{"id":1028,"date":"2023-11-27T18:28:58","date_gmt":"2023-11-27T18:28:58","guid":{"rendered":"https:\/\/integralsandseries.in\/?p=1028"},"modified":"2025-11-09T16:58:19","modified_gmt":"2025-11-09T16:58:19","slug":"integral-representations-of-the-reciprocal-beta-function","status":"publish","type":"post","link":"https:\/\/integralsandseries.in\/?p=1028","title":{"rendered":"Integral representations of the reciprocal beta function"},"content":{"rendered":"\n<p>\nIn this post, we&#8217;ll prove a very interesting identity:\n$$ \\int_0^\\pi \\left(\\sin(\\theta) \\right)^{\\alpha-1} e^{i \\beta \\theta} \\; d\\theta = \\frac{\\pi e^{\\frac{i \\pi}{2} \\beta}}{\\alpha 2^{\\alpha-1} B \\left(\\frac{\\alpha+\\beta+1}{2}, \\frac{\\alpha-\\beta+1}{2} \\right)} $$\nwhere $\\beta + 1> \\alpha > 0$ and $B(x,y) = \\frac{\\Gamma(x)\\Gamma(y)}{\\Gamma(x+y)}$ is the <a href=\"https:\/\/mathworld.wolfram.com\/BetaFunction.html\">Beta function<\/a>.\n<\/p>\n<p>\nThe idea is to integrate the principal branch of $f(z) = (1-z^2)^{\\alpha-1} (-iz)^{\\beta &#8211; \\alpha}$ around the following contour:\n<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"2500\" height=\"1281\" src=\"https:\/\/integralsandseries.in\/wp-content\/uploads\/2023\/11\/contour11.png\" alt=\"\" class=\"wp-image-1043\" style=\"width:607px;height:auto\" srcset=\"https:\/\/integralsandseries.in\/wp-content\/uploads\/2023\/11\/contour11.png 2500w, https:\/\/integralsandseries.in\/wp-content\/uploads\/2023\/11\/contour11-300x154.png 300w, https:\/\/integralsandseries.in\/wp-content\/uploads\/2023\/11\/contour11-1024x525.png 1024w, https:\/\/integralsandseries.in\/wp-content\/uploads\/2023\/11\/contour11-768x394.png 768w, https:\/\/integralsandseries.in\/wp-content\/uploads\/2023\/11\/contour11-1536x787.png 1536w, https:\/\/integralsandseries.in\/wp-content\/uploads\/2023\/11\/contour11-2048x1049.png 2048w\" sizes=\"auto, (max-width: 2500px) 100vw, 2500px\" \/><\/figure>\n<\/div>\n\n\n<p>\nwhere\n<ul>\n  <li>$C_\\epsilon$ is an arc parameterized by $e^{it}$, where $\\varphi_\\epsilon\\leq t \\leq \\pi &#8211; \\varphi_\\epsilon$ and $\\varphi_\\epsilon = \\text{arctan}\\left( \\frac{\\epsilon \\sqrt{4-\\epsilon^2}}{2-\\epsilon^2}\\right)$.<\/li>\n  <li>$I_{1,\\epsilon}, \\; I_{2,\\epsilon}, \\; I_{3,\\epsilon}$ are circular indents of radius $\\epsilon$ around the branch points -1, 1 and 0, respectively.<\/li>\n  <li>$S_{1,\\epsilon}$ is a line joining $-1+\\epsilon$ and $-\\epsilon$.<\/li>\n  <li>$S_{2,\\epsilon}$ is a line joining $\\epsilon$ and $1-\\epsilon$.<\/li>\n<\/ul>\n<\/p>\n\n\n\n\n<p>\nThe condition $\\beta + 1 > \\alpha > 0 $ ensures that:\n$$ \\lim_{\\epsilon\\to 0^+}\\int_{I_{1,\\epsilon}}f(z)\\; dz = \\lim_{\\epsilon\\to 0^+}\\int_{I_{2,\\epsilon}}f(z)\\; dz = \\lim_{\\epsilon\\to 0^+}\\int_{I_{3,\\epsilon}}f(z)\\; dz = 0 $$\nWe&#8217;ll only prove that the integral around $I_{3,\\epsilon}$ tends to 0 as $\\epsilon \\to 0^+$. The proof is quite similar for $I_{1,\\epsilon}$ and $I_{2,\\epsilon}$. We have\n$$ \\begin{aligned} \\left| \\int_{I_{3,\\epsilon}} f(z) \\; dz\\right| &#038;= \\left|-i\\epsilon \\int_0^\\pi e^{it} f(\\epsilon e^{it})\\; dt\\right| \\\\ &#038;\\leq \\epsilon\\int_0^\\pi |f(\\epsilon e^{it})| \\; dt \\\\  &#038;\\leq \\epsilon^{\\beta-\\alpha+1} \\int_0^\\pi |1-\\epsilon^2 e^{2it}|^{\\alpha-1} \\; dt \\\\ &#038;\\leq \\pi \\epsilon^{\\beta-\\alpha+1}(1+\\epsilon^2)^{\\alpha-1} \\end{aligned} $$\nHence, $\\left| \\int_{I_{3,\\epsilon}} f(z) \\; dz\\right| \\to 0$ as $\\epsilon\\to 0^+$. Next, by Cauchy&#8217;s theorem we have:\n$$ \\begin{aligned}\n\\int_{-1}^1 f(z) \\; dz &#038;= -i\\int_0^\\pi e^{it}f(e^{it})\\; dt\n\\end{aligned} $$\nThe integral on the left hand side is:\n$$ \\begin{aligned}\\int_{-1}^1 f(z) \\; dz &#038;= 2\\cos\\left( (\\beta-\\alpha)\\frac{\\pi}{2}\\right)\\int_0^1 (1-z^2)^{\\alpha-1} z^{\\beta-\\alpha}\\; dz \\\\ &#038;= \\cos\\left( (\\beta-\\alpha)\\frac{\\pi}{2}\\right) \\int_0^1 (1-w)^{\\alpha-1} w^{\\frac{\\beta-\\alpha-1}{2}}\\; dw \\\\ &#038;= \\cos\\left( (\\beta-\\alpha)\\frac{\\pi}{2}\\right) B\\left(\\alpha, \\frac{\\beta-\\alpha+1}{2} \\right) \\\\ &#038;= \\cos\\left( (\\beta-\\alpha)\\frac{\\pi}{2}\\right) \\frac{\\Gamma(\\alpha) \\Gamma\\left(\\frac{\\beta-\\alpha+1}{2} \\right)}{\\Gamma\\left(\\frac{\\beta+\\alpha+1}{2} \\right)} \\\\ &#038;= \\frac{\\pi \\Gamma(\\alpha)}{\\Gamma\\left(\\frac{\\beta+\\alpha+1}{2} \\right) \\Gamma\\left(\\frac{-\\beta+\\alpha+1}{2} \\right)} \\\\\n&#038;= \\frac{\\pi}{\\alpha B \\left(\\frac{\\alpha+\\beta+1}{2}, \\frac{\\alpha-\\beta+1}{2} \\right)}\\end{aligned} $$\nThe integral on the right hand side is:\n<\/p>\n<p>\n$$ \\begin{aligned}-i\\int_0^\\pi e^{it}f(e^{it})\\; dt &#038;= \\int_0^\\pi (1-e^{2it})^{\\alpha-1}(-i e^{it})^{\\beta-\\alpha+1} \\; dt \\\\\n&#038;= \\int_0^\\pi (2\\sin(t))^{\\alpha-1}e^{i\\beta t -\\frac{i\\pi}{2}\\beta}\\; dt \\\\\n&#038;= e^{-\\frac{i\\pi}{2}\\beta} 2^{\\alpha-1} \\int_0^\\pi (\\sin (t))^{\\alpha-1}e^{i\\beta t}\\; dt\n\\end{aligned} $$\n<\/p>\n\n\n\n<p>Therefore, we have:<\/p>\n<p>\n$$ \\begin{aligned}e^{-\\frac{i\\pi}{2}\\beta} 2^{\\alpha-1} \\int_0^\\pi (\\sin (t))^{\\alpha-1}e^{i\\beta t}\\; dt &#038;=  \\frac{\\pi}{\\alpha B \\left(\\frac{\\alpha+\\beta+1}{2}, \\frac{\\alpha-\\beta+1}{2} \\right)} \\\\ \\implies \\int_0^\\pi \\left(\\sin(t) \\right)^{\\alpha-1} e^{i \\beta t} \\; dt &#038;= \\frac{\\pi e^{\\frac{i \\pi}{2} \\beta}}{\\alpha 2^{\\alpha-1} B \\left(\\frac{\\alpha+\\beta+1}{2}, \\frac{\\alpha-\\beta+1}{2} \\right)}\\end{aligned}$$\nWe can use a similar technique to show that:\n$$\\int_{-\\frac{\\pi}{2}}^{\\frac{\\pi}{2}}\\left(\\cos(t)\\right)^{\\alpha-1}e^{i\\beta t}\\; dt = \\frac{\\pi}{\\alpha 2^{\\alpha-1} B \\left(\\frac{\\alpha+\\beta+1}{2}, \\frac{\\alpha-\\beta+1}{2} \\right)}, \\quad \\beta+1 > \\alpha > 0 $$\n<\/p>\n\n\n\n<p>One can further argue using the <a href=\"https:\/\/en.wikipedia.org\/wiki\/Identity_theorem\">identity theorem<\/a> that these results still hold even if we relax the requirement: $\\beta + 1> \\alpha$.<\/p><p> Many interesting logarithmic integrals can be derived using these identities by differentiating both sides with respect to the parameters $\\alpha$ and $\\beta$. One such example is:<\/p>\n$$ \\int_0^{\\frac{\\pi}{2}} \\theta^2 \\log^2\\left(\\cos \\theta\\right)d\\theta = \\frac{11\\pi^5}{1440}+\\frac{\\pi^3}{24}\\log^2(2)+\\frac{\\pi}{2} \\log(2)\\zeta(3) $$\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>In this post, we&#8217;ll prove a very interesting identity: $$ \\int_0^\\pi \\left(\\sin(\\theta) \\right)^{\\alpha-1} e^{i \\beta \\theta} \\; d\\theta = \\frac{\\pi e^{\\frac{i \\pi}{2} \\beta}}{\\alpha 2^{\\alpha-1} B \\left(\\frac{\\alpha+\\beta+1}{2}, \\frac{\\alpha-\\beta+1}{2} \\right)} $$ where $\\beta + 1> \\alpha > 0$ and $B(x,y) = \\frac{\\Gamma(x)\\Gamma(y)}{\\Gamma(x+y)}$ &hellip; <a href=\"https:\/\/integralsandseries.in\/?p=1028\">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":[11,6],"tags":[],"class_list":["post-1028","post","type-post","status-publish","format-standard","hentry","category-beta-function","category-logarithmic-integrals"],"_links":{"self":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/1028","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=1028"}],"version-history":[{"count":58,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/1028\/revisions"}],"predecessor-version":[{"id":1138,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/1028\/revisions\/1138"}],"wp:attachment":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1028"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1028"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1028"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}