{"id":1615,"date":"2021-07-22T16:53:05","date_gmt":"2021-07-22T16:53:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/?post_type=chapter&#038;p=1615"},"modified":"2022-03-19T04:51:28","modified_gmt":"2022-03-19T04:51:28","slug":"a-comparison-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/a-comparison-theorem\/","title":{"raw":"A Comparison Theorem","rendered":"A Comparison Theorem"},"content":{"raw":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Use the comparison theorem to determine whether a definite integral is convergent<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165043273053\">It is not always easy or even possible to evaluate an improper integral directly; however, by comparing it with another carefully chosen integral, it may be possible to determine its convergence or divergence. To see this, consider two continuous functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] satisfying [latex]0\\le f\\left(x\\right)\\le g\\left(x\\right)[\/latex] for [latex]x\\ge a[\/latex] (Figure 5). In this case, we may view integrals of these functions over intervals of the form [latex]\\left[a,t\\right][\/latex] as areas, so we have the relationship<\/p>\r\n\r\n<div id=\"fs-id1165043385358\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0\\le {\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx\\le {\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx\\text{ for }t\\ge a[\/latex].<\/div>\r\n&nbsp;\r\n<figure id=\"CNX_Calc_Figure_07_07_005\"><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\/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]<\/figure>\r\n<p id=\"fs-id1165042706165\">Thus, if<\/p>\r\n\r\n<div id=\"fs-id1165042706168\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx=\\text{+}\\infty [\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042359605\">then<\/p>\r\n<p id=\"fs-id1165042359608\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }g\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx=\\text{+}\\infty [\/latex] as well. That is, if the area of the region between the graph of [latex]f\\left(x\\right)[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex] is infinite, then the area of the region between the graph of [latex]g\\left(x\\right)[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex] is infinite too.<\/p>\r\n<p id=\"fs-id1165042387260\">On the other hand, if<\/p>\r\n<p id=\"fs-id1165042387263\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }g\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx=L[\/latex] for some real number [latex]L[\/latex], then<\/p>\r\n<p id=\"fs-id1165043314585\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] must converge to some value less than or equal to [latex]L[\/latex], since [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] increases as [latex]t[\/latex] increases and [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx\\le L[\/latex] for all [latex]t\\ge a[\/latex].<\/p>\r\n<p id=\"fs-id1165042851612\">If the area of the region between the graph of [latex]g\\left(x\\right)[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex] is finite, then the area of the region between the graph of [latex]f\\left(x\\right)[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex] is also finite.<\/p>\r\n<p id=\"fs-id1165043207372\">These conclusions are summarized in the following theorem.<\/p>\r\n\r\n<div id=\"fs-id1165043207376\" class=\"theorem\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">A Comparison Theorem<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1165043207383\">Let [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] be continuous over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex]. Assume that [latex]0\\le f\\left(x\\right)\\le g\\left(x\\right)[\/latex] for [latex]x\\ge a[\/latex].<\/p>\r\n\r\n<ol id=\"fs-id1165042558470\" type=\"i\">\r\n \t<li>If [latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx=\\text{+}\\infty [\/latex], then [latex]{\\displaystyle\\int }_{a}^{+\\infty }g\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx=\\text{+}\\infty [\/latex].