{"id":773,"date":"2025-06-20T17:11:53","date_gmt":"2025-06-20T17:11:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=773"},"modified":"2025-07-22T13:59:15","modified_gmt":"2025-07-22T13:59:15","slug":"improper-integrals-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/improper-integrals-learn-it-1\/","title":{"raw":"Improper Integrals: Learn It 1","rendered":"Improper Integrals: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate integrals over infinite intervals<\/li>\r\n \t<li>Find integrals when there's an infinite discontinuity inside your interval<\/li>\r\n \t<li>Use the comparison theorem to determine if an improper integral converges<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Improper Integrals<\/h2>\r\n<p class=\"whitespace-normal break-words\">Is the area between the graph of [latex]f(x) = \\frac{1}{x}[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][1, +\\infty)[\/latex] finite or infinite? What about the volume if we revolve this region around the [latex]x[\/latex]-axis?<\/p>\r\n<p class=\"whitespace-normal break-words\">The answer might surprise you: the area is infinite, but the volume is finite. This paradox introduces us to <strong>improper integrals<\/strong>\u2014integrals that extend over infinite intervals or involve functions with discontinuities.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>improper integrals<\/h3>\r\nIntegrals that either have infinite limits of integration or involve functions with discontinuities within the interval of integration.\r\n\r\n<\/section>In this section, we'll explore how to evaluate these special integrals using limits, discovering when they converge to finite values and when they diverge to infinity.\r\n<h2 data-type=\"title\">Integrating over an Infinite Interval<\/h2>\r\n<p class=\"whitespace-normal break-words\">How do we make sense of an integral like [latex]\\int_a^{+\\infty} f(x)\u00a0 dx[\/latex]? We can't simply plug in infinity as our upper limit.<\/p>\r\n<p class=\"whitespace-normal break-words\">Instead, we use a limit approach:<\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">First, integrate from [latex]a[\/latex] to some finite value [latex]t[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Then examine what happens as [latex]t[\/latex] approaches infinity<\/li>\r\n<\/ol>\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Think of it this way: We're asking \"What happens to the area under the curve as we extend our region farther and farther to the right?<\/section>Figure 1 below illustrates this concept. As [latex]t[\/latex] increases, the shaded region under [latex]f(x)[\/latex] gets wider. The improper integral [latex]\\int_a^{+\\infty} f(x) dx[\/latex] represents the limit of these areas as [latex]t \\to +\\infty[\/latex].\r\n<figure id=\"CNX_Calc_Figure_07_07_001\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233844\/CNX_Calc_Figure_07_07_001.jpg\" alt=\"This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and decreasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. The region in the first curve is small, and progressively gets wider under the second and third graph as t moves further to the right away from a on the x-axis.\" width=\"975\" height=\"165\" data-media-type=\"image\/jpeg\" \/> Figure 1. To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.[\/caption]<\/figure>\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>improper integrals with infinite limits<\/h3>\r\n<ul>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Type 1: Upper limit approaches [latex]+\\infty[\/latex]<\/strong>\r\n<p style=\"padding: 0px 15px 0px 15px;\">Let [latex]f(x)[\/latex] be continuous over [latex][a, +\\infty)[\/latex]. Then:<\/p>\r\n\r\n<center>[latex]\\int_a^{+\\infty} f(x)\u00a0 dx = \\lim_{t \\to +\\infty} \\int_a^t f(x)\u00a0 dx[\/latex]<\/center><\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Type 2: Lower limit approaches [latex]-\\infty[\/latex]\r\n<\/strong>\r\n<p style=\"padding: 0px 15px 0px 15px;\">Let [latex]f(x)[\/latex] be continuous over [latex](-\\infty, b][\/latex]. Then:<\/p>\r\n\r\n<center>[latex]\\int_{-\\infty}^b f(x)\u00a0 dx = \\lim_{t \\to -\\infty} \\int_t^b f(x) dx[\/latex]<\/center><\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Type 3: Both limits are infinite<\/strong>\r\n<p style=\"padding: 0px 15px 0px 15px;\">Let [latex]f(x)[\/latex] be continuous over [latex](-\\infty, +\\infty)[\/latex]. Then:<\/p>\r\n\r\n<center>[latex]\\int_{-\\infty}^{+\\infty} f(x)\u00a0 dx = \\int_{-\\infty}^0 f(x) dx + \\int_0^{+\\infty} f(x)\u00a0 dx[\/latex]<\/center><\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Note<\/strong>: You can split at any point [latex]a[\/latex], not just [latex]0[\/latex].