{"id":476,"date":"2020-08-01T09:15:41","date_gmt":"2020-08-01T09:15:41","guid":{"rendered":"https:\/\/integralsandseries.in\/?p=476"},"modified":"2025-11-10T17:59:17","modified_gmt":"2025-11-10T17:59:17","slug":"evaluating-very-nasty-logarithmic-integrals-part-i","status":"publish","type":"post","link":"https:\/\/integralsandseries.in\/?p=476","title":{"rendered":"Evaluating very nasty logarithmic integrals: Part I"},"content":{"rendered":"\n

\nIn this post, we’ll continue our exploration of logarithmic integrals and Euler sums. We’ll also discuss the contour integration method for evaluating Euler sums. It is recommended that the reader goes through the previous posts, (A)<\/a> and (B)<\/a>, before reading this post.<\/p>\n

Integral #1<\/h2>\n

\nOur first integral is the following:\n$$\n\\begin{aligned}\nI &= \\int_0^1 \\frac{\\log(t)\\log(1-t)\\log(1+t)}{t}dt\n\\end{aligned}\n$$\nWe will use the integral $J = \\int_0^1 \\frac{\\log(t)\\log^2(1-t)}{t}dt$ as a starting point. Make the substitution $t\\mapsto t^2$ in $J$ to obtain:\n$$\n\\begin{aligned}\nJ &= 2\\int_0^1 \\frac{\\log(t^2) \\log^2(1-t^2)}{t}dt \\\\\n&= 4\\int_0^1 \\frac{\\log(t)\\log^2(1-t)}{t}dt + 4\\int_0^1 \\frac{\\log(t)\\log^2(1+t)}{t}dt + 8\\int_0^1 \\frac{\\log(t)\\log(1-t) \\log(1+t)}{t}dt \\\\\n&= 4J + 4\\int_0^1 \\frac{\\log(t)\\log^2(1+t)}{t}dt + 8I\n\\end{aligned}\n$$\n<\/p>\n

\nSolving the above equation for $I$ gives us:\n$$\nI = -\\frac{3}{8}J – \\frac{1}{2} \\int_0^1 \\frac{\\log(t)\\log^2(1+t)}{t}dt \\quad \\color{blue}{\\cdots (1)}\n$$\nApplying integration by parts and using the generating function of the Harmonic number yields:\n$$\n\\begin{aligned}\n\\int_0^1 \\frac{\\log(t)\\log^2(1+t)}{t}dt &= -\\int_0^1 \\frac{\\log^2(t)\\log(1+t)}{1+t}dt \\\\\n&= -\\sum_{n=1}^\\infty (-1)^{n+1} H_n \\int_0^1 t^{n} \\log^2(t) \\; dt \\\\\n&= -2\\sum_{n=1}^\\infty (-1)^{n+1} \\frac{H_n}{(n+1)^3} \\\\\n&= -2\\sum_{n=0}^\\infty (-1)^{n+1} \\frac{H_{n+1}-\\frac{1}{n+1}}{(n+1)^3} \\\\\n&= 2\\sum_{n=1}^\\infty (-1)^{n+1} \\frac{H_n}{n^3} – 2\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}}{n^4} \\\\\n&= 2\\text{Li}_4(-1) + 2\\sum_{n=1}^\\infty (-1)^{n+1} \\frac{H_n}{n^3} \\quad \\color{blue}{\\cdots (2)}\n\\end{aligned}\n$$\nA similar calculation shows that:\n$$\nJ = 2\\zeta(4) – 2\\sum_{n=1}^\\infty \\frac{H_n}{n^3} \\quad \\color{blue}{\\cdots (3)}\n$$\nFrom (A)<\/a>, we know how to evaluate $\\sum_{n=1}^\\infty (-1)^{n+1} \\frac{H_n}{n^3}$ and $\\sum_{n=1}^\\infty \\frac{H_n}{n^3}$. \n$$\n\\begin{aligned}\n\\sum_{n=1}^\\infty \\frac{H_n}{n^3} &= \\frac{\\pi^4}{72} \\\\\n\\sum_{n=1}^\\infty (-1)^{n+1}\\frac{H_n}{n^3} &= \\frac{11\\pi^4}{360}+\\frac{\\pi^2}{12}\\log^2(2)-\\frac{\\log^4(2)}{12}-\\frac{7}{4}\\log(2)\\zeta(3) -2\\text{Li}_4\\left(\\frac{1}{2}\\right)\n\\end{aligned}\n$$\nFinally, putting everything together gives us:\n$$ \\boxed{I = -\\frac{3\\pi^4}{160}-\\frac{\\pi^2}{12}\\log^2(2)+\\frac{\\log^4(2)}{12}+\\frac{7}{4}\\log(2)\\zeta(3) +2\\text{Li}_4\\left(\\frac{1}{2}\\right)} $$\n