<\/li>\r\n \t<li>If [latex]{\\displaystyle\\int }_{a}^{+\\infty }g\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx=L[\/latex], where [latex]L[\/latex] is a real number, then [latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx=M[\/latex] for some real number [latex]M\\le L[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165039554188\" data-type=\"example\">\r\n<div id=\"fs-id1165039554190\" data-type=\"exercise\">\r\n<div id=\"fs-id1165039554192\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n\r\nSince utilizing this comparison theorem requires the use of inequalities, it is helpful to note the following fact when comparing functions.\r\n<div class=\"textbox examples\">\r\n<h3>Recall: Algebra of inequalities<\/h3>\r\nIf [latex] 0 &lt; f(x) \\le g(x) [\/latex], then [latex] \\frac{1}{f(x)} \\ge \\frac{1}{g(x)} [\/latex]\r\n\r\nIn other words, a smaller denominator corresponds to a larger fraction, and vice versa.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Applying the Comparison Theorem<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042601863\" data-type=\"example\">\r\n<div id=\"fs-id1165042601865\" data-type=\"exercise\">\r\n<div id=\"fs-id1165042601868\" data-type=\"problem\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Applying the Comparison Theorem<\/h3>\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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse a comparison to show that [latex]{\\displaystyle\\int }_{e}^{+\\infty }\\frac{\\text{ln}x}{x}dx[\/latex] diverges.\r\n[reveal-answer q=\"907753222\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"907753222\"]\r\n[latex]\\frac{1}{x}\\le \\frac{\\text{ln}x}{x}[\/latex] on [latex]\\left[e,\\text{+}\\infty \\right)[\/latex]\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"6657883\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"6657883\"]\r\nSince [latex]{\\displaystyle\\int }_{e}^{+\\infty }\\frac{1}{x}dx=\\text{+}\\infty [\/latex], [latex]{\\displaystyle\\int }_{e}^{+\\infty }\\frac{\\text{ln}x}{x}dx[\/latex] diverges\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8QzQO748O58?controls=0&amp;start=1783&amp;end=1837&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>\r\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.7ImproperIntegrals1783to1837_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.7 Improper Integrals\" here (opens in new window)<\/a>.\r\n<div class=\"textbox tryit\">\r\n<h3 style=\"text-align: center;\" data-type=\"title\">ACTIVITY: Laplace Transforms<\/h3>\r\n<p id=\"fs-id1165042854865\">In the last few modules, we have looked at several ways to use integration for solving real-world problems. For this next project, we are going to explore a more advanced application of integration: integral transforms. Specifically, we describe the <span class=\"no-emphasis\" data-type=\"term\">Laplace transform<\/span> and some of its properties. The Laplace transform is used in engineering and physics to simplify the computations needed to solve some problems. It takes functions expressed in terms of time and <em data-effect=\"italics\">transforms<\/em> them to functions expressed in terms of frequency. It turns out that, in many cases, the computations needed to solve problems in the frequency domain are much simpler than those required in the time domain.<\/p>\r\n<p id=\"fs-id1165042854884\">The Laplace transform is defined in terms of an integral as<\/p>\r\n\r\n<div id=\"fs-id1165042854888\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{f\\left(t\\right)\\right\\}=F\\left(s\\right)={\\displaystyle\\int }_{0}^{\\infty }{e}^{\\text{-}st}f\\left(t\\right)dt[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043389093\">Note that the input to a Laplace transform is a function of time, [latex]f\\left(t\\right)[\/latex], and the output is a function of frequency, [latex]F\\left(s\\right)[\/latex]. Although many real-world examples require the use of complex numbers (involving the imaginary number [latex]i=\\sqrt{-1}[\/latex]), in this project we limit ourselves to functions of real numbers.