<\/p>\r\n[latex]\\\\[\/latex]\r\n<p class=\"whitespace-normal break-words\"><strong>Convergence vs. Divergence<\/strong><\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Converges<\/strong>: The limit exists and equals a finite number<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Diverges<\/strong>: The limit doesn't exist or approaches [latex]\\pm\\infty[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\">For Type 3 integrals, <strong>both<\/strong> parts must converge for the whole integral to converge.<\/p>\r\n\r\n<\/section>Now let's tackle the question from our introduction: Is the area between [latex]f(x) = \\frac{1}{x}[\/latex] and the [latex]x[\/latex]-axis over [latex][1, +\\infty)[\/latex] finite or infinite?\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165043085741\" data-type=\"problem\">\r\n<p id=\"fs-id1165043346878\">Determine whether the area between the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] is finite or infinite.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558899\"]\r\n<div id=\"fs-id1165042330814\" data-type=\"solution\">\r\n<p id=\"fs-id1165043350067\">We first do a quick sketch of the region in question, as shown in the following graph.<\/p>\r\n\r\n<figure id=\"CNX_Calc_Figure_07_07_002\"><figcaption><\/figcaption>[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233847\/CNX_Calc_Figure_07_07_002.jpg\" alt=\"This figure is the graph of the function y = 1\/x. It is a decreasing function with a vertical asymptote at the y-axis. In the first quadrant there is a shaded region under the curve bounded by x = 1 and x = 4.\" width=\"487\" height=\"351\" data-media-type=\"image\/jpeg\" \/> Figure 2. We can find the area between the curve [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the x-axis on an infinite interval.[\/caption]<\/figure>\r\n<p id=\"fs-id1165042925668\">We can see that the area of this region is given by [latex]A={\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{x}dx[\/latex]. Then we have<\/p>\r\n\r\n<div id=\"fs-id1165043131589\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill A&amp; ={\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{x}dx\\hfill &amp; &amp; &amp; \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}\\frac{1}{x}dx\\hfill &amp; &amp; &amp; \\text{Rewrite the improper integral as a limit.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\text{ln}|x||{}_{\\begin{array}{c}\\\\ 1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{ln}|t|-\\text{ln}1\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\text{+}\\infty .\\hfill &amp; &amp; &amp; \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043117138\">Since the improper integral diverges to [latex]+\\infty [\/latex], the area of the region is infinite.<\/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-id1165042925771\" data-type=\"problem\">\r\n<p id=\"fs-id1165042677398\">Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] about the [latex]x[\/latex] -axis.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558898\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558898\"]\r\n<div id=\"fs-id1165042582834\" data-type=\"solution\">\r\n<p id=\"fs-id1165042978442\">The solid is shown in Figure 3. Using the disk method, we see that the volume <em data-effect=\"italics\">V<\/em> is<\/p>\r\n\r\n<div id=\"fs-id1165042321060\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]V=\\pi {\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{2}}dx[\/latex].<\/div>\r\n&nbsp;\r\n<figure id=\"CNX_Calc_Figure_07_07_003\"><figcaption><\/figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/4175\/2019\/04\/11233850\/CNX_Calc_Figure_07_07_003.jpg\" alt=\"This figure is the graph of the function y = 1\/x. It is a decreasing function with a vertical asymptote at the y-axis. The graph shows a solid that has been generated by rotating the curve in the first quadrant around the x-axis.\" width=\"731\" height=\"351\" data-media-type=\"image\/jpeg\" \/> Figure 3. The solid of revolution can be generated by rotating an infinite area about the x-axis.