{"id":679,"date":"2020-09-07T18:58:20","date_gmt":"2020-09-07T18:58:20","guid":{"rendered":"https:\/\/integralsandseries.in\/?p=679"},"modified":"2025-11-10T17:45:55","modified_gmt":"2025-11-10T17:45:55","slug":"evaluating-very-nasty-logarithmic-integrals-part-iii","status":"publish","type":"post","link":"https:\/\/integralsandseries.in\/?p=679","title":{"rendered":"Evaluating very nasty logarithmic integrals: Part III"},"content":{"rendered":"\n<p>\nThis is part 3 of our series on very nasty logarithmic integrals. Please have a look at <a href=\"https:\/\/integralsandseries.in\/?p=476\">part 1<\/a> and <a href=\"https:\/\/integralsandseries.in\/?p=532\">part 2<\/a> before reading this post.\n<\/p>\n<h2>Integral #5<\/h2>\n<p>\nThe first integral that we will evaluate in this post is the following:\n$$\nI_1 = \\int_0^1 \\frac{\\log^2(x) \\arctan(x)}{1+x^2}dx\n$$\nOf course, one can use brute force methods to find a closed form anti-derivative in terms of polylogarithms. Instead, a more elegant solution is possible by contour integration.\n<\/p>\n<p>\nWe&#8217;ll integrate the principal branch of $f(z) = \\frac{\\arctan(z)}{1+z^2}\\left(\\text{arctanh}^2(z) + \\frac{\\pi^2}{16} \\right)$ around the following contour:\n<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"334\" height=\"319\" src=\"https:\/\/integralsandseries.in\/wp-content\/uploads\/2020\/09\/contour.png\" alt=\"\" class=\"wp-image-682\" srcset=\"https:\/\/integralsandseries.in\/wp-content\/uploads\/2020\/09\/contour.png 334w, https:\/\/integralsandseries.in\/wp-content\/uploads\/2020\/09\/contour-300x287.png 300w\" sizes=\"auto, (max-width: 334px) 100vw, 334px\" \/><\/figure>\n<\/div>\n\n\n<p>\nwhere \n<ul>\n<li>$\\gamma_{3,\\epsilon}$ is an arc parameterized by $e^{it}$, where $\\arctan\\left(\\frac{\\epsilon \\sqrt{4-\\epsilon^2}}{2-\\epsilon^2} \\right)\\leq t \\leq \\frac{\\pi}{2} &#8211; \\arctan\\left(\\frac{\\epsilon \\sqrt{4-\\epsilon^2}}{2-\\epsilon^2} \\right)$.<\/li>\n<li>$\\gamma_{2,\\epsilon}$ and $\\gamma_{4,\\epsilon}$ are circular indents of radius $\\epsilon$ around the branch points at $z=1$ and $z=i$, respectively.\n<\/li>\n<li>$\\gamma_{1,\\epsilon}$ is a straight line joining $0$ and $1-\\epsilon$.<\/li>\n<li>$\\gamma_{4,\\epsilon}$ is a straight line joining $(1-\\epsilon)i$ and $0$.