{"id":398,"date":"2020-07-14T20:12:16","date_gmt":"2020-07-14T20:12:16","guid":{"rendered":"https:\/\/integralsandseries.in\/?p=398"},"modified":"2025-11-13T15:28:32","modified_gmt":"2025-11-13T15:28:32","slug":"the-contour-integration-approach-to-infinite-series","status":"publish","type":"post","link":"https:\/\/integralsandseries.in\/?p=398","title":{"rendered":"The Contour Integration approach to Infinite Series"},"content":{"rendered":"\n<p>\nIn this post, we will evaluate the series $\\sum_{n=0}^\\infty \\frac{\\cot\\left(\\frac{2n+1}{2}\\pi\\sqrt{2} \\right)}{(2n+1)^3}$ using contour integration. Let&#8217;s define $f:\\mathbb{C}\\to \\mathbb{C}$ as\n$$ f(z) = \\frac{\\pi \\tan(\\pi z)\\tan(\\pi z \\theta)}{z^3} $$\nwhere the parameter $\\theta$ is a positive irrational number. Let $C_N$ denote the positively oriented square with vertices $(N+1)(1+i),\\; (N+1)(-1+i),\\; (N+1)(-1-i)$ and $(N+1)(1-i)$. With some effort, one can show that:\n$$\n\\begin{aligned}\n\\left|\\int_{C_N} f(z) dz\\right| &amp;\\leq \\frac{4\\pi}{(N+1)^2} \\left(|\\tan((N+1)\\pi \\theta)| + \\frac{1}{|\\tanh((N+1)\\pi) \\tanh((N+1)\\pi \\theta)|}\\right)\n\\end{aligned}\n$$\nBy <a href=\"https:\/\/integralsandseries.in\/?p=37\">Weyl&#8217;s equidistribution theorem<\/a>, the sequence $\\{(N+1)\\theta \\}_{N=1}^\\infty $ is equidistributed modulo 1. Therefore, we can choose a subsequence $\\{(N_k+1) \\theta\\}_{k=1}^\\infty $ such that $|\\tan(\\pi (N_k+1) \\theta)|$ remains bounded. It follows that:\n$$ \\lim_{k\\to \\infty}\\int_{C_{N_k}} f(z)\\; dz = 0 $$\nOn the other hand, Residue theorem gives us:\n$$\n\\frac{1}{2\\pi i}\\int_{C_{N_k}} f(z) dz = \\substack{\\displaystyle \\text{Res} \\\\ z=0}f(z) + \\sum_{i=-{N_k}}^{N_k} \\substack{\\displaystyle \\text{Res} \\\\ z=\\frac{2i+1}{2}}f(z) + \\sum_{|j+\\frac{1}{2}|\\leq \\theta (N_k+1)} \\substack{\\displaystyle \\text{Res} \\\\ z=\\frac{2j+1}{2\\theta}}f(z) \n$$\n<\/p>\n\n\n\n<p>\nThis means that the sum of residues of $f(z)$ at it&#8217;s poles is equal to zero. A simple calculation shows that:\n$$\n\\begin{aligned}\n\\substack{\\displaystyle \\text{Res} \\\\ z=0}f(z) &amp;= \\pi^3 \\theta \\\\\n\\substack{\\displaystyle \\text{Res} \\\\ z=\\frac{2n+1}{2}}f(z)  &amp;= -8\\frac{\\tan\\left(\\frac{\\pi \\theta}{2}(2n+1) \\right)}{(2n+1)^3}, \\quad n\\in \\{0,1,2,\\cdots\\}\\\\\n\\substack{\\displaystyle \\text{Res} \\\\ z=\\frac{2n+1}{2\\theta}}f(z)  &amp;= -8\\theta^2 \\frac{\\tan\\left(\\frac{\\pi}{2\\theta}(2n+1) \\right)}{(2n+1)^3} , \\quad n\\in \\{0,1,2,\\cdots\\}\n\\end{aligned}\n$$\nFinally, putting everything together gives us the relation:\n$$\n\\sum_{n=0}^\\infty \\frac{\\tan\\left(\\frac{\\pi \\theta}{2}(2n+1) \\right)}{(2n+1)^3} + \\theta^2 \\sum_{n=0}^\\infty \\frac{\\tan\\left(\\frac{\\pi}{2\\theta}(2n+1) \\right)}{(2n+1)^3} = \\frac{\\pi^3 \\theta}{16}\n$$\nThe final result is obtained by substituting $\\theta = \\sqrt{2}+1$ in the above equation. \n$$\\boxed{\\sum_{n=0}^\\infty \\frac{\\cot\\left(\\frac{2n+1}{2}\\pi\\sqrt{2} \\right)}{(2n+1)^3} = -\\frac{\\pi^3}{32\\sqrt{2}}} $$\n<\/p>\n\n\n","protected":false},"excerpt":{"rendered":"<p>In this post, we will evaluate the series $\\sum_{n=0}^\\infty \\frac{\\cot\\left(\\frac{2n+1}{2}\\pi\\sqrt{2} \\right)}{(2n+1)^3}$ using contour integration. Let&#8217;s define $f:\\mathbb{C}\\to \\mathbb{C}$ as $$ f(z) = \\frac{\\pi \\tan(\\pi z)\\tan(\\pi z \\theta)}{z^3} $$ where the parameter $\\theta$ is a positive irrational number. Let $C_N$ denote &hellip; <a href=\"https:\/\/integralsandseries.in\/?p=398\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_bbp_topic_count":0,"_bbp_reply_count":0,"_bbp_total_topic_count":0,"_bbp_total_reply_count":0,"_bbp_voice_count":0,"_bbp_anonymous_reply_count":0,"_bbp_topic_count_hidden":0,"_bbp_reply_count_hidden":0,"_bbp_forum_subforum_count":0,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-398","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"_links":{"self":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/398","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=398"}],"version-history":[{"count":38,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/398\/revisions"}],"predecessor-version":[{"id":1166,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/398\/revisions\/1166"}],"wp:attachment":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=398"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=398"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=398"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}