{"id":775,"date":"2025-06-20T17:11:59","date_gmt":"2025-06-20T17:11:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=775"},"modified":"2025-07-22T14:04:56","modified_gmt":"2025-07-22T14:04:56","slug":"improper-integrals-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/improper-integrals-learn-it-3\/","title":{"raw":"Improper Integrals: Learn It 3","rendered":"Improper Integrals: Learn It 3"},"content":{"raw":"<h2>A Comparison Theorem<\/h2>\r\n<p class=\"whitespace-normal break-words\">Sometimes evaluating an improper integral directly is difficult or impossible. In these cases, we can often determine whether the integral converges or diverges by comparing it to another integral whose behavior we already know.<\/p>\r\n<p class=\"whitespace-normal break-words\">The key insight is simple: if one function is always larger than another, then the area under the larger function must be at least as big as the area under the smaller function.<\/p>\r\n<p class=\"whitespace-normal break-words\">Consider two continuous functions [latex]f(x)[\/latex] and [latex]g(x)[\/latex] where [latex]0 \\leq f(x) \\leq g(x)[\/latex] for [latex]x \\geq a[\/latex]. Figure 5 shows this relationship visually.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233859\/CNX_Calc_Figure_07_07_005.jpg\" alt=\"This figure has two graphs. The graphs are f(x) and g(x). The first graph f(x) is a decreasing, non-negative function with a horizontal asymptote at the x-axis. It has a sharper bend in the curve compared to g(x). The graph of g(x) is a decreasing, non-negative function with a horizontal asymptote at the x-axis.\" width=\"487\" height=\"276\" data-media-type=\"image\/jpeg\" \/> Figure 5. If [latex]0\\le f\\left(x\\right)\\le g\\left(x\\right)[\/latex] for [latex]x\\ge a[\/latex], then for [latex]t\\ge a[\/latex], [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx\\le {\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx[\/latex].[\/caption]\r\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]f(x)[\/latex] is always below [latex]g(x)[\/latex], we have:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]0 \\leq \\int_a^t f(x)\u00a0 dx \\leq \\int_a^t g(x)\u00a0 dx \\text{ for } t \\geq a[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This inequality gives us two powerful conclusions:<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>If the smaller function diverges, the larger one must diverge too.<\/strong>\r\nIf [latex]\\int_a^{+\\infty} f(x)\u00a0 dx = +\\infty[\/latex], then [latex]\\int_a^{+\\infty} g(x)\u00a0 dx = +\\infty[\/latex] as well.<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>If the larger function converges, the smaller one must converge too.<\/strong>\r\nIf [latex]\\int_a^{+\\infty} g(x) dx = L[\/latex] (finite), then [latex]\\int_a^{+\\infty} f(x)\u00a0 dx[\/latex] converges to some value [latex]M \\leq L[\/latex].<\/li>\r\n<\/ul>\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>comparison test for improper integrals<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous over [latex][a, +\\infty)[\/latex] with [latex]0 \\leq f(x) \\leq g(x)[\/latex] for [latex]x \\geq a[\/latex].<\/p>\r\n\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Case 1<\/strong>:\r\n<p style=\"padding: 0px 15px 0px 15px;\">If [latex]\\int_a^{+\\infty} f(x) , dx = +\\infty[\/latex], then [latex]\\int_a^{+\\infty} g(x) , dx = +\\infty[\/latex]<\/p>\r\n<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Case 2<\/strong>:\r\n<p style=\"padding: 0px 15px 0px 15px;\">If [latex]\\int_a^{+\\infty} g(x) , dx[\/latex] converges to [latex]L[\/latex], then [latex]\\int_a^{+\\infty} f(x) , dx[\/latex] converges to some [latex]M \\leq L[\/latex]<\/p>\r\n<\/li>\r\n<\/ul>\r\n<\/section>Since utilizing this comparison theorem requires the use of inequalities, it is helpful to note the following fact when comparing functions.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Inequality with Reciprocals\r\n[latex]\\\\[\/latex]\r\n<\/strong>If [latex] 0 &lt; f(x) \\le g(x) [\/latex], then [latex] \\frac{1}{f(x)} \\ge \\frac{1}{g(x)} [\/latex][latex]\\\\[\/latex]Remember: smaller denominators create larger fractions.\r\n\r\n<\/section>This comparison technique is especially useful when dealing with rational functions, exponential functions, or other expressions where direct integration is challenging.