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  "body": "\n## Example\n\nLet $\\omega \u003e 0$, and define the [[Hilbert transform|NSCEBXM7ZM]] of a bounded continuous function $f$ by the [[principal value|0SYR10HD4R]] at $0$ and at $\\infty$\n$$Hf(x) = \\frac{1}{\\pi} \\lim_{\\varepsilon \\to 0^+,\\, R \\to \\infty} \\int_{\\varepsilon \u003c \\abs{u} \u003c R} \\frac{f(x - u)}{u}\\, du.$$\nThen\n$$H[\\cos(\\omega \\cdot)](x) = \\sin(\\omega x), \\qquad H[\\sin(\\omega \\cdot)](x) = -\\cos(\\omega x).$$\nCheck for the cosine: $\\cos(\\omega(x - u)) = \\cos(\\omega x) \\cos(\\omega u) + \\sin(\\omega x) \\sin(\\omega u)$.\nOn the symmetric set $\\varepsilon \u003c \\abs{u} \u003c R$, the odd function $\\cos(\\omega u)/u$ integrates to $0$.\nThe even function $\\sin(\\omega u)/u$ integrates to $2 \\int_{\\varepsilon}^{R} \\frac{\\sin(\\omega u)}{u}\\, du$, which tends to $2 \\cdot \\frac{\\pi}{2} = \\pi$.\nSo $H[\\cos(\\omega \\cdot)](x) = \\frac{1}{\\pi} \\cdot \\pi \\sin(\\omega x) = \\sin(\\omega x)$.\nThe same computation with $\\sin(\\omega(x - u)) = \\sin(\\omega x)\\cos(\\omega u) - \\cos(\\omega x)\\sin(\\omega u)$ gives $-\\cos(\\omega x)$.\nFor $\\omega \u003c 0$ the signs change, since $\\int_0^\\infty \\frac{\\sin(\\omega u)}{u}\\, du = -\\frac{\\pi}{2}$.\n\n## Remarks\n\nThe result agrees with the Fourier multiplier $-i \\sgn(\\omega)$ of the Hilbert transform: $H[e^{i\\omega x}] = -i \\sgn(\\omega)\\, e^{i\\omega x}$.\nApplying $H$ twice gives $H^2 \\cos = -\\cos$, an instance of $H^2 = -\\id$.\nThe function $\\cos(\\omega x) + i \\sin(\\omega x) = e^{i\\omega x}$ is the analytic signal of $\\cos(\\omega x)$.\n\n## Explanations\n\n### L1 human\n\nThe Hilbert transform shifts every wave by a quarter of a period.\nA cosine becomes a sine, and a sine becomes minus a cosine.\nThis is the property that makes it useful in signal processing, where it turns a real oscillation into its complex partner.\n",
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