<\/p>\r\n<p id=\"fs-id1165043389151\">Let\u2019s start with a simple example. Here we calculate the Laplace transform of [latex]f\\left(t\\right)=t[\/latex] . We have<\/p>\r\n\r\n<div id=\"fs-id1165039566620\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{t\\right\\}={\\displaystyle\\int }_{0}^{\\infty }t{e}^{\\text{-}st}dt[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165039566673\">This is an improper integral, so we express it in terms of a limit, which gives<\/p>\r\n\r\n<div id=\"fs-id1165039566676\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{t\\right\\}={\\displaystyle\\int }_{0}^{\\infty }t{e}^{\\text{-}st}dt=\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}t{e}^{\\text{-}st}dt[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165039566031\">Now we use integration by parts to evaluate the integral. Note that we are integrating with respect to <em data-effect=\"italics\">t<\/em>, so we treat the variable <em data-effect=\"italics\">s<\/em> as a constant. We have<\/p>\r\n\r\n<div id=\"fs-id1165039566045\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill u&amp; =\\hfill &amp; t\\hfill &amp; &amp; &amp; \\hfill dv&amp; =\\hfill &amp; {e}^{\\text{-}st}dt\\hfill \\\\ \\hfill du&amp; =\\hfill &amp; dt\\hfill &amp; &amp; &amp; \\hfill v&amp; =\\hfill &amp; -\\frac{1}{s}{e}^{\\text{-}st}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042323365\">Then we obtain<\/p>\r\n\r\n<div id=\"fs-id1165042672302\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}t{e}^{\\text{-}st}dt&amp; =\\underset{z\\to \\infty }{\\text{lim}}\\left[{\\left[-\\frac{t}{s}{e}^{\\text{-}st}\\right]|}_{0}^{z}+\\frac{1}{s}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}dt\\right]\\hfill \\\\ &amp; =\\underset{z\\to \\infty }{\\text{lim}}\\left[\\left[-\\frac{z}{s}{e}^{\\text{-}sz}+\\frac{0}{s}{e}^{-0s}\\right]+\\frac{1}{s}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}dt\\right]\\hfill \\\\ &amp; =\\underset{z\\to \\infty }{\\text{lim}}\\left[\\left[-\\frac{z}{s}{e}^{\\text{-}sz}+0\\right]-\\frac{1}{s}{\\left[\\frac{{e}^{\\text{-}st}}{s}\\right]|}_{0}^{z}\\right]\\hfill \\\\ &amp; =\\underset{z\\to \\infty }{\\text{lim}}\\left[\\left[-\\frac{z}{s}{e}^{\\text{-}sz}\\right]-\\frac{1}{{s}^{2}}\\left[{e}^{\\text{-}sz}-1\\right]\\right]\\hfill \\\\ &amp; =\\underset{z\\to \\infty }{\\text{lim}}\\left[-\\frac{z}{s{e}^{sz}}\\right]-\\underset{z\\to \\infty }{\\text{lim}}\\left[\\frac{1}{{s}^{2}{e}^{sz}}\\right]+\\underset{z\\to \\infty }{\\text{lim}}\\frac{1}{{s}^{2}}\\hfill \\\\ &amp; =0 - 0+\\frac{1}{{s}^{2}}\\hfill \\\\ &amp; =\\frac{1}{{s}^{2}}.\\hfill \\end{array}[\/latex]<\/div>\r\n&nbsp;\r\n<ol id=\"fs-id1165042659369\" type=\"1\">\r\n \t<li>Calculate the Laplace transform of [latex]f\\left(t\\right)=1[\/latex].<\/li>\r\n \t<li>Calculate the Laplace transform of [latex]f\\left(t\\right)={e}^{-3t}[\/latex].<\/li>\r\n \t<li>Calculate the Laplace transform of [latex]f\\left(t\\right)={t}^{2}[\/latex]. (Note, you will have to integrate by parts twice.)<span data-type=\"newline\">\r\n<\/span>\r\nLaplace transforms are often used to solve differential equations. Differential equations are not covered in detail until later in this book; but, for now, let\u2019s look at the relationship between the Laplace transform of a function and the Laplace transform of its derivative.<span data-type=\"newline\">\r\n<\/span>\r\nLet\u2019s start with the definition of the Laplace transform. We have<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165042692735\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{f\\left(t\\right)\\right\\}={\\displaystyle\\int }_{0}^{\\infty }{e}^{\\text{-}st}f\\left(t\\right)dt=\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}f\\left(t\\right)dt[\/latex].<\/div><\/li>\r\n \t<li style=\"text-align: left;\">Use integration by parts to evaluate [latex]\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}f\\left(t\\right)dt[\/latex]. (Let [latex]u=f\\left(t\\right)[\/latex] and [latex]dv={e}^{\\text{-}st}dt.[\/latex])<span data-type=\"newline\">\r\n<\/span>\r\nAfter integrating by parts and evaluating the limit, you should see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165039566792\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{f\\left(t\\right)\\right\\}=\\frac{f\\left(0\\right)}{s}+\\frac{1}{s}\\left[L\\left\\{{f}^{\\prime }\\left(t\\right)\\right\\}\\right][\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThen,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165042467980\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{{f}^{\\prime }\\left(t\\right)\\right\\}=sL\\left\\{f\\left(t\\right)\\right\\}-f\\left(0\\right)[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nThus, differentiation in the time domain simplifies to multiplication by <em data-effect=\"italics\">s<\/em> in the frequency domain.