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  "body": "\n## Proof\n\n1. Let $a$, $R$, $C_R$, $g$, $f$ and $M_R$ be as in the statement. By the definition of the [[contour integral|00FWNP9TXY]], $$\\int_{C_R} f(z) \\, dz = \\int_0^\\pi e^{iaRe^{i\\theta}} \\, g(Re^{i\\theta}) \\, iRe^{i\\theta} \\, d\\theta.$$\n2. For real $t$, $\\abs{e^{it}} = \\abs{\\cos t + i \\sin t} = 1$, by [[Euler's formula|04R3CY41VJ]] and the [[Pythagorean identity|03H7WZ4T3K]].\n3. We have $iaRe^{i\\theta} = -aR\\sin\\theta + iaR\\cos\\theta$. By the [[product of exponentials|0WM7DJCM6G]], step 2 and the [[multiplicativity of the modulus|0HYNZE940T]], $\\abs{e^{iaRe^{i\\theta}}} = e^{-aR\\sin\\theta}$.\n4. So the modulus of the integrand in step 1 is at most $e^{-aR\\sin\\theta} M_R R$, since $\\abs{iRe^{i\\theta}} = R$.\n5. For a continuous complex function $k$ on $[0, \\pi]$, write $\\int_0^\\pi k = \\abs{\\int_0^\\pi k} \\, e^{i\\alpha}$ with $\\alpha$ real. Then $\\abs{\\int_0^\\pi k} = \\int_0^\\pi \\operatorname{Re}(e^{-i\\alpha} k) \\le \\int_0^\\pi \\abs{k}$, by [[monotonicity of the integral|0MBWQKM8QR]].\n6. By steps 4 and 5, $\\abs{\\int_{C_R} f(z) \\, dz} \\le R M_R \\int_0^\\pi e^{-aR\\sin\\theta} \\, d\\theta$.\n7. The [[substitution|0D37QFPEZE]] $\\theta \\mapsto \\pi - \\theta$ and $\\sin(\\pi - \\theta) = \\sin\\theta$ give $\\int_{\\pi/2}^{\\pi} e^{-aR\\sin\\theta} \\, d\\theta = \\int_0^{\\pi/2} e^{-aR\\sin\\theta} \\, d\\theta$. So the integral in step 6 is $2 \\int_0^{\\pi/2} e^{-aR\\sin\\theta} \\, d\\theta$.\n8. Jordan's inequality: $\\sin\\theta \\ge 2\\theta/\\pi$ for $\\theta \\in [0, \\pi/2]$. Let $h(\\theta) = \\sin\\theta - 2\\theta/\\pi$. Then $h(0) = h(\\pi/2) = 0$, and $h'(\\theta) = \\cos\\theta - 2/\\pi$ is strictly decreasing on $[0, \\pi/2]$.\n9. By the [[mean value theorem|0BARQY52N5]] there is $\\xi \\in (0, \\pi/2)$ with $h'(\\xi) = 0$. So $h' \\ge 0$ on $[0, \\xi]$ and $h' \\le 0$ on $[\\xi, \\pi/2]$.\n10. By [[nonnegative derivative implies increasing|026BJBQNGJ]], applied to $h$ on $[0, \\xi]$ and to $-h$ on $[\\xi, \\pi/2]$, we get $h \\ge h(0) = 0$ and $h \\ge h(\\pi/2) = 0$. This proves Jordan's inequality.\n11. Since $a R \u003e 0$ and the [[exponential function is increasing|0KN2V7840E]], step 10 gives $e^{-aR\\sin\\theta} \\le e^{-2aR\\theta/\\pi}$ on $[0, \\pi/2]$.\n12. By monotonicity of the integral, $\\int_0^{\\pi/2} e^{-aR\\sin\\theta} \\, d\\theta \\le \\int_0^{\\pi/2} e^{-2aR\\theta/\\pi} \\, d\\theta = \\frac{\\pi}{2aR} \\bigl( 1 - e^{-aR} \\bigr) \\le \\frac{\\pi}{2aR}$.\n13. Combining steps 6, 7 and 12, $\\abs{\\int_{C_R} f(z) \\, dz} \\le R M_R \\cdot 2 \\cdot \\frac{\\pi}{2aR} = \\frac{\\pi}{a} M_R$.\n\n## Idea\n\nOn the upper semicircle, $\\abs{e^{iaz}} = e^{-aR\\sin\\theta}$ is very small except near the two ends, where $\\sin\\theta$ is close to $0$. Jordan's inequality $\\sin\\theta \\ge 2\\theta/\\pi$ bounds this factor by an exponential that integrates to about $\\pi/(2aR)$. This factor $1/R$ cancels the length $\\pi R$ of the semicircle.\n",
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