[\/caption]<\/figure>\r\n<p id=\"fs-id1165043112190\">Then we have<\/p>\r\n\r\n<div id=\"fs-id1165043259904\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill V&amp; =\\pi {\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{2}}dx\\hfill &amp; &amp; &amp; \\\\ &amp; =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}\\frac{1}{{x}^{2}}dx\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}-\\frac{1}{x}|{}_{\\begin{array}{c}\\\\ 1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{-}\\frac{1}{t}+1\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\pi .\\hfill &amp; &amp; &amp; \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042788783\">The improper integral converges to [latex]\\pi [\/latex]. Therefore, the volume of the solid of revolution is [latex]\\pi [\/latex].<\/p>\r\nIn conclusion, although the area of the region between the [latex]x[\/latex]-axis and the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] is infinite, the volume of the solid generated by revolving this region about the [latex]x[\/latex]-axis is finite. The solid generated is known as <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">Gabriel\u2019s Horn<\/em><\/span>.\r\n\r\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit <a href=\"https:\/\/en.wikipedia.org\/wiki\/Gabriel%27s_Horn\" target=\"_blank\" rel=\"noopener\">this website to read more about Gabriel\u2019s Horn<\/a>.<\/section><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>Because improper integrals require evaluating limits at infinity, at times we may be required to use L'H\u00f4pital's Rule to evaluate a limit.\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: L'H\u00f4pital's Rule<\/strong>\r\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval [latex]\\left(a, \\infty \\right) [\/latex] for some value of [latex] a [\/latex]. If either:<\/p>\r\n\r\n<ol>\r\n \t<li>[latex]\\underset{x\\to \\infty}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to \\infty}{\\lim}g(x)=0[\/latex]<\/li>\r\n \t<li>[latex]\\underset{x\\to \\infty}{\\lim}f(x)= \\infty [\/latex] (or [latex] -\\infty [\/latex]) and\u00a0[latex]\\underset{x\\to \\infty}{\\lim}g(x)= \\infty [\/latex] (or [latex] -\\infty [\/latex]), then<\/li>\r\n<\/ol>\r\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to \\infty}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex]<\/div>\r\n&nbsp;\r\n<div><\/div>\r\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<div id=\"fs-id1165042608027\" data-type=\"problem\">\r\n<p id=\"fs-id1165042318435\">Evaluate [latex]{\\displaystyle\\int }_{-\\infty }^{+\\infty }x{e}^{x}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\r\n\r\n<\/div>\r\n[reveal-answer q=\"44558896\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558896\"]\r\n<div id=\"fs-id1165042960104\" data-type=\"solution\">\r\n<p id=\"fs-id1165042705512\">Start by splitting up the integral:<\/p>\r\n\r\n<div id=\"fs-id1165042705515\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx={\\displaystyle\\int }_{\\text{-}\\infty }^{0}x{e}^{x}dx+{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042519801\">If either [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}x{e}^{x}dx[\/latex] or [latex]{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex] diverges, then [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx[\/latex] diverges. Compute each integral separately. For the first integral,<\/p>\r\n\r\n<div id=\"fs-id1165039560750\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{-\\infty }^{0}x{e}^{x}dx}&amp; ={\\underset{t\\to -\\infty}\\lim}{\\displaystyle\\int _{t}^{0}x{e}^{x}dx}\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; ={\\underset{t\\to -\\infty}\\lim}\\left(x{e}^{x}-{e}^{x}\\right)\\Biggr|_{t}^{0} \\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Use integration by parts to find the} \\hfill \\\\ \\text{antiderivative. (Here } u=x \\text{ and } dv=e_{dv}^{x}\\text{.)}\\end{array} \\\\ &amp; ={\\underset{t\\to -\\infty}\\lim}\\left(-1-t{e}^{t}+{e}^{t}\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =-1.\\hfill &amp; &amp; &amp; \\begin{array}{c}\\text{Evaluate the limit.}\\mathit{\\text{Note:}} {\\underset{t\\to -\\infty}\\lim}t{e}^{t}\\text{is}\\hfill \\\\ \\text{indeterminate of the form}0\\cdot \\infty .