<span data-type=\"newline\">\r\n<\/span>\r\nThe final thing we look at in this project is how the Laplace transforms of [latex]f\\left(t\\right)[\/latex] and its antiderivative are related. Let [latex]g\\left(t\\right)={\\displaystyle\\int }_{0}^{t}f\\left(u\\right)du[\/latex]. Then,<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165039564752\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{g\\left(t\\right)\\right\\}={\\displaystyle\\int }_{0}^{\\infty }{e}^{\\text{-}st}g\\left(t\\right)dt=\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}g\\left(t\\right)dt[\/latex].<\/div>\r\n&nbsp;<\/li>\r\n \t<li style=\"text-align: left;\">Use integration by parts to evaluate [latex]\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}g\\left(t\\right)dt[\/latex]. (Let [latex]u=g\\left(t\\right)[\/latex] and [latex]dv={e}^{\\text{-}st}dt[\/latex]. Note, by the way, that we have defined [latex]g\\left(t\\right)[\/latex], [latex]du=f\\left(t\\right)dt.[\/latex] <span data-type=\"newline\">\r\n<\/span>\r\nAs you might expect, you should see that<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1165042657860\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{g\\left(t\\right)\\right\\}=\\frac{1}{s}\\cdot L\\left\\{f\\left(t\\right)\\right\\}[\/latex].<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>\r\nIntegration in the time domain simplifies to division by <em data-effect=\"italics\">s<\/em> in the frequency domain.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section id=\"fs-id1165042657932\" class=\"key-concepts\" data-depth=\"1\"><\/section>\r\n<div data-type=\"glossary\"><\/div>","rendered":"<div class=\"textbox learning-objectives\" data-type=\"abstract\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Use the comparison theorem to determine whether a definite integral is convergent<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165043273053\">It is not always easy or even possible to evaluate an improper integral directly; however, by comparing it with another carefully chosen integral, it may be possible to determine its convergence or divergence. To see this, consider two continuous functions [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] satisfying [latex]0\\le f\\left(x\\right)\\le g\\left(x\\right)[\/latex] for [latex]x\\ge a[\/latex] (Figure 5). In this case, we may view integrals of these functions over intervals of the form [latex]\\left[a,t\\right][\/latex] as areas, so we have the relationship<\/p>\n<div id=\"fs-id1165043385358\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]0\\le {\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx\\le {\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx\\text{ for }t\\ge a[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<figure id=\"CNX_Calc_Figure_07_07_005\"><figcaption><\/figcaption><div style=\"width: 497px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165042706165\">Thus, if<\/p>\n<div id=\"fs-id1165042706168\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx=\\text{+}\\infty[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042359605\">then<\/p>\n<p id=\"fs-id1165042359608\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }g\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx=\\text{+}\\infty[\/latex] as well. That is, if the area of the region between the graph of [latex]f\\left(x\\right)[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex] is infinite, then the area of the region between the graph of [latex]g\\left(x\\right)[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex] is infinite too.<\/p>\n<p id=\"fs-id1165042387260\">On the other hand, if<\/p>\n<p id=\"fs-id1165042387263\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }g\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx=L[\/latex] for some real number [latex]L[\/latex], then<\/p>\n<p id=\"fs-id1165043314585\">[latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] must converge to some value less than or equal to [latex]L[\/latex], since [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx[\/latex] increases as [latex]t[\/latex] increases and [latex]{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx\\le L[\/latex] for all [latex]t\\ge a[\/latex].