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165039554192\" data-type=\"problem\">\r\n<p id=\"fs-id1165039554198\">Use a comparison to show that [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{x{e}^{x}}dx[\/latex] converges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558869\"]\r\n<div id=\"fs-id1165039554239\" data-type=\"solution\">\r\n<p id=\"fs-id1165039554241\">We can see that on the interval where [latex] x \\ge 1 [\/latex]<\/p>\r\n\r\n<div id=\"fs-id1165039554244\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0\\le \\frac{1}{x{e}^{x}}\\le \\frac{1}{{e}^{x}}={e}^{\\text{-}x}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042859832\">so if [latex]{\\displaystyle\\int }_{1}^{+\\infty }{e}^{\\text{-}x}dx[\/latex] converges, then so does [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{x{e}^{x}}dx[\/latex]. To evaluate [latex]{\\displaystyle\\int }_{1}^{+\\infty }{e}^{\\text{-}x}dx[\/latex], first rewrite it as a limit:<\/p>\r\n\r\n<div id=\"fs-id1165042634862\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int }_{1}^{+\\infty }{e}^{\\text{-}x}dx&amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}{e}^{\\text{-}x}dx\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{-}{e}^{\\text{-}x}\\right)|\\begin{array}{c}t\\\\ 1\\end{array}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{-}{e}^{\\text{-}t}+{e}^{1}\\right)\\hfill \\\\ &amp; ={e}^{1}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043188461\">Since [latex]{\\displaystyle\\int }_{1}^{+\\infty }{e}^{\\text{-}x}dx[\/latex] converges, so does [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{x{e}^{x}}dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042601868\" data-type=\"problem\">\r\n<p id=\"fs-id1165042601873\">Use the comparison theorem to show that [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{p}}dx[\/latex] diverges for all [latex]p&lt;1[\/latex].<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n<div id=\"fs-id1165042601923\" data-type=\"solution\">\r\n<p id=\"fs-id1165042601925\">For [latex]p&lt;1[\/latex], [latex]\\frac{1}{x}\\le \\frac{1}{\\left({x}^{p}\\right)}[\/latex] over [latex]\\left[1,\\text{+}\\infty \\right)[\/latex]. In the example: Finding an area, we showed that [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{x}dx=\\text{+}\\infty [\/latex]. Therefore, [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{p}}dx[\/latex] diverges for all [latex]p&lt;1[\/latex].<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<h2>A Comparison Theorem<\/h2>\n<p class=\"whitespace-normal break-words\">Sometimes evaluating an improper integral directly is difficult or impossible. In these cases, we can often determine whether the integral converges or diverges by comparing it to another integral whose behavior we already know.<\/p>\n<p class=\"whitespace-normal break-words\">The key insight is simple: if one function is always larger than another, then the area under the larger function must be at least as big as the area under the smaller function.<\/p>\n<p class=\"whitespace-normal break-words\">Consider two continuous functions [latex]f(x)[\/latex] and [latex]g(x)[\/latex] where [latex]0 \\leq f(x) \\leq g(x)[\/latex] for [latex]x \\geq a[\/latex]. Figure 5 shows this relationship visually.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233859\/CNX_Calc_Figure_07_07_005.jpg\" alt=\"This figure has two graphs. The graphs are f(x) and g(x). The first graph f(x) is a decreasing, non-negative function with a horizontal asymptote at the x-axis. It has a sharper bend in the curve compared to g(x). The graph of g(x) is a decreasing, non-negative function with a horizontal asymptote at the x-axis.\" width=\"487\" height=\"276\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 5. If [latex]0\\le f\\left(x\\right)\\le g\\left(x\\right)[\/latex] for [latex]x\\ge a[\/latex], then for [latex]t\\ge a[\/latex], [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx\\le {\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx[\/latex].<\/figcaption><\/figure>\n<p class=\"whitespace-pre-wrap break-words\">Since [latex]f(x)[\/latex] is always below [latex]g(x)[\/latex], we have:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]0 \\leq \\int_a^t f(x)\u00a0 dx \\leq \\int_a^t g(x)\u00a0 dx \\text{ for } t \\geq a[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This inequality gives us two powerful conclusions:<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>If the smaller function diverges, the larger one must diverge too.<\/strong><br \/>\nIf [latex]\\int_a^{+\\infty} f(x)\u00a0 dx = +\\infty[\/latex], then [latex]\\int_a^{+\\infty} g(x)\u00a0 dx = +\\infty[\/latex] as well.<\/li>\n<li class=\"whitespace-normal break-words\"><strong>If the larger function converges, the smaller one must converge too.<\/strong><br \/>\nIf [latex]\\int_a^{+\\infty} g(x) dx = L[\/latex] (finite), then [latex]\\int_a^{+\\infty} f(x)\u00a0 dx[\/latex] converges to some value [latex]M \\leq L[\/latex].<\/li>\n<\/ul>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>comparison test for improper integrals<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]f(x)[\/latex] and [latex]g(x)[\/latex] be continuous over [latex][a, +\\infty)[\/latex] with [latex]0 \\leq f(x) \\leq g(x)[\/latex] for [latex]x \\geq a[\/latex].