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  "body": "\n## Example\n\nThe improper integral of $\\frac{\\sin x}{x}$ over $(0, \\infty)$ converges, and\n$$\\int_0^\\infty \\frac{\\sin x}{x} \\, dx = \\frac{\\pi}{2}.$$\nThe integrand extends continuously to $x = 0$ with value $1$. So only the upper limit is improper.\n[[Integration by parts|08Y1BBQTK9]] gives $\\int_1^b \\frac{\\sin x}{x} \\, dx = \\cos 1 - \\frac{\\cos b}{b} - \\int_1^b \\frac{\\cos x}{x^2} \\, dx$. The last integral converges absolutely, so the limit exists.\n\n## Remarks\n\nThe integral does not converge [[absolutely|03GP7KMBYM]]: $\\int_0^\\infty \\frac{|\\sin x|}{x} \\, dx = \\infty$.\nSo $\\frac{\\sin x}{x}$ is not Lebesgue integrable on $(0, \\infty)$.\nThe value $\\pi/2$ needs other methods, such as differentiation under the integral sign or contour integration.\n\n## Explanations\n\n### L1 human\n\nThe area under $\\frac{\\sin x}{x}$ alternates in sign, and the pieces shrink.\nThe positive and negative pieces cancel to give the value $\\pi/2$.\n",
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