<\/p>\n<p id=\"fs-id1165042851612\">If the area of the region between the graph of [latex]g\\left(x\\right)[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex] is finite, then the area of the region between the graph of [latex]f\\left(x\\right)[\/latex] and the <em data-effect=\"italics\">x<\/em>-axis over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex] is also finite.<\/p>\n<p id=\"fs-id1165043207372\">These conclusions are summarized in the following theorem.<\/p>\n<div id=\"fs-id1165043207376\" class=\"theorem\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\" data-type=\"title\">A Comparison Theorem<\/h3>\n<hr \/>\n<p id=\"fs-id1165043207383\">Let [latex]f\\left(x\\right)[\/latex] and [latex]g\\left(x\\right)[\/latex] be continuous over [latex]\\left[a,\\text{+}\\infty \\right)[\/latex]. Assume that [latex]0\\le f\\left(x\\right)\\le g\\left(x\\right)[\/latex] for [latex]x\\ge a[\/latex].<\/p>\n<ol id=\"fs-id1165042558470\" type=\"i\">\n<li>If [latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx=\\text{+}\\infty[\/latex], then [latex]{\\displaystyle\\int }_{a}^{+\\infty }g\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx=\\text{+}\\infty[\/latex].<\/li>\n<li>If [latex]{\\displaystyle\\int }_{a}^{+\\infty }g\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}g\\left(x\\right)dx=L[\/latex], where [latex]L[\/latex] is a real number, then [latex]{\\displaystyle\\int }_{a}^{+\\infty }f\\left(x\\right)dx=\\underset{t\\to \\text{+}\\infty }{\\text{lim}}{\\displaystyle\\int }_{a}^{t}f\\left(x\\right)dx=M[\/latex] for some real number [latex]M\\le L[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165039554188\" data-type=\"example\">\n<div id=\"fs-id1165039554190\" data-type=\"exercise\">\n<div id=\"fs-id1165039554192\" data-type=\"problem\">\n<div data-type=\"title\">\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<div class=\"textbox examples\">\n<h3>Recall: Algebra of inequalities<\/h3>\n<p>If [latex]0 < f(x) \\le g(x)[\/latex], then [latex]\\frac{1}{f(x)} \\ge \\frac{1}{g(x)}[\/latex]\n\nIn other words, a smaller denominator corresponds to a larger fraction, and vice versa.\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Applying the Comparison Theorem<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558869\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042601863\" data-type=\"example\">\n<div id=\"fs-id1165042601865\" data-type=\"exercise\">\n<div id=\"fs-id1165042601868\" data-type=\"problem\">\n<div data-type=\"title\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Applying the Comparison Theorem<\/h3>\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\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44558859\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use a comparison to show that [latex]{\\displaystyle\\int }_{e}^{+\\infty }\\frac{\\text{ln}x}{x}dx[\/latex] diverges.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q907753222\">Hint<\/span><\/p>\n<div id=\"q907753222\" class=\"hidden-answer\" style=\"display: none\">\n[latex]\\frac{1}{x}\\le \\frac{\\text{ln}x}{x}[\/latex] on [latex]\\left[e,\\text{+}\\infty \\right)[\/latex]\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q6657883\">Show Solution<\/span><\/p>\n<div id=\"q6657883\" class=\"hidden-answer\" style=\"display: none\">\nSince [latex]{\\displaystyle\\int }_{e}^{+\\infty }\\frac{1}{x}dx=\\text{+}\\infty[\/latex], [latex]{\\displaystyle\\int }_{e}^{+\\infty }\\frac{\\text{ln}x}{x}dx[\/latex] diverges\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/8QzQO748O58?controls=0&amp;start=1783&amp;end=1837&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<p class=\"p1\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus+II\/Transcripts\/3.7ImproperIntegrals1783to1837_transcript.html\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.7 Improper Integrals&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox tryit\">\n<h3 style=\"text-align: center;\" data-type=\"title\">ACTIVITY: Laplace Transforms<\/h3>\n<p id=\"fs-id1165042854865\">In the last few modules, we have looked at several ways to use integration for solving real-world problems. For this next project, we are going to explore a more advanced application of integration: integral transforms. Specifically, we describe the <span class=\"no-emphasis\" data-type=\"term\">Laplace transform<\/span> and some of its properties. The Laplace transform is used in engineering and physics to simplify the computations needed to solve some problems. It takes functions expressed in terms of time and <em data-effect=\"italics\">transforms<\/em> them to functions expressed in terms of frequency. It turns out that, in many cases, the computations needed to solve problems in the frequency domain are much simpler than those required in the time domain.