<\/p>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Case 1<\/strong>:\n<p style=\"padding: 0px 15px 0px 15px;\">If [latex]\\int_a^{+\\infty} f(x) , dx = +\\infty[\/latex], then [latex]\\int_a^{+\\infty} g(x) , dx = +\\infty[\/latex]<\/p>\n<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Case 2<\/strong>:\n<p style=\"padding: 0px 15px 0px 15px;\">If [latex]\\int_a^{+\\infty} g(x) , dx[\/latex] converges to [latex]L[\/latex], then [latex]\\int_a^{+\\infty} f(x) , dx[\/latex] converges to some [latex]M \\leq L[\/latex]<\/p>\n<\/li>\n<\/ul>\n<\/section>\n<p>Since utilizing this comparison theorem requires the use of inequalities, it is helpful to note the following fact when comparing functions.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: Inequality with Reciprocals<br \/>\n[latex]\\\\[\/latex]<br \/>\n<\/strong>If [latex]0 < f(x) \\le g(x)[\/latex], then [latex]\\frac{1}{f(x)} \\ge \\frac{1}{g(x)}[\/latex][latex]\\\\[\/latex]Remember: smaller denominators create larger fractions.\n\n<\/section>\n<p>This comparison technique is especially useful when dealing with rational functions, exponential functions, or other expressions where direct integration is challenging.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165039554192\" data-type=\"problem\">\n<p id=\"fs-id1165039554198\">Use a comparison to show that [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{x{e}^{x}}dx[\/latex] converges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558869\">Show Solution<\/button><\/p>\n<div id=\"q44558869\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165039554239\" data-type=\"solution\">\n<p id=\"fs-id1165039554241\">We can see that on the interval where [latex]x \\ge 1[\/latex]<\/p>\n<div id=\"fs-id1165039554244\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0\\le \\frac{1}{x{e}^{x}}\\le \\frac{1}{{e}^{x}}={e}^{\\text{-}x}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042859832\">so if [latex]{\\displaystyle\\int }_{1}^{+\\infty }{e}^{\\text{-}x}dx[\/latex] converges, then so does [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{x{e}^{x}}dx[\/latex]. To evaluate [latex]{\\displaystyle\\int }_{1}^{+\\infty }{e}^{\\text{-}x}dx[\/latex], first rewrite it as a limit:<\/p>\n<div id=\"fs-id1165042634862\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\int }_{1}^{+\\infty }{e}^{\\text{-}x}dx& =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}{e}^{\\text{-}x}dx\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{-}{e}^{\\text{-}x}\\right)|\\begin{array}{c}t\\\\ 1\\end{array}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{-}{e}^{\\text{-}t}+{e}^{1}\\right)\\hfill \\\\ & ={e}^{1}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043188461\">Since [latex]{\\displaystyle\\int }_{1}^{+\\infty }{e}^{\\text{-}x}dx[\/latex] converges, so does [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{x{e}^{x}}dx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042601868\" data-type=\"problem\">\n<p id=\"fs-id1165042601873\">Use the comparison theorem to show that [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{p}}dx[\/latex] diverges for all [latex]p<1[\/latex].<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558859\">Show Solution<\/button><\/p>\n<div id=\"q44558859\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042601923\" data-type=\"solution\">\n<p id=\"fs-id1165042601925\">For [latex]p<1[\/latex], [latex]\\frac{1}{x}\\le \\frac{1}{\\left({x}^{p}\\right)}[\/latex] over [latex]\\left[1,\\text{+}\\infty \\right)[\/latex]. In the example: Finding an area, we showed that [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{x}dx=\\text{+}\\infty[\/latex]. Therefore, [latex]{\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{p}}dx[\/latex] diverges for all [latex]p<1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":17,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":667,"module-header":"- Select Header -","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/775"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":8,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/775\/revisions"}],"predecessor-version":[{"id":1366,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/775\/revisions\/1366"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/667"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/775\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=775"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=775"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=775"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=775"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}