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  "body": "\n## Proof\n\nWe compute the moments of the standard normal law and expand the cosine in its Taylor polynomials.\n\n1. Let $\\mu \\in \\R$, $\\sigma^2 \u003e 0$, and $\\sigma = \\sqrt{\\sigma^2} \u003e 0$. Let $X \\sim \\mathcal{N}(\\mu, \\sigma^2)$ be [[normal|07RXBVTVZB]], and fix $t \\in \\R$.\n2. Let $f = \\varphi_{\\mu, \\sigma^2}$ and $\\varphi = \\varphi_{0,1}$ be the [[normal densities|027JF1FBB0]]. By definition, $f$ is a [[probability density function|04YG690N5V]] of $X$, and $\\int_\\R \\varphi = 1$.\n3. By definition, the [[characteristic function|0MYDNRT7PP]] is $\\varphi_X(t) = E[\\cos(tX)] + i \\, E[\\sin(tX)]$. Both expectations exist, since the variables are bounded.\n4. By the [[law of the unconscious statistician|01ZWX875AT]], $E[g(X)] = \\int_\\R g(x) f(x) \\, dx$ for each bounded continuous $g$. These integrals converge absolutely as [[improper integrals|0VCTBJMEJF]].\n5. For all real $z$, $\\sigma f(\\mu + \\sigma z) = \\varphi(z)$. With [[substitution|0D37QFPEZE]] $x = \\mu + \\sigma z$ on bounded intervals, step 4 gives $E[g(X)] = \\int_\\R g(\\mu + \\sigma z) \\varphi(z) \\, dz$.\n\nFirst, the integrals $\\int_\\R \\abs{z}^k \\varphi(z) \\, dz$ converge.\n\n6. Let $k \\ge 0$ be an integer. By the [[exponential series|06JZGMYTYD]], $e^{z^2/4} \\ge (z^2/4)^k / k!$. So $\\abs{z}^k e^{-z^2/4} \\le k! \\, 4^k$ for $\\abs{z} \\ge 1$, and it is at most $1$ for $\\abs{z} \u003c 1$.\n7. Since $(\\abs{z}/2 - 1)^2 \\ge 0$, we have $z^2/4 \\ge \\abs{z} - 1$. So $e^{-z^2/4} \\le e \\cdot e^{-\\abs{z}}$.\n8. By steps 6 and 7, $\\abs{z}^k \\varphi(z) \\le C_k e^{-\\abs{z}}$ with $C_k = k! \\, 4^k e$. By the [[comparison test|0FR1BDKK10]] and the [[integral of $e^{-x}$|039JAF7ZAN]], $\\int_\\R \\abs{z}^k \\varphi(z) \\, dz$ converges.\n9. Step 8 with $k + 1$ in place of $k$ gives $\\abs{z}^k \\varphi(z) \\le C_{k+1} e^{-\\abs{z}}/\\abs{z}$ for $z \\ne 0$. So $\\abs{z}^k \\varphi(z) \\to 0$ as $\\abs{z} \\to \\infty$.\n\nNext, the even moments.\n\n10. For $k \\ge 0$, let $m_{2k} = \\int_\\R z^{2k} \\varphi(z) \\, dz$. Then $m_0 = 1$ by step 2.\n11. By the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]], $\\varphi'(z) = -z \\varphi(z)$.\n12. [[Integration by parts|08Y1BBQTK9]] with $u(z) = z^{2k+1}$ and $v(z) = -\\varphi(z)$ gives, for $R \u003e 0$,\n$$\\int_{-R}^{R} z^{2k+2} \\varphi(z) \\, dz = -2R^{2k+1} \\varphi(R) + (2k + 1) \\int_{-R}^{R} z^{2k} \\varphi(z) \\, dz.$$\n13. Let $R \\to \\infty$. By step 9, the boundary term tends to $0$. So $m_{2k+2} = (2k + 1) \\, m_{2k}$.\n14. By induction on $k$, $m_{2k} = (2k)!/(2^k k!)$. Indeed, $(2k+1) (2k)!