<\/li>\n<\/ul>\nNote that $f$ is analytic on $|z| &lt; 1$. It is easy to see that\n$$\n\\begin{aligned}\n\\lim_{\\epsilon\\to 0^+} \\int_{\\gamma_{2,\\epsilon}}f(z)\\; dz &amp;= 0 \\\\\n\\lim_{\\epsilon\\to 0^+} \\int_{\\gamma_{4,\\epsilon}}f(z)\\; dz &amp;= 0\n\\end{aligned}\n$$\nOn $\\gamma_{1,\\epsilon}$, we have\n$$\n\\begin{aligned}\n\\lim_{\\epsilon\\to 0^+} \\int_{\\gamma_{1,\\epsilon}} f(z)\\; dz &#038;= \\int_0^1\\frac{\\arctan(x)}{1+x^2}\\left(\\text{arctanh}^2(x) + \\frac{\\pi^2}{16} \\right)dx \\\\\n&#038;= \\int_0^1\\frac{\\arctan(x) \\text{arctanh}^2(x) }{1+x^2}dx +  \\frac{\\pi^2}{16} \\frac{\\arctan^2(x)}{2}\\Big|_0^1 \\\\\n&#038;= \\frac{1}{4}\\int_0^1 \\frac{\\arctan\\left(\\frac{1-x}{1+x} \\right)\\log^2(x)}{1+x^2}dx +  \\frac{\\pi^4}{512} \\quad \\left(x\\mapsto \\frac{1-x}{1+x} \\right) \\\\\n&#038;= \\frac{1}{4}\\int_0^1 \\frac{\\left(\\frac{\\pi}{4}-\\arctan(x) \\right)\\log^2(x)}{1+x^2}dx +  \\frac{\\pi^4}{512} \\\\\n&#038;= -\\frac{I_1}{4} +\\frac{\\pi}{16}\\int_0^1 \\frac{\\log^2(x)}{1+x^2}dx +\\frac{\\pi^4}{512} \\\\\n&#038;= -\\frac{I_1}{4} + \\frac{3\\pi^4}{512}\\quad \\color{blue}{\\cdots (1)}\n\\end{aligned}\n$$\nHere, we used the fact that $\\int_0^1 \\frac{\\log^2(x)}{1+x^2}dx\n = \\frac{\\pi^3}{16}$. Similarly, on $\\gamma_{4,\\epsilon}$ we have\n$$\n\\begin{aligned}\n\\lim_{\\epsilon\\to 0^+} \\int_{\\gamma_{5,\\epsilon}} f(z) \\; dz &#038;= \\int_i^0 \\frac{\\arctan(x)}{1+x^2}\\left(\\text{arctanh}^2(x) + \\frac{\\pi^2}{16} \\right)dx \\\\\n&#038;= \\int_0^1 \\frac{\\text{arctanh}(x)}{1-x^2}\\left(\\frac{\\pi^2}{16}-\\arctan^2(x) \\right)dx \\quad \\left(x\\mapsto i x \\right) \\\\\n&#038;= \\int_0^1\\frac{\\arctan(x) \\text{arctanh}^2(x) }{1+x^2}dx \\quad (\\text{IBP}) \\\\\n&#038;= -\\frac{I_1}{4}+\\frac{\\pi^4}{256} \\quad \\color{blue}{\\cdots (2)}\n\\end{aligned}\n$$\n<\/p>\n\n\n\n<p>\nFor the integral over $\\gamma_{3,\\epsilon}$, we will take advantage of the following identities:\n$$\n\\begin{aligned}\n\\arctan(e^{i\\theta}) &amp;= \\frac{\\pi}{4}+\\frac{i}{2} \\log\\left(\\frac{1+\\tan \\frac{\\theta}{2}}{1-\\tan \\frac{\\theta}{2}} \\right), \\quad 0\\leq \\theta \\leq \\frac{\\pi}{2} \\\\\n\\text{arctanh}(e^{i\\theta}) &amp;= -\\frac{1}{2}\\log \\left(\\tan\\frac{\\theta}{2} \\right)+\\frac{i\\pi}{4}, \\quad 0\\leq \\theta \\leq \\frac{\\pi}{2}\\\\\n\\end{aligned}\n$$\nWe have\n$$\n\\begin{aligned}\n\\lim_{\\epsilon\\to 0^+} \\int_{\\gamma_{3,\\epsilon}}f(z) \\; dz &amp;= i \\int_0^{\\frac{\\pi}{2}}f(e^{i\\theta}) e^{i\\theta} \\; d\\theta \\\\\n&amp;= \\frac{i}{2} \\int_0^{\\frac{\\pi}{2}}\\frac{\\left( \\frac{\\pi}{4}+\\frac{i}{2} \\log\\left(\\frac{1+\\tan \\frac{\\theta}{2}}{1-\\tan \\frac{\\theta}{2}} \\right)\\right)\\left( -\\frac{i\\pi}{4}\\log \\left(\\tan\\frac{\\theta}{2} \\right)+\\frac{1}{4}\\log^2\\left(\\tan \\frac{\\theta}{2} \\right) \\right)}{\\cos \\theta}d\\theta\n\\end{aligned}\n$$\nThe