{"id":1,"date":"2020-01-24T07:47:14","date_gmt":"2020-01-24T07:47:14","guid":{"rendered":"http:\/\/integralsandseries.in\/?p=1"},"modified":"2025-11-16T09:56:17","modified_gmt":"2025-11-16T09:56:17","slug":"hello-world","status":"publish","type":"post","link":"https:\/\/integralsandseries.in\/?p=1","title":{"rendered":"Simplification of Incomplete Beta Functions"},"content":{"rendered":"\n<p>\nIn this post, we will derive closed form solutions for two incomplete beta integrals.\n$$\n\\begin{aligned}\n\\int_0^1\\frac{dx}{\\sqrt{1-x}\\sqrt[6]{9-x}\\sqrt[3]x}&amp;=\\frac\\pi{\\sqrt3} \\quad\\quad (1)\\\\\n\\int_0^1\\frac{dx}{\\sqrt{1-x}\\sqrt[4]x\\sqrt[4]{2-x\\sqrt3}} &amp;= \\frac{2\\sqrt{2}\\pi}{3\\sqrt[8]{3}} \\quad\\quad (2)\n\\end{aligned}\n$$\nThese problems are from a <a href=\"https:\/\/math.stackexchange.com\/questions\/541135\/conjecture-int-01-frac-mathrm-dx-sqrt1-x-sqrt4x-sqrt42-x-sqrt\">math.stackexchange.com<\/a> question posted by <a href=\"https:\/\/math.stackexchange.com\/users\/19661\/vladimir-reshetnikov?tab=questions\">Vladimir Reshetnikov<\/a>.\n<\/p>\n<p><b>Alternate forms<\/b><\/p>\n<p>\nThe equations (1) and (2) can be expressed in terms of the incomplete beta function using some elementary hypergeometric identities. This makes our problems equivalent to proving that\n$$\\begin{aligned}\nB\\left(\\frac{1}{9};\\frac{1}{6},\\frac{1}{3}\\right) &amp;= \\frac{\\sqrt{3}}{2^{\\frac{4}{3}}\\pi}\\Gamma\\left(\\frac{1}{3}\\right)^3 \\quad\\quad (3)\\\\\nB\\left(\\frac{\\sqrt{3}}{2};\\frac{1}{4},\\frac{1}{4}\\right) &amp;= \\frac{2}{3\\sqrt{\\pi}}\\Gamma\\left(\\frac{1}{4}\\right)^2 \\quad\\quad (4)\n\\end{aligned}\n$$\n<b>Proof of (3)<\/b>\n$$ B\\left(\\frac{1}{9};\\frac{1}{6},\\frac{1}{3}\\right)  = \\int_0^{\\frac{1}{9}}x^{-\\frac{5}{6}}\\left( 1-x\\right)^{-\\frac{2}{3}} dx $$\n\nSubstitute $ x=\\frac{y^3}{1+y^3}$ in the above integral to get\n\n$$ B\\left(\\frac{1}{9};\\frac{1}{6},\\frac{1}{3}\\right) =3\\int_0^{\\frac{1}{2}}\\frac{1}{\\sqrt{y+y^4}}dy $$\n\nWe can now transform this integral into a beta integral using the substitution $y=\\frac{1-z}{2+z}$.\n\n$$B\\left(\\frac{1}{9};\\frac{1}{6},\\frac{1}{3}\\right)  =3\\int_0^1\\frac{1}{\\sqrt{1-z^3}}dz = B\\left( \\frac{1}{3},\\frac{1}{2}\\right)=\\frac{\\sqrt{3}}{2^{\\frac{4}{3}}\\pi}\\Gamma\\left(\\frac{1}{3}\\right)^3 $$ \nThis proves equations (1) and (3).\n<\/p>\n<p>\n<b>Proof of (4)<\/b>\n<\/p>\n<p>\nThe proof of integrals (2) and (4) is more involved. Using the substitution $x=\\frac{4t^2}{1+4t^2}$, the integral transforms into the Weierstrass form:\n\n$$ B\\left(\\frac{\\sqrt{3}}{2};\\frac{1}{4},\\frac{1}{4}\\right) =\\int_0^{\\frac{\\sqrt{3}}{2}}x^{-\\frac{3}{4}}(1-x)^{-\\frac{3}{4}}dx=2\\sqrt{2} \\int_0^\\alpha \\frac{dt}{\\sqrt{4t^3+t}}$$ \nwhere $\\alpha=\\frac{\\sqrt{3+2\\sqrt{3}}}{2}$.  