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  "body": "\n## Proof\n\nWe find an antiderivative of $1/x$ on each half line and apply the fundamental theorem of calculus.\n\n1. Let $f(x) = 1/x$ for $x \\ne 0$. The function $x \\mapsto x$ is differentiable and nonzero at each $x \\ne 0$. By the [[reciprocal rule|0QD4E5JYPT]], $f$ is differentiable at each $x \\ne 0$.\n2. [[Differentiable functions are continuous|07ZMX4ANJS]], so $f$ is continuous on $(0, \\infty)$ and on $(-\\infty, 0)$.\n3. Let $I$ be one of the open intervals $(0, \\infty)$ or $(-\\infty, 0)$. Let $F \\colon I \\to \\R$ be differentiable with $F' = f$ on $I$. We claim that $\\int_a^b \\frac{dx}{x} = F(b) - F(a)$ for all $a, b \\in I$.\n4. Proof of the claim for $a \u003c b$. The interval $[a, b]$ lies in $I$. By step 2, $f$ is [[Riemann integrable|06SESV7GQY]] on $[a, b]$. Also $F$ is differentiable on $I$, so it is continuous on $[a, b]$ by the result cited in step 2. The [[second fundamental theorem of calculus|0XYAFCN0PT]] gives $\\int_a^b f = F(b) - F(a)$.\n5. Proof of the claim for $a = b$ and $a \u003e b$. By the convention for the [[definite integral|021WEM680W]], $\\int_a^a f = 0 = F(a) - F(a)$. For $a \u003e b$, $\\int_a^b f = -\\int_b^a f = -(F(a) - F(b)) = F(b) - F(a)$, by step 4.\n6. Case $a, b \u003e 0$. Take $I = (0, \\infty)$ and $F = \\ln$, the [[natural logarithm|0044CPHJBV]]. By the [[derivative of the logarithm|07X1QA6BRW]], $F' = f$ on $I$.\n7. By steps 3 and 6, $\\int_a^b \\frac{dx}{x} = \\ln b - \\ln a$. By the [[logarithm of a quotient|0J7PHRBNNS]], this equals $\\ln \\frac{b}{a}$.\n8. Case $a, b \u003c 0$. Take $I = (-\\infty, 0)$ and $F(x) = \\ln(-x)$. Let $\\varphi(x) = -x$. By the [[derivative of a polynomial|0MH1WK89PP]], $\\varphi'(x) = -1$, and $\\varphi$ maps $I$ into $(0, \\infty)$.\n9. By the [[chain rule|0RXX2PT1MQ]] and the derivative of the logarithm, $F$ is differentiable on $I$ and $F'(x) = \\frac{1}{-x} \\cdot (-1) = \\frac{1}{x}$.\n10. By steps 3 and 9, $\\int_a^b \\frac{dx}{x} = \\ln(-b) - \\ln(-a)$. Since $-a, -b \u003e 0$, the logarithm of a quotient gives $\\ln \\frac{-b}{-a} = \\ln \\frac{b}{a}$.\n11. For the last claim, let $x \u003e 0$ and apply step 7 with $a = 1$ and $b = x$. This gives $\\int_1^x \\frac{dt}{t} = \\ln \\frac{x}{1} = \\ln x$.\n\n## Idea\n\nOn the positive half line, $\\ln x$ has derivative $1/x$. On the negative half line, $\\ln(-x)$ has derivative $1/x$ by the chain rule. The fundamental theorem of calculus then gives a difference of logarithms, which is the logarithm of the quotient $b/a$.\n",
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