\\text{Thus,}\\hfill \\\\ {\\underset{t\\to -\\infty}\\lim}t{e}^{t}={\\underset{t\\to -\\infty}\\lim}\\frac{t}{{e}^{\\text{-}t}}={\\underset{t\\to -\\infty}\\lim}\\frac{-1}{{e}^{-t}}={\\underset{t\\to -\\infty}\\lim}-{e}^{t}=0\\text{by}\\hfill \\\\ \\text{L'H}\\hat{o}\\text{pital's Rule.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043354391\">The first improper integral converges. For the second integral,<\/p>\r\n\r\n<div id=\"fs-id1165043354394\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{0}^{+\\infty }x{e}^{x}dx}&amp; =\\underset{t\\to +\\infty }{\\text{lim}}{\\displaystyle\\int _{0}^{t}x{e}^{x}dx}\\hfill &amp; &amp; &amp; \\text{Rewrite as a limit.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(x{e}^{x}-{e}^{x}\\right)\\Biggr|_{t}^{0} \\hfill &amp; &amp; &amp; \\text{Find the antiderivative.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(t{e}^{t}-{e}^{t}+1\\right)\\hfill &amp; &amp; &amp; \\text{Evaluate the antiderivative.}\\hfill \\\\ &amp; =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\left(t - 1\\right){e}^{t}+1\\right)\\hfill &amp; &amp; &amp; \\text{Rewrite.} (t e^t-e^t \\text{ is indeterminate.)} \\hfill \\\\ &amp; =\\text{+}\\infty .\\hfill &amp; &amp; &amp; \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042910269\">Thus, [latex]{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex] diverges. Since this integral diverges, [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx[\/latex] diverges as well.<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]25530[\/ohm_question]<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate integrals over infinite intervals<\/li>\n<li>Find integrals when there&#8217;s an infinite discontinuity inside your interval<\/li>\n<li>Use the comparison theorem to determine if an improper integral converges<\/li>\n<\/ul>\n<\/section>\n<h2>Improper Integrals<\/h2>\n<p class=\"whitespace-normal break-words\">Is the area between the graph of [latex]f(x) = \\frac{1}{x}[\/latex] and the [latex]x[\/latex]-axis over the interval [latex][1, +\\infty)[\/latex] finite or infinite? What about the volume if we revolve this region around the [latex]x[\/latex]-axis?<\/p>\n<p class=\"whitespace-normal break-words\">The answer might surprise you: the area is infinite, but the volume is finite. This paradox introduces us to <strong>improper integrals<\/strong>\u2014integrals that extend over infinite intervals or involve functions with discontinuities.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>improper integrals<\/h3>\n<p>Integrals that either have infinite limits of integration or involve functions with discontinuities within the interval of integration.<\/p>\n<\/section>\n<p>In this section, we&#8217;ll explore how to evaluate these special integrals using limits, discovering when they converge to finite values and when they diverge to infinity.<\/p>\n<h2 data-type=\"title\">Integrating over an Infinite Interval<\/h2>\n<p class=\"whitespace-normal break-words\">How do we make sense of an integral like [latex]\\int_a^{+\\infty} f(x)\u00a0 dx[\/latex]? We can&#8217;t simply plug in infinity as our upper limit.<\/p>\n<p class=\"whitespace-normal break-words\">Instead, we use a limit approach:<\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\">First, integrate from [latex]a[\/latex] to some finite value [latex]t[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Then examine what happens as [latex]t[\/latex] approaches infinity<\/li>\n<\/ol>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Think of it this way: We&#8217;re asking &#8220;What happens to the area under the curve as we extend our region farther and farther to the right?<\/section>\n<p>Figure 1 below illustrates this concept. As [latex]t[\/latex] increases, the shaded region under [latex]f(x)[\/latex] gets wider. The improper integral [latex]\\int_a^{+\\infty} f(x) dx[\/latex] represents the limit of these areas as [latex]t \\to +\\infty[\/latex].<\/p>\n<figure id=\"CNX_Calc_Figure_07_07_001\">\n<figure style=\"width: 975px\" 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\/11233844\/CNX_Calc_Figure_07_07_001.jpg\" alt=\"This figure has three graphs. All the graphs have the same curve, which is f(x). The curve is non-negative, only in the first quadrant, and decreasing. Under all three curves is a shaded region bounded by a on the x-axis an t on the x-axis. The region in the first curve is small, and progressively gets wider under the second and third graph as t moves further to the right away from a on the x-axis.