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  "body": "\n## Proof\n\nWe substitute $x = \\mu + \\sigma z$ and use that $z \\varphi(z)$ is an odd function.\n\n1. Let $\\mu \\in \\R$, $\\sigma^2 \u003e 0$, and $\\sigma = \\sqrt{\\sigma^2} \u003e 0$. Let $X$ have the [[normal distribution|07RXBVTVZB]] $\\mathcal{N}(\\mu, \\sigma^2)$.\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 $X$.\n3. By the [[law of the unconscious statistician|01ZWX875AT]], $X$ is integrable if $\\int_{\\R} \\abs{x} f(x) \\, dx \u003c \\infty$. In that case $E[X] = \\int_{\\R} x f(x) \\, dx$.\n4. For a continuous function $h \\ge 0$ on $\\R$, the integral $\\int_\\R h \\, dx$ is the limit of $\\int_{a}^{b} h \\, dx$ as $a \\to -\\infty$ and $b \\to \\infty$. So we may compute with [[improper integrals|0VCTBJMEJF]].\n5. For all real $z$, $\\sigma f(\\mu + \\sigma z) = \\varphi(z)$. So, for continuous $g$ and $R \u003e 0$, [[substitution|0D37QFPEZE]] with $x = \\mu + \\sigma z$ gives\n$$\\int_{\\mu - \\sigma R}^{\\mu + \\sigma R} g(x) f(x) \\, dx = \\int_{-R}^{R} g(\\mu + \\sigma z) \\, \\varphi(z) \\, dz.$$\n6. By the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential function|0EX9GFJ0B3]], the function $-\\varphi$ has derivative $z \\varphi(z)$.\n7. By the [[fundamental theorem of calculus|0XYAFCN0PT]], $\\int_0^R z \\varphi(z) \\, dz = \\varphi(0) - \\varphi(R) \\le \\varphi(0) = 1/\\sqrt{2\\pi}$.\n8. The function $\\abs{z} \\varphi(z)$ is even. So $\\int_{-R}^{R} \\abs{z} \\varphi(z) \\, dz = 2 \\int_0^R z \\varphi(z) \\, dz \\le 2/\\sqrt{2\\pi}$.\n9. Since $\\varphi$ is the density of $\\mathcal{N}(0, 1)$, its integral over $\\R$ is $1$. So $\\int_{-R}^R \\varphi(z) \\, dz \\le 1$.\n10. By step 5 with $g(x) = \\abs{x}$ and $\\abs{\\mu + \\sigma z} \\le \\abs{\\mu} + \\sigma \\abs{z}$, we get $\\int_{\\mu - \\sigma R}^{\\mu + \\sigma R} \\abs{x} f(x) \\, dx \\le \\abs{\\mu} + 2\\sigma/\\sqrt{2\\pi}$.\n11. This bound does not depend on $R$. By steps 3 and 4, $\\int_\\R \\abs{x} f(x) \\, dx \u003c \\infty$, so $X$ is integrable.\n12. Write $x f(x)$ as the difference of its positive part and its negative part. Both parts are continuous, nonnegative and at most $\\abs{x} f(x)$.\n13. Applying step 4 to each part gives $E[X] = \\lim_{R \\to \\infty} \\int_{\\mu - \\sigma R}^{\\mu + \\sigma R} x f(x) \\, dx$.\n14. By step 5 with $g(x) = x$, the integral in step 13 equals $\\mu \\int_{-R}^{R} \\varphi(z) \\, dz + \\sigma \\int_{-R}^{R} z \\varphi(z) \\, dz$.\n15. The function $z \\varphi(z)$ is odd. Substituting $z = -u$ on $[-R, 0]$ shows that $\\int_{-R}^{0} z \\varphi(z) \\, dz = -\\int_0^R u \\varphi(u) \\, du$. So $\\int_{-R}^{R} z \\varphi(z) \\, dz = 0$.\n16. As $R \\to \\infty$, $\\int_{-R}^{R} \\varphi(z) \\, dz \\to 1$ by step 9.\n17. By steps 13 to 16, the [[expected value|0DCT0QZVQ1]] is $E[X] = \\mu \\cdot 1 + \\sigma \\cdot 0 = \\mu$. This is the stated integral.\n\n## Idea\n\nThe substitution $x = \\mu + \\sigma z$ reduces everything to the standard normal density $\\varphi$. The integral of $\\varphi$ is $1$, and the integral of $z \\varphi(z)$ is $0$ because the function is odd. What is left is $\\mu$.\n",
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