Let $\\wp(z)$ denote the Weierstrass Elliptic function with invariants $g_2=-1$ and $g_3=0$. $\\wp(z)$ satisfies the differential equation: \n$$\\wp'(z)^2=4\\wp(z)^3+\\wp(z)$$\nThe half periods are $\\omega_1=\\frac{1+i}{4}L$ and $\\omega_2=\\frac{-1+i}{4}L$ where $L=\\frac{1}{\\sqrt{2\\pi}}\\Gamma\\left(\\frac{1}{4}\\right)^2$ is the Lemniscate constant. Now, we can express our integral as follows: \n$$ B\\left(\\frac{\\sqrt{3}}{2};\\frac{1}{4},\\frac{1}{4}\\right) =2\\sqrt{2}\\left(\\wp^{-1}(0)-\\wp^{-1}(\\alpha) \\right)$$ \nNote that the following identity is easy to prove using the beta function: $$ \\int_0^\\infty \\frac{dt}{\\sqrt{4t^3+t}}=\\frac{1}{2\\sqrt{2\\pi}}\\Gamma\\left(\\frac{1}{4}\\right)^2=\\frac{L}{2}$$\nTherefore, $\\wp^{-1}(0)=\\frac{L}{2}$. To prove the claim we only need to prove that $\\wp\\left(\\frac{L}{6}\\right)=\\alpha$. Using the addition theorem of $\\wp(z)$ and the fact that $\\wp\\left(\\frac{L}{2}\\right)=0$ we get: \n$$\n\\begin{aligned}\n\\wp\\left(\\frac{L}{6}\\right) &amp;= \\wp\\left(\\frac{L}{2}-\\frac{L}{3}\\right)\\\\\n&amp;= \\frac{1}{4}\\left(\\frac{\\wp&#8217;\\left(\\frac{L}{3}\\right)}{\\wp\\left(\\frac{L}{3}\\right)} \\right)^2-\\wp\\left(\\frac{L}{3}\\right) \\\\\n&amp;= \\frac{4\\wp\\left( \\frac{L}{3}\\right)^3+\\wp\\left(\\frac{L}{3}\\right)}{4\\wp\\left(\\frac{L}{3}\\right)^2}-\\wp\\left(\\frac{L}{3}\\right)\\\\\n&amp;= \\frac{1}{4\\wp\\left(\\frac{L}{3}\\right)}\\quad\\quad\\quad (5)\n\\end{aligned}$$ \nOn the other hand, by the duplication theorem:\n$$\n\\begin{aligned} \\wp\\left(\\frac{L}{3}\\right)&amp;= \\frac{1}{4}\\frac{\\left(6\\wp\\left(\\frac{L}{6}\\right)^2+\\frac{1}{2}\\right)^2}{4\\wp\\left(\\frac{L}{6}\\right)^3+\\wp\\left(\\frac{L}{6}\\right)}-2\\wp\\left(\\frac{L}{6}\\right)\\quad\\quad\\quad ({6}) \\end{aligned}\n$$\nCombine equations (5) and (6) to get:\n\n$$\\frac{1+4\\wp\\left(\\frac{L}{6}\\right)^2}{\\left(6\\wp\\left(\\frac{L}{6}\\right)^2+\\frac{1}{2} \\right)^2-32\\wp\\left(\\frac{L}{6}\\right)^4-8\\wp\\left(\\frac{L}{6}\\right)^2}=1 \\\\\n\\implies 16\\wp\\left(\\frac{L}{6}\\right)^4-24\\wp\\left(\\frac{L}{6}\\right)^2-3=0 $$\nSince $\\wp\\left( \\frac{L}{6}\\right)$ is positive, we conclude that \n$$ \\wp\\left(\\frac{L}{6}\\right)=\\frac{\\sqrt{3+2\\sqrt{3}}}{2} $$\nThis completes our proof of equation (4) and (2). <\/p>\n","protected":false},"excerpt":{"rendered":"<p>In this post, we derive closed form solutions for two incomplete beta integrals using properties of the Weierstrass P function. These problems are from a <a href=\"https:\/\/math.stackexchange.com\/questions\/541135\/conjecture-int-01-frac-mathrm-dx-sqrt1-x-sqrt4x-sqrt42-x-sqrt\">math.stackexchange.com<\/a> question posted by <a href=\"https:\/\/math.stackexchange.com\/users\/19661\/vladimir-reshetnikov?tab=questions\">Vladimir Reshetnikov<\/a>. <a href=\"https:\/\/integralsandseries.in\/?p=1\">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":[3],"tags":[],"class_list":["post-1","post","type-post","status-publish","format-standard","hentry","category-elliptic-integrals"],"_links":{"self":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/1","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=1"}],"version-history":[{"count":10,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/1\/revisions"}],"predecessor-version":[{"id":1176,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/1\/revisions\/1176"}],"wp:attachment":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}