\" width=\"975\" height=\"165\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 1. To integrate a function over an infinite interval, we consider the limit of the integral as the upper limit increases without bound.<\/figcaption><\/figure>\n<\/figure>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>improper integrals with infinite limits<\/h3>\n<ul>\n<li class=\"whitespace-normal break-words\"><strong>Type 1: Upper limit approaches [latex]+\\infty[\/latex]<\/strong>\n<p style=\"padding: 0px 15px 0px 15px;\">Let [latex]f(x)[\/latex] be continuous over [latex][a, +\\infty)[\/latex]. Then:<\/p>\n<div style=\"text-align: center;\">[latex]\\int_a^{+\\infty} f(x)\u00a0 dx = \\lim_{t \\to +\\infty} \\int_a^t f(x)\u00a0 dx[\/latex]<\/div>\n<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Type 2: Lower limit approaches [latex]-\\infty[\/latex]<br \/>\n<\/strong><\/p>\n<p style=\"padding: 0px 15px 0px 15px;\">Let [latex]f(x)[\/latex] be continuous over [latex](-\\infty, b][\/latex]. Then:<\/p>\n<div style=\"text-align: center;\">[latex]\\int_{-\\infty}^b f(x)\u00a0 dx = \\lim_{t \\to -\\infty} \\int_t^b f(x) dx[\/latex]<\/div>\n<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Type 3: Both limits are infinite<\/strong>\n<p style=\"padding: 0px 15px 0px 15px;\">Let [latex]f(x)[\/latex] be continuous over [latex](-\\infty, +\\infty)[\/latex]. Then:<\/p>\n<div style=\"text-align: center;\">[latex]\\int_{-\\infty}^{+\\infty} f(x)\u00a0 dx = \\int_{-\\infty}^0 f(x) dx + \\int_0^{+\\infty} f(x)\u00a0 dx[\/latex]<\/div>\n<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Note<\/strong>: You can split at any point [latex]a[\/latex], not just [latex]0[\/latex].<\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Convergence vs. Divergence<\/strong><\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-1.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Converges<\/strong>: The limit exists and equals a finite number<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Diverges<\/strong>: The limit doesn&#8217;t exist or approaches [latex]\\pm\\infty[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\">For Type 3 integrals, <strong>both<\/strong> parts must converge for the whole integral to converge.<\/p>\n<\/section>\n<p>Now let&#8217;s tackle the question from our introduction: Is the area between [latex]f(x) = \\frac{1}{x}[\/latex] and the [latex]x[\/latex]-axis over [latex][1, +\\infty)[\/latex] finite or infinite?<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165043085741\" data-type=\"problem\">\n<p id=\"fs-id1165043346878\">Determine whether the area between the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] is finite or infinite.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558899\">Show Solution<\/button><\/p>\n<div id=\"q44558899\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042330814\" data-type=\"solution\">\n<p id=\"fs-id1165043350067\">We first do a quick sketch of the region in question, as shown in the following graph.<\/p>\n<figure id=\"CNX_Calc_Figure_07_07_002\"><figcaption><\/figcaption><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\/11233847\/CNX_Calc_Figure_07_07_002.jpg\" alt=\"This figure is the graph of the function y = 1\/x. It is a decreasing function with a vertical asymptote at the y-axis. In the first quadrant there is a shaded region under the curve bounded by x = 1 and x = 4.\" width=\"487\" height=\"351\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 2. We can find the area between the curve [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the x-axis on an infinite interval.<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1165042925668\">We can see that the area of this region is given by [latex]A={\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{x}dx[\/latex]. Then we have<\/p>\n<div id=\"fs-id1165043131589\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill A& ={\\displaystyle\\int }_{1}^{\\infty }\\frac{1}{x}dx\\hfill & & & \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}\\frac{1}{x}dx\\hfill & & & \\text{Rewrite the improper integral as a limit.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\text{ln}|x||{}_{\\begin{array}{c}\\\\ 1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{ln}|t|-\\text{ln}1\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\text{+}\\infty .