real part of the above integral is\n$$\n\\begin{aligned}\n\\text{Re}\\left[\\lim_{\\epsilon\\to 0^+} \\int_{\\gamma_{3,\\epsilon}}f(z) \\; dz \\right] &amp;= \\int_0^\\frac{\\pi}{2} \\frac{\\frac{1}{16}\\log^2\\left(\\tan\\frac{\\theta}{2} \\right)\\log\\left(\\frac{1-\\tan \\frac{\\theta}{2}}{1+\\tan \\frac{\\theta}{2}} \\right)+\\frac{\\pi^2}{32}\\log\\left(\\tan\\frac{\\theta}{2} \\right)}{\\cos \\theta} d\\theta \\\\\n&amp;= \\int_0^1 \\frac{\\frac{1}{8}\\log^2\\left(u \\right)\\log\\left(\\frac{1-u}{1+u} \\right)+\\frac{\\pi^2}{16}\\log\\left(u \\right)}{1-u^2} du \\quad \\left(u= \\tan\\frac{\\theta}{2} \\right) \\\\\n&amp;= -\\frac{1}{8} \\int_0^1 \\frac{\\log^2(u)\\log\\left(\\frac{1+u}{1-u}\\right)}{1-u^2} + \\frac{\\pi^2}{16}\\left(-\\frac{\\pi^2}{8} \\right) \\\\\n&amp;= \\frac{1}{16}\\int_0^1 \\frac{\\log(u) \\log^2\\left(\\frac{1-u}{1+u} \\right)}{u}du &#8211; \\frac{\\pi^4}{128} \\quad \\color{blue}{\\cdots (3)}\n\\end{aligned}\n$$\nUsing the results from <a href=\"https:\/\/integralsandseries.in\/?p=476\">part 1<\/a>, we get\n$$\n\\begin{aligned}\n\\int_0^1 \\frac{\\log(u)\\log^2\\left(\\frac{1-u}{1+u} \\right)}{u}du &amp;= \\frac{7}{4}\\int_0^1 \\frac{\\log(u)\\log^2(1-u)}{u}du + 2\\int_0^1 \\frac{\\log(u)\\log^2(1+u)}{u}du \\\\\n&amp;= \\frac{7}{4}\\left(2\\zeta(4) &#8211; 2\\sum_{n=1}^\\infty \\frac{H_n}{n^3} \\right)+2\\left(2\\text{Li}_4(-1)+2\\sum_{n=1}^\\infty\\frac{(-1)^{n+1} H_n}{n^3} \\right) \\\\\n&amp;= -\\frac{7\\pi^4}{144}+4\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}H_n}{n^3}\n\\end{aligned}\n$$\nBy Cauchy&#8217;s integral theorem, the sum of (1), (2) and (3) is equal to zero. Therefore,\n$$\n\\begin{aligned}\n&amp;\\; -\\frac{I_1}{2} + \\frac{5\\pi^4}{512}+\\frac{1}{16}\\left(-\\frac{7\\pi^4}{144}+4 \\sum_{n=1}^\\infty \\frac{(-1)^{n+1}H_n}{n^3}\\right)-\\frac{\\pi^4}{128} = 0 \\\\\n&amp;\\implies I_1 = -\\frac{5\\pi^4}{2304}+\\frac{1}{2}\\sum_{n=1}^\\infty (-1)^{n+1}\\frac{H_n}{n^3} \\\\\n&amp;\\implies I_1 = -\\frac{5\\pi^4}{2304}+\\frac{1}{2}\\left(\\frac{11\\pi^4}{360}+\\frac{\\pi^2}{12}\\log^2(2)-\\frac{\\log^4(2)}{12} -\\frac{7}{4}\\log(2)\\zeta(3)-2\\text{Li}_4\\left(\\frac{1}{2} \\right)\\right) \\\\\n&amp;\\implies \\boxed{I_1 = \\frac{151 \\pi ^4}{11520}+\\frac{\\pi ^2}{24}  \\log ^2(2)-\\frac{\\log ^4(2)}{24} -\\frac{7}{8} \\zeta (3) \\log (2)-\\text{Li}_4\\left(\\frac{1}{2}\\right)}\n\\end{aligned}\n$$\nRefer to <a href=\"https:\/\/integralsandseries.in\/?p=349\">this post<\/a> for the evaluation of $\\sum_{n=1}^\\infty (-1)^{n+1}\\frac{H_n}{n^3}$.