].\n\n## Explanations\n\n### L1 human\n\nA hard real integral can become easy when you view the real line as part of a closed curve in the complex plane.\nThe residue theorem gives the integral over the closed curve from a few residues.\nThen you show that the extra part of the curve contributes nothing in the limit.\n",
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        "locator": "Using the method of residues",
        "url": "https://en.wikipedia.org/wiki/Contour_integration"
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        "locator": "Example 3a – trigonometric integrals, the general procedure",
        "url": "https://en.wikipedia.org/wiki/Contour_integration"
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  "msc": [
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  "name": "Evaluation of real integrals by residues",
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