\\hfill & & & \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043117138\">Since the improper integral diverges to [latex]+\\infty[\/latex], the area of the region is infinite.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042925771\" data-type=\"problem\">\n<p id=\"fs-id1165042677398\">Find the volume of the solid obtained by revolving the region bounded by the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] about the [latex]x[\/latex] -axis.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558898\">Show Solution<\/button><\/p>\n<div id=\"q44558898\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042582834\" data-type=\"solution\">\n<p id=\"fs-id1165042978442\">The solid is shown in Figure 3. Using the disk method, we see that the volume <em data-effect=\"italics\">V<\/em> is<\/p>\n<div id=\"fs-id1165042321060\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]V=\\pi {\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{2}}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<figure id=\"CNX_Calc_Figure_07_07_003\"><figcaption><\/figcaption><figure style=\"width: 731px\" 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\/11233850\/CNX_Calc_Figure_07_07_003.jpg\" alt=\"This figure is the graph of the function y = 1\/x. It is a decreasing function with a vertical asymptote at the y-axis. The graph shows a solid that has been generated by rotating the curve in the first quadrant around the x-axis.\" width=\"731\" height=\"351\" data-media-type=\"image\/jpeg\" \/><figcaption class=\"wp-caption-text\">Figure 3. The solid of revolution can be generated by rotating an infinite area about the x-axis.<\/figcaption><\/figure>\n<\/figure>\n<p id=\"fs-id1165043112190\">Then we have<\/p>\n<div id=\"fs-id1165043259904\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill V& =\\pi {\\displaystyle\\int }_{1}^{+\\infty }\\frac{1}{{x}^{2}}dx\\hfill & & & \\\\ & =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{1}^{t}\\frac{1}{{x}^{2}}dx\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}-\\frac{1}{x}|{}_{\\begin{array}{c}\\\\ 1\\end{array}}^{\\begin{array}{c}t\\\\ \\end{array}}\\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\pi \\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\text{-}\\frac{1}{t}+1\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\pi .\\hfill & & & \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042788783\">The improper integral converges to [latex]\\pi[\/latex]. Therefore, the volume of the solid of revolution is [latex]\\pi[\/latex].<\/p>\n<p>In conclusion, although the area of the region between the [latex]x[\/latex]-axis and the graph of [latex]f\\left(x\\right)=\\frac{1}{x}[\/latex] over the interval [latex]\\left[1,\\text{+}\\infty \\right)[\/latex] is infinite, the volume of the solid generated by revolving this region about the [latex]x[\/latex]-axis is finite. The solid generated is known as <span class=\"no-emphasis\" data-type=\"term\"><em data-effect=\"italics\">Gabriel\u2019s Horn<\/em><\/span>.<\/p>\n<section class=\"textbox interact\" aria-label=\"Interact\">Visit <a href=\"https:\/\/en.wikipedia.org\/wiki\/Gabriel%27s_Horn\" target=\"_blank\" rel=\"noopener\">this website to read more about Gabriel\u2019s Horn<\/a>.<\/section>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<p>Because improper integrals require evaluating limits at infinity, at times we may be required to use L&#8217;H\u00f4pital&#8217;s Rule to evaluate a limit.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\"><strong>Recall: L&#8217;H\u00f4pital&#8217;s Rule<\/strong><\/p>\n<p id=\"fs-id1165043276140\">Suppose [latex]f[\/latex] and [latex]g[\/latex] are differentiable functions over an open interval [latex]\\left(a, \\infty \\right)[\/latex] for some value of [latex]a[\/latex]. If either:<\/p>\n<ol>\n<li>[latex]\\underset{x\\to \\infty}{\\lim}f(x)=0[\/latex] and [latex]\\underset{x\\to \\infty}{\\lim}g(x)=0[\/latex]<\/li>\n<li>[latex]\\underset{x\\to \\infty}{\\lim}f(x)= \\infty[\/latex] (or [latex]-\\infty[\/latex]) and\u00a0[latex]\\underset{x\\to \\infty}{\\lim}g(x)= \\infty[\/latex] (or [latex]-\\infty[\/latex]), then<\/li>\n<\/ol>\n<div id=\"fs-id1165043020148\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\infty}{\\lim}\\dfrac{f(x)}{g(x)}=\\underset{x\\to \\infty}{\\lim}\\dfrac{f^{\\prime}(x)}{g^{\\prime}(x)}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<div><\/div>\n<p id=\"fs-id1165043035673\">assuming the limit on the right exists or is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<div id=\"fs-id1165042608027\" data-type=\"problem\">\n<p id=\"fs-id1165042318435\">Evaluate [latex]{\\displaystyle\\int }_{-\\infty }^{+\\infty }x{e}^{x}dx[\/latex]. State whether the improper integral converges or diverges.