\n<\/p>\n\n\n\n<h2>Integral #6<\/h2>\n<p>\nThe next integral that we&#8217;ll evaluate is \n$$\nI_2 = \\int_0^1 \\frac{\\log^3(x) \\arctan(x)}{1+x^2}dx \\quad \\color{blue}{\\cdots (4)}\n$$\nUsing the transformation $x\\mapsto \\frac{1}{x}$, we write the integral as:\n$$\n\\begin{aligned}\nI_2 &amp;= -\\int_1^\\infty\\frac{\\log^3(x)\\left(\\frac{\\pi}{2}-\\arctan(x) \\right)}{1+x^2}dx \\quad \\color{blue}{\\cdots (5)}\n\\end{aligned}\n$$\nAdding up equations (4) and (5) and dividing both sides by 2 gives us:\n$$\n\\begin{aligned}\nI_2 &amp;=  -\\frac{\\pi}{4}\\int_1^\\infty \\frac{\\log^3(x)}{1+x^2}dx + \\frac{1}{2}\\int_0^\\infty \\frac{\\log^3(x) \\arctan(x)}{1+x^2}dx\\\\\n&amp;= \\frac{\\pi}{4}\\int_0^1 \\frac{\\log^3(x)}{1+x^2}dx + \\frac{1}{2}\\int_0^\\infty \\frac{\\log^3(x) \\arctan(x)}{1+x^2}dx\n\\end{aligned}\n$$\nNote that\n$$\n\\begin{aligned}\n\\int_0^1 \\frac{\\log^3(x)}{1+x^2}dx &amp;= \\sum_{n=0}^\\infty (-1)^n \\int_0^1 x^{2n}\\log^3(x) \\; dx \\\\\n&amp;= -6\\sum_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)^4} \\\\\n&amp;= -6\\beta(4)\n\\end{aligned}\n$$\nSo, we have\n$$ I_2 = &#8211; \\frac{3\\pi}{2}\\beta(4) + \\frac{1}{2}\\int_0^\\infty \\frac{\\log^3(x) \\arctan(x)}{1+x^2}dx \\quad \\color{blue}{\\cdots (6)} $$\nTo evaluate the integral in the above equation, we will use the Feynman technique. Define the function $F:[0,\\infty) \\to \\mathbb{R}$ as \n$F(s) = \\int_0^\\infty \\frac{\\log^3(x)\\arctan(s x)}{1+x^2}dx$. Now, we have\n$$ F'(s) = \\int_0^\\infty \\frac{x \\log^3(x)}{(1+x^2)(1+s^2 x^2)}dx $$\nBefore evaluating $F'(s)$, we will evaluate the simpler integral $\\int_0^\\infty \\frac{x \\log(x)}{(1+x^2)(1+s^2 x^2)}dx$. To do this, integrate the principal branch of $g(z) = \\frac{z\\log^2(-z)}{(1+z^2)(1+s^2 z^2)}$ around the following &#8220;key hole&#8221; contour:\n<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"360\" height=\"375\" src=\"https:\/\/integralsandseries.in\/wp-content\/uploads\/2020\/09\/contour1.png\" alt=\"\" class=\"wp-image-757\" srcset=\"https:\/\/integralsandseries.in\/wp-content\/uploads\/2020\/09\/contour1.png 360w, https:\/\/integralsandseries.in\/wp-content\/uploads\/2020\/09\/contour1-288x300.png 288w\" sizes=\"auto, (max-width: 360px) 100vw, 360px\" \/><\/figure>\n<\/div>\n\n\n<p>\nwhere:\n<ul>\n<li>$\\gamma_{1,\\epsilon, R}$ is a line joining the points $\\sqrt{R^2-\\epsilon^2}-i\\epsilon$ and $-i\\epsilon$.<\/li>\n<li>$\\gamma_{2,\\epsilon, R}$ is a line joining the points $i\\epsilon$ and $\\sqrt{R^2-\\epsilon^2}+i\\epsilon$.<\/li>\n<li>$\\gamma_{3,\\epsilon}$ is parameterized by $\\epsilon e^{-i t}$ where $\\frac{\\pi}{2} \\leq t \\leq \\frac{3\\pi}{2}$.