<\/p>\n<p id=\"fs-id1165042854884\">The Laplace transform is defined in terms of an integral as<\/p>\n<div id=\"fs-id1165042854888\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{f\\left(t\\right)\\right\\}=F\\left(s\\right)={\\displaystyle\\int }_{0}^{\\infty }{e}^{\\text{-}st}f\\left(t\\right)dt[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043389093\">Note that the input to a Laplace transform is a function of time, [latex]f\\left(t\\right)[\/latex], and the output is a function of frequency, [latex]F\\left(s\\right)[\/latex]. Although many real-world examples require the use of complex numbers (involving the imaginary number [latex]i=\\sqrt{-1}[\/latex]), in this project we limit ourselves to functions of real numbers.<\/p>\n<p id=\"fs-id1165043389151\">Let\u2019s start with a simple example. Here we calculate the Laplace transform of [latex]f\\left(t\\right)=t[\/latex] . We have<\/p>\n<div id=\"fs-id1165039566620\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{t\\right\\}={\\displaystyle\\int }_{0}^{\\infty }t{e}^{\\text{-}st}dt[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165039566673\">This is an improper integral, so we express it in terms of a limit, which gives<\/p>\n<div id=\"fs-id1165039566676\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{t\\right\\}={\\displaystyle\\int }_{0}^{\\infty }t{e}^{\\text{-}st}dt=\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}t{e}^{\\text{-}st}dt[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165039566031\">Now we use integration by parts to evaluate the integral. Note that we are integrating with respect to <em data-effect=\"italics\">t<\/em>, so we treat the variable <em data-effect=\"italics\">s<\/em> as a constant. We have<\/p>\n<div id=\"fs-id1165039566045\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cccccccc}\\hfill u& =\\hfill & t\\hfill & & & \\hfill dv& =\\hfill & {e}^{\\text{-}st}dt\\hfill \\\\ \\hfill du& =\\hfill & dt\\hfill & & & \\hfill v& =\\hfill & -\\frac{1}{s}{e}^{\\text{-}st}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042323365\">Then we obtain<\/p>\n<div id=\"fs-id1165042672302\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\begin{array}{cc}\\hfill \\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}t{e}^{\\text{-}st}dt& =\\underset{z\\to \\infty }{\\text{lim}}\\left[{\\left[-\\frac{t}{s}{e}^{\\text{-}st}\\right]|}_{0}^{z}+\\frac{1}{s}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}dt\\right]\\hfill \\\\ & =\\underset{z\\to \\infty }{\\text{lim}}\\left[\\left[-\\frac{z}{s}{e}^{\\text{-}sz}+\\frac{0}{s}{e}^{-0s}\\right]+\\frac{1}{s}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}dt\\right]\\hfill \\\\ & =\\underset{z\\to \\infty }{\\text{lim}}\\left[\\left[-\\frac{z}{s}{e}^{\\text{-}sz}+0\\right]-\\frac{1}{s}{\\left[\\frac{{e}^{\\text{-}st}}{s}\\right]|}_{0}^{z}\\right]\\hfill \\\\ & =\\underset{z\\to \\infty }{\\text{lim}}\\left[\\left[-\\frac{z}{s}{e}^{\\text{-}sz}\\right]-\\frac{1}{{s}^{2}}\\left[{e}^{\\text{-}sz}-1\\right]\\right]\\hfill \\\\ & =\\underset{z\\to \\infty }{\\text{lim}}\\left[-\\frac{z}{s{e}^{sz}}\\right]-\\underset{z\\to \\infty }{\\text{lim}}\\left[\\frac{1}{{s}^{2}{e}^{sz}}\\right]+\\underset{z\\to \\infty }{\\text{lim}}\\frac{1}{{s}^{2}}\\hfill \\\\ & =0 - 0+\\frac{1}{{s}^{2}}\\hfill \\\\ & =\\frac{1}{{s}^{2}}.\\hfill \\end{array}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<ol id=\"fs-id1165042659369\" type=\"1\">\n<li>Calculate the Laplace transform of [latex]f\\left(t\\right)=1[\/latex].<\/li>\n<li>Calculate the Laplace transform of [latex]f\\left(t\\right)={e}^{-3t}[\/latex].<\/li>\n<li>Calculate the Laplace transform of [latex]f\\left(t\\right)={t}^{2}[\/latex]. (Note, you will have to integrate by parts twice.)<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nLaplace transforms are often used to solve differential equations. Differential equations are not covered in detail until later in this book; but, for now, let\u2019s look at the relationship between the Laplace transform of a function and the Laplace transform of its derivative.