{"id":234112,"date":"2025-06-13T18:21:26","date_gmt":"2025-06-13T18:21:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234112"},"modified":"2025-06-13T18:21:29","modified_gmt":"2025-06-13T18:21:29","slug":"use-the-keyhole-contour-in-figure-2-17-in-the-manner-of-examples-9-and-10-to-compute-the-following","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/use-the-keyhole-contour-in-figure-2-17-in-the-manner-of-examples-9-and-10-to-compute-the-following\/","title":{"rendered":"Use the \\&#8221;keyhole\\&#8221; contour in Figure 2.17 in the manner of Examples 9 and 10 to compute the following"},"content":{"rendered":"\n<pre id=\"preorder-ask-header-text\" class=\"wp-block-preformatted\">Use the \\\"keyhole\\\" contour in Figure 2.17 in the manner of Examples 9 and 10 to compute the following integrals.&nbsp;\u22c513.\u222b0\u221ex\u03b1x2+3x+2dx,0&lt;\u03b1&lt;1&nbsp;\u22c514.\u222b0\u221exx2+2x+5dx&nbsp;Answer:&nbsp;\u03c0225\u22121&nbsp;\u22c515.\u222b0\u221ex\u03bbx2+2xcos\u2061\u03c9+1dx,\u2212\u03c0&lt;\u03c9&lt;\u03c0,0&lt;\u03bb&lt;1<\/pre>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/image-456.png\" alt=\"\" class=\"wp-image-234113\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To compute the integral I = \u222b\u2080^\u221e [x^\u03b1 \/ (x\u00b2 + 3x + 2)] dx, for 0 &lt; \u03b1 &lt; 1, the method of contour integration using a keyhole contour is employed.<\/p>\n\n\n\n<p>First, consider the complex function f(z) = z^\u03b1 \/ (z\u00b2 + 3x + 2). A branch of the multi-valued function z^\u03b1 = e^(\u03b1 log z) is chosen with the branch cut along the positive real axis, such that the argument of z is in the range 0 \u2264 arg(z) &lt; 2\u03c0. The function f(z) is integrated over a keyhole contour C, which consists of a large circle C_R of radius R, a small circle C_\u03b5 of radius \u03b5 around the origin, and two line segments, L+ and L-, running along the top and bottom of the positive real axis, respectively.<\/p>\n\n\n\n<p>The singularities of f(z) are the poles determined by the roots of the denominator z\u00b2 + 3z + 2 = (z+1)(z+2) = 0. The simple poles are at z\u2081 = -1 and z\u2082 = -2. Both poles lie inside the contour C for R &gt; 2. The residues at these poles are calculated as follows:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At z = -1 = e^(i\u03c0):<br>Res(f, -1) = lim_(z\u2192-1) (z+1)f(z) = [z^\u03b1 \/ (z+2)]_(z=-1) = (-1)^\u03b1 \/ 1 = (e^(i\u03c0))^\u03b1 = e^(i\u03c0\u03b1).<\/li>\n\n\n\n<li>At z = -2 = 2e^(i\u03c0):<br>Res(f, -2) = lim_(z\u2192-2) (z+2)f(z) = [z^\u03b1 \/ (z+1)]_(z=-2) = (-2)^\u03b1 \/ (-1) = -(2e^(i\u03c0))^\u03b1 = -2^\u03b1 e^(i\u03c0\u03b1).<\/li>\n<\/ul>\n\n\n\n<p>By the Residue Theorem, the integral around the closed contour is:<br>\u222e_C f(z) dz = 2\u03c0i [Res(f, -1) + Res(f, -2)] = 2\u03c0i (e^(i\u03c0\u03b1) &#8211; 2^\u03b1 e^(i\u03c0\u03b1)) = 2\u03c0i (1 &#8211; 2^\u03b1)e^(i\u03c0\u03b1).<\/p>\n\n\n\n<p>The contour integral is also the sum of the integrals over its four segments. The condition 0 &lt; \u03b1 &lt; 1 ensures that the integrals over the large circle C_R and the small circle C_\u03b5 vanish as R \u2192 \u221e and \u03b5 \u2192 0.<br>The integral along L+ (where arg(z)=0) becomes the desired real integral I. The integral along L- (where arg(z)=2\u03c0) becomes -e^(i2\u03c0\u03b1)I.<\/p>\n\n\n\n<p>Equating the results gives:<br>I &#8211; e^(i2\u03c0\u03b1)I = 2\u03c0i (1 &#8211; 2^\u03b1)e^(i\u03c0\u03b1)<br>I (1 &#8211; e^(i2\u03c0\u03b1)) = 2\u03c0i (1 &#8211; 2^\u03b1)e^(i\u03c0\u03b1)<\/p>\n\n\n\n<p>Solving for I:<br>I = [2\u03c0i (1 &#8211; 2^\u03b1)e^(i\u03c0\u03b1)] \/ [1 &#8211; e^(i2\u03c0\u03b1)]<\/p>\n\n\n\n<p>Using the identity 1 &#8211; e^(i2\u03b8) = e^(i\u03b8)(e^(-i\u03b8) &#8211; e^(i\u03b8)) = -2i sin(\u03b8)e^(i\u03b8), the expression simplifies:<br>I = [2\u03c0i (1 &#8211; 2^\u03b1)e^(i\u03c0\u03b1)] \/ [-2i sin(\u03c0\u03b1)e^(i\u03c0\u03b1)] = -[\u03c0(1 &#8211; 2^\u03b1)] \/ sin(\u03c0\u03b1)<\/p>\n\n\n\n<p>Thus, the value of the integral is:<\/p>\n\n\n\n<p><strong>Answer:<\/strong><br>\u222b\u2080^\u221e (x^\u03b1 \/ (x\u00b2 + 3x + 2)) dx =&nbsp;<strong>\u03c0(2^\u03b1 &#8211; 1) \/ sin(\u03c0\u03b1)<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-470.jpeg\" alt=\"\" class=\"wp-image-234114\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Use the \\&#8221;keyhole\\&#8221; contour in Figure 2.17 in the manner of Examples 9 and 10 to compute the following integrals.&nbsp;\u22c513.\u222b0\u221ex\u03b1x2+3x+2dx,0&lt;\u03b1&lt;1&nbsp;\u22c514.\u222b0\u221exx2+2x+5dx&nbsp;Answer:&nbsp;\u03c0225\u22121&nbsp;\u22c515.\u222b0\u221ex\u03bbx2+2xcos\u2061\u03c9+1dx,\u2212\u03c0&lt;\u03c9&lt;\u03c0,0&lt;\u03bb&lt;1 The Correct Answer and Explanation is: To compute the integral I = \u222b\u2080^\u221e [x^\u03b1 \/ (x\u00b2 + 3x + 2)] dx, for 0 &lt; \u03b1 &lt; 1, the method of contour integration using a keyhole [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-234112","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234112","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=234112"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234112\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=234112"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=234112"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=234112"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}