<\/li>\n<li>$\\gamma_{4,R}$ is parameterized by $R e^{i t}$ where $\\arctan\\left(\\frac{\\epsilon}{\\sqrt{R^2-\\epsilon^2}} \\right) \\leq t \\leq 2\\pi &#8211; \\arctan\\left(\\frac{\\epsilon}{\\sqrt{R^2-\\epsilon^2}} \\right)$.<\/li>\n<\/ul>\nAs $\\epsilon\\to 0^+$ and $R\\to \\infty$, the integrals along $\\gamma_{3,\\epsilon}$ and $\\gamma_{4,R}$ tend to 0. Therefore, we are only left with the integrals above and below the branch cut. The residues of $g(z)$ at it&#8217;s poles is given by:\n$$\n\\begin{aligned}\n\\mathop{\\text{Res}}\\limits_{z=i} \\; g(z) &#038;= -\\frac{\\pi^2}{8(1-s^2)} \\\\\n\\mathop{\\text{Res}}\\limits_{z=-i} \\; g(z) &#038;= -\\frac{\\pi^2}{8(1-s^2)} \\\\\n\\mathop{\\text{Res}}\\limits_{z=i\/s} \\; g(z) &#038;= -\\frac{\\left(\\log(s)+\\frac{i\\pi}{2} \\right)^2}{2(1-s^2)} \\\\\n\\mathop{\\text{Res}}\\limits_{z=-i\/s} \\; g(z) &#038;= -\\frac{\\left(\\log(s)-\\frac{i\\pi}{2} \\right)^2}{2(1-s^2)}\n\\end{aligned}\n$$\nNow, the residue theorem gives us:\n$$\n\\begin{aligned}\n\\int_0^\\infty \\frac{x\\left((\\log (x)-i\\pi)^2 &#8211; (\\log (x)+i\\pi)^2 \\right)}{(1+x^2)(1+s^2 x^2)}dx &#038;= 2i\\pi \\Big(\\mathop{\\text{Res}}\\limits_{z=i} \\; g(z)+ \\mathop{\\text{Res}}\\limits_{z=-i}\\; g(z) + \\mathop{\\text{Res}}\\limits_{z=i\/s}\\; g(z) \\\\ &#038;\\quad + \\mathop{\\text{Res}}\\limits_{z=-i\/s}\\; g(z) \\Big) \\\\\n\\implies -4i\\pi \\int_0^\\infty \\frac{x\\log(x)}{(1+x^2)(1+s^2 x^2)}dx &#038;= -2i\\pi \\frac{\\log^2(s)}{1-s^2} \\\\\n\\implies \\int_0^\\infty \\frac{x\\log(x)}{(1+x^2)(1+s^2 x^2)}dx &#038;= \\frac{\\log^2(s)}{2(1-s^2)} \\quad \\color{blue}{\\cdots (7)}\n\\end{aligned}\n$$\nTo evaluate $F'(s)$, we integrate the principal branch of $h(z) = \\frac{z \\log^4(-z)}{(1+z^2)(1+s^2 z^2)}$ around the same contour. This time, the residues are:\n$$\n\\begin{aligned}\n\\mathop{\\text{Res}}\\limits_{z=i} \\; h(z) &#038;= \\frac{\\pi^4}{32(1-s^2)} \\\\\n\\mathop{\\text{Res}}\\limits_{z=-i} \\; h(z) &#038;= \\frac{\\pi^4}{32(1-s^2)} \\\\\n\\mathop{\\text{Res}}\\limits_{z=i\/s} \\; h(z) &#038;= -\\frac{\\left(\\log(s)+\\frac{i\\pi}{2} \\right)^4}{2(1-s^2)} \\\\\n\\mathop{\\text{Res}}\\limits_{z=-i\/s} \\; h(z) &#038;= -\\frac{\\left(\\log(s)-\\frac{i\\pi}{2} \\right)^4}{2(1-s^2)}\n\\end{aligned}\n$$\nThe residue theorem gives us:\n$$\n\\begin{aligned}\n\\int_0^\\infty \\frac{x\\left((\\log (x)-i\\pi)^4 &#8211; (\\log (x)+i\\pi)^4 \\right)}{(1+x^2)(1+s^2 x^2)}dx &#038;= 2i\\pi \\Big(\\mathop{\\text{Res}}\\limits_{z=i} \\; h(z)+ \\mathop{\\text{Res}}\\limits_{z=-i}\\; h(z) + \\mathop{\\text{Res}}\\limits_{z=i\/s}\\; h(z) \\\\ &#038;\\quad + \\mathop{\\text{Res}}\\limits_{z=-i\/s}\\; h(z) \\Big) \\\\\n\\implies -8i\\pi \\int_0^\\infty \\frac{x\\left(\\log^3(x) -\\pi^2 \\log(x) \\right)}{(1+x^2)(1+s^2 x^2)}dx &#038;= 2i\\pi \\frac{\\frac{3\\pi^2}{2}\\log^2(s)-\\log^4(s)}{1-s^2} \\\\\n\\implies \\int_0^\\infty \\frac{x\\log^3(x)}{(1+x^2)(1+s^2 x^2)}dx &#038;= \\frac{\\pi^2\\log^2(s)+2\\log^4(s)}{8(1-s^2)} \\quad \\color{blue}{\\cdots (8)}\n\\end{aligned}\n$$\nNote that we used equation (7) to the get the above result. We can now calculate our original integral as follows:\n$$\n\\begin{aligned}\nI_2 &#038;= -\\frac{3\\pi}{2}\\beta(4) + \\frac{1}{2}\\int_0^1 F'(s) ds \\\\\n&#038;= -\\frac{3\\pi}{2}\\beta(4) + \\frac{1}{2}\\int_0^1 \\frac{\\pi^2\\log^2(s)+2\\log^4(s)}{8(1-s^2)} ds \\\\\n&#038;= -\\frac{3\\pi}{2}\\beta(4) + \\frac{\\pi^2}{16}\\int_0^1\\frac{\\log^2(s)}{1-s^2}ds+\\frac{1}{8}\\int_0^1 \\frac{\\log^4(s)}{1-s^2} ds \\\\\n&#038;= -\\frac{3\\pi}{2}\\beta(4) + \\frac{\\pi^2}{16}\\sum_{k=0}^\\infty\\int_0^1 s^{2k}\\log^2(s)ds+\\frac{1}{8}\\sum_{k=0}^\\infty\\int_0^1 s^{2k}\\log^4(s) ds \\\\\n&#038;= -\\frac{3\\pi}{2}\\beta(4) + \\frac{\\pi^2}{8}\\sum_{k=0}^\\infty\\frac{1}{(2k+1)^3}+3\\sum_{k=0}^\\infty\\frac{1}{(2k+1)^5} \\\\\n&#038;= \\boxed{-\\frac{3\\pi}{2}\\beta(4) + \\frac{7\\pi^2}{64}\\zeta(3) +\\frac{93}{32}\\zeta(5)}\n\\end{aligned}\n$$\nInterestingly, $I_2$ can be reduced into an Euler sum which can be evaluated using contour integration.\n<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>This is part 3 of our series on very nasty logarithmic integrals. Please have a look at part 1 and part 2 before reading this post. Integral #5 The first integral that we will evaluate in this post is the &hellip; <a href=\"https:\/\/integralsandseries.in\/?p=679\">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":[6],"tags":[4,5],"class_list":["post-679","post","type-post","status-publish","format-standard","hentry","category-logarithmic-integrals","tag-euler-sums","tag-polylogarithm"],"_links":{"self":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/679","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=679"}],"version-history":[{"count":87,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/679\/revisions"}],"predecessor-version":[{"id":1153,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/679\/revisions\/1153"}],"wp:attachment":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=679"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=679"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=679"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}