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nLet\u2019s start with the definition of the Laplace transform. We have<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165042692735\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{f\\left(t\\right)\\right\\}={\\displaystyle\\int }_{0}^{\\infty }{e}^{\\text{-}st}f\\left(t\\right)dt=\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}f\\left(t\\right)dt[\/latex].<\/div>\n<\/li>\n<li style=\"text-align: left;\">Use integration by parts to evaluate [latex]\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}f\\left(t\\right)dt[\/latex]. (Let [latex]u=f\\left(t\\right)[\/latex] and [latex]dv={e}^{\\text{-}st}dt.[\/latex])<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nAfter integrating by parts and evaluating the limit, you should see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165039566792\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{f\\left(t\\right)\\right\\}=\\frac{f\\left(0\\right)}{s}+\\frac{1}{s}\\left[L\\left\\{{f}^{\\prime }\\left(t\\right)\\right\\}\\right][\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThen,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165042467980\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{{f}^{\\prime }\\left(t\\right)\\right\\}=sL\\left\\{f\\left(t\\right)\\right\\}-f\\left(0\\right)[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThus, differentiation in the time domain simplifies to multiplication by <em data-effect=\"italics\">s<\/em> in the frequency domain.<span data-type=\"newline\"><br \/>\n<\/span><br \/>\nThe final thing we look at in this project is how the Laplace transforms of [latex]f\\left(t\\right)[\/latex] and its antiderivative are related. Let [latex]g\\left(t\\right)={\\displaystyle\\int }_{0}^{t}f\\left(u\\right)du[\/latex]. Then,<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165039564752\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{g\\left(t\\right)\\right\\}={\\displaystyle\\int }_{0}^{\\infty }{e}^{\\text{-}st}g\\left(t\\right)dt=\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}g\\left(t\\right)dt[\/latex].<\/div>\n<p>&nbsp;<\/li>\n<li style=\"text-align: left;\">Use integration by parts to evaluate [latex]\\underset{z\\to \\infty }{\\text{lim}}{\\displaystyle\\int }_{0}^{z}{e}^{\\text{-}st}g\\left(t\\right)dt[\/latex]. (Let [latex]u=g\\left(t\\right)[\/latex] and [latex]dv={e}^{\\text{-}st}dt[\/latex]. Note, by the way, that we have defined [latex]g\\left(t\\right)[\/latex], [latex]du=f\\left(t\\right)dt.[\/latex] <span data-type=\"newline\"><br \/>\n<\/span><br \/>\nAs you might expect, you should see that<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165042657860\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]L\\left\\{g\\left(t\\right)\\right\\}=\\frac{1}{s}\\cdot L\\left\\{f\\left(t\\right)\\right\\}[\/latex].<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span><br \/>\nIntegration in the time domain simplifies to division by <em data-effect=\"italics\">s<\/em> in the frequency domain.<\/li>\n<\/ol>\n<\/div>\n<section id=\"fs-id1165042657932\" class=\"key-concepts\" data-depth=\"1\"><\/section>\n<div data-type=\"glossary\"><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1615\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.7 Improper Integrals. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":26,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"3.7 Improper Integrals\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1615","chapter","type-chapter","status-publish","hentry"],"part":158,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1615","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1615\/revisions"}],"predecessor-version":[{"id":2074,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1615\/revisions\/2074"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/158"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1615\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1615"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1615"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1615"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1615"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}