/(2^k k!) = (2k+2)!/(2^{k+1} (k+1)!)$, since $(2k+2)/(2(k+1)) = 1$.\n\nNow the standard normal case.\n\n15. For real $s$, let $S(s) = \\int_\\R \\sin(sz) \\varphi(z) \\, dz$ and $C(s) = \\int_\\R \\cos(sz) \\varphi(z) \\, dz$. Both converge absolutely by step 8 with $k = 0$.\n16. The function $\\sin(sz) \\varphi(z)$ is odd in $z$. Substituting $z = -u$ on $(-\\infty, 0]$ shows that $S(s) = 0$.\n17. Fix $N \\ge 0$. By [[Taylor's theorem with the Lagrange remainder|0H5ZAZCX2T]] for $\\cos$ at $0$, of degree $2N + 1$, we get for real $y$\n$$\\abs{\\cos y - \\sum_{k=0}^{N} \\frac{(-1)^k y^{2k}}{(2k)!}} \\le \\frac{y^{2N+2}}{(2N+2)!}.$$\nHere we used that all derivatives of $\\cos$ are $\\pm\\sin$ or $\\pm\\cos$, so they are bounded by $1$.\n18. Put $y = sz$, multiply by $\\varphi(z) \u003e 0$, and integrate. By [[monotonicity of the integral|0MBWQKM8QR]] on bounded intervals and a limit, step 14 gives\n$$\\abs{C(s) - \\sum_{k=0}^{N} \\frac{(-1)^k s^{2k} m_{2k}}{(2k)!}} \\le \\frac{s^{2N+2} m_{2N+2}}{(2N+2)!} = \\frac{(s^2/2)^{N+1}}{(N+1)!}.$$\n19. By step 14, the sum in step 18 equals $\\sum_{k=0}^{N} (-s^2/2)^k / k!$. By the [[exponential series|06JZGMYTYD]], it tends to $e^{-s^2/2}$ as $N \\to \\infty$.\n20. The right side of step 18 is a term of the convergent series for $e^{s^2/2}$. So it tends to $0$ as $N \\to \\infty$.\n21. By steps 18 to 20, $C(s) = e^{-s^2/2}$ for all real $s$.\n\nFinally, the general case.\n\n22. By step 5 and the [[angle addition formulas|028XEQCE88]], $E[\\cos(tX)] = \\cos(\\mu t) \\, C(\\sigma t) - \\sin(\\mu t) \\, S(\\sigma t)$.\n23. In the same way, $E[\\sin(tX)] = \\sin(\\mu t) \\, C(\\sigma t) + \\cos(\\mu t) \\, S(\\sigma t)$.\n24. By steps 16 and 21, $E[\\cos(tX)] = \\cos(\\mu t) \\, e^{-\\sigma^2 t^2/2}$ and $E[\\sin(tX)] = \\sin(\\mu t) \\, e^{-\\sigma^2 t^2/2}$.\n25. By step 3 and [[Euler's formula|04R3CY41VJ]], $\\varphi_X(t) = (\\cos(\\mu t) + i \\sin(\\mu t)) \\, e^{-\\sigma^2 t^2/2} = e^{i\\mu t} e^{-\\sigma^2 t^2/2}$.\n26. By the [[product of exponentials|0WM7DJCM6G]], $e^{w} e^{w'} = e^{w + w'}$ for complex $w$ and $w'$. So $\\varphi_X(t) = \\exp\\left(i\\mu t - \\tfrac{1}{2}\\sigma^2 t^2\\right)$.\n27. For $\\mu = 0$ and $\\sigma^2 = 1$, this gives $\\varphi(t) = e^{-t^2/2}$ for the standard normal distribution.\n\n## Idea\n\nThe sine part vanishes because the standard normal density is symmetric. For the cosine part, expand $\\cos(sz)$ in its Taylor polynomials and integrate term by term. The even moments $(2k)!/(2^k k!)$ turn the cosine series into the series of $e^{-s^2/2}$, and the Taylor remainder controls the error.\n",
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