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44558896\">Show Solution<\/button><\/p>\n<div id=\"q44558896\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"fs-id1165042960104\" data-type=\"solution\">\n<p id=\"fs-id1165042705512\">Start by splitting up the integral:<\/p>\n<div id=\"fs-id1165042705515\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx={\\displaystyle\\int }_{\\text{-}\\infty }^{0}x{e}^{x}dx+{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042519801\">If either [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{0}x{e}^{x}dx[\/latex] or [latex]{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex] diverges, then [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx[\/latex] diverges. Compute each integral separately. For the first integral,<\/p>\n<div id=\"fs-id1165039560750\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{-\\infty }^{0}x{e}^{x}dx}& ={\\underset{t\\to -\\infty}\\lim}{\\displaystyle\\int _{t}^{0}x{e}^{x}dx}\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & ={\\underset{t\\to -\\infty}\\lim}\\left(x{e}^{x}-{e}^{x}\\right)\\Biggr|_{t}^{0} \\hfill & & & \\begin{array}{c}\\text{Use integration by parts to find the} \\hfill \\\\ \\text{antiderivative. (Here } u=x \\text{ and } dv=e_{dv}^{x}\\text{.)}\\end{array} \\\\ & ={\\underset{t\\to -\\infty}\\lim}\\left(-1-t{e}^{t}+{e}^{t}\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =-1.\\hfill & & & \\begin{array}{c}\\text{Evaluate the limit.}\\mathit{\\text{Note:}} {\\underset{t\\to -\\infty}\\lim}t{e}^{t}\\text{is}\\hfill \\\\ \\text{indeterminate of the form}0\\cdot \\infty .\\text{Thus,}\\hfill \\\\ {\\underset{t\\to -\\infty}\\lim}t{e}^{t}={\\underset{t\\to -\\infty}\\lim}\\frac{t}{{e}^{\\text{-}t}}={\\underset{t\\to -\\infty}\\lim}\\frac{-1}{{e}^{-t}}={\\underset{t\\to -\\infty}\\lim}-{e}^{t}=0\\text{by}\\hfill \\\\ \\text{L'H}\\hat{o}\\text{pital's Rule.}\\hfill \\end{array}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043354391\">The first improper integral converges. For the second integral,<\/p>\n<div id=\"fs-id1165043354394\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{ccccc}\\hfill {\\displaystyle\\int _{0}^{+\\infty }x{e}^{x}dx}& =\\underset{t\\to +\\infty }{\\text{lim}}{\\displaystyle\\int _{0}^{t}x{e}^{x}dx}\\hfill & & & \\text{Rewrite as a limit.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(x{e}^{x}-{e}^{x}\\right)\\Biggr|_{t}^{0} \\hfill & & & \\text{Find the antiderivative.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(t{e}^{t}-{e}^{t}+1\\right)\\hfill & & & \\text{Evaluate the antiderivative.}\\hfill \\\\ & =\\underset{t\\to \\text{+}\\infty }{\\text{lim}}\\left(\\left(t - 1\\right){e}^{t}+1\\right)\\hfill & & & \\text{Rewrite.} (t e^t-e^t \\text{ is indeterminate.)} \\hfill \\\\ & =\\text{+}\\infty .\\hfill & & & \\text{Evaluate the limit.}\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042910269\">Thus, [latex]{\\displaystyle\\int }_{0}^{+\\infty }x{e}^{x}dx[\/latex] diverges. Since this integral diverges, [latex]{\\displaystyle\\int }_{\\text{-}\\infty }^{+\\infty }x{e}^{x}dx[\/latex] diverges as well.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm25530\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=25530&theme=lumen&iframe_resize_id=ohm25530&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":15,"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\/773"}],"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":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/773\/revisions"}],"predecessor-version":[{"id":1339,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/773\/revisions\/1339"}],"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\/773\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=773"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=773"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=773"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=773"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}