\n\n\n\n

Integral #2<\/h2>\n

\nThe next integral on our list is\n$$ K = \\int_0^1 \\frac{\\log(x)\\log(1+x^2)\\arctan(x)}{x}dx $$\nThis integral was originally posted by the user FDP on math.stackexchange.com. My solution<\/a> is posted there as well. \n<\/p>\n

\nUsing integration by parts gives us:\n$$\nK = – \\frac{1}{2}\\int_0^1 \\frac{\\log^2(x)\\log(1+x^2)}{1+x^2}dx – \\int_0^1 \\frac{x \\log^2(x) \\arctan(x)}{1+x^2}dx\n$$\nNow, we can use the following series expansions to reduce $K$ into Euler sums:\n$$\n\\begin{aligned}\n\\frac{\\arctan (x)}{1+x^2} &= \\sum_{n=0}^\\infty (-1)^n \\tilde{H}_n x^{2n+1} , \\quad |x| < 1\\\\\n\\frac{\\log(1+x^2)}{1+x^2} &= \\sum_{n=1}^\\infty (-1)^{n+1} H_n x^{2n} , \\quad |x| < 1 \n\\end{aligned}\n$$\nwhere $\\tilde{H}_n = \\sum_{i=0}^n \\frac{1}{2i+1}$. This gives us:\n$$\n\\begin{aligned}\nK &= -\\sum_{n=0}^\\infty (-1)^n \\tilde{H}_n\\int_0^1 x^{2n+2} \\log^2(x)\\; dx-\\frac{1}{2}\\sum_{n=1}^\\infty (-1)^{n+1} H_n\\int_0^1 x^{2n}\\log^2(x)\\; dx \\\\\n&= -2\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{(2n+3)^3} – \\sum_{n=1}^\\infty \\frac{(-1)^{n+1} H_n}{(2n+1)^3} \\\\\n&= -2\\sum_{n=1}^\\infty \\frac{(-1)^{n+1} H_{2n}}{(2n+1)^3} \\quad \\color{blue}{\\cdots (4)}\n\\end{aligned}\n$$\nWe’ll employ the contour integration method to evaluate the above Euler sum. It is a very powerful tool that can handle a large class of Euler sums (see, for e.g. Euler sums and contour integral representations by Philippe Flajolet and Bruno Salvy<\/a>). \n<\/p>\n

\nWe’ll integrate the function $f(z) = \\pi \\csc(\\pi z) \\frac{\\gamma+\\psi_0(-2z+1)}{(-2z+1)^3}$ around the positively oriented square, $C_N$, with vertices $\\pm \\left(N+\\frac{1}{4} \\right)\\pm \\left(N+\\frac{1}{4} \\right)i$. It is easy to see that \n$$\n\\lim_{N\\to \\infty}\\int_{C_N}f(z)\\; dz = 0\n$$\nHence, the sum of all residues of $f(z)$ at its poles is equal to $0$. A straightforward computation shows that the residues are:\n$$\n\\begin{aligned}\n\\mathop{\\text{Res}}\\limits_{z=-n} f(z) &= (-1)^n \\frac{H_{2n}}{(2n+1)^3} , \\quad n\\in \\{0,1,2,\\cdots\\} \\\\\n\\mathop{\\text{Res}}\\limits_{z=\\frac{2n+1}{2}} f(z) &= \\frac{(-1)^{n+1} \\pi}{16 n^3} , \\quad n\\in \\{1,2,3,\\cdots\\} \\\\\n\\mathop{\\text{Res}}\\limits_{z=n} f(z) &= \\frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3}- 3\\frac{(-1)^{n+1}}{(2n-1)^4}, \\quad n\\in \\{1,2,3,\\cdots \\}\\\\\n\\mathop{\\text{Res}}\\limits_{z=\\frac{1}{2}} f(z) &= \\frac{\\pi \\zeta(3)}{2}\n\\end{aligned}\n$$\nThe list of local expansions of basic kernels given on page 6 of the above mentioned paper are quite useful for carrying out these computations. Now, adding up all the residues gives us:\n$$\n\\begin{aligned}\n\\frac{\\pi \\zeta(3)}{2}+\\sum_{n=1}^\\infty \\frac{(-1)^n H_{2n}}{(2n+1)^3} + \\frac{\\pi}{16}\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}}{n^3} + \\sum_{n=1}\n^\\infty \\frac{(-1)^{n+1}H_{2n-1}}{(2n-1)^3} -3\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}}{(2n-1)^4}&= 0\\\\\n\\implies \\frac{\\pi \\zeta(3)}{2}+\\sum_{n=1}^\\infty \\frac{(-1)^n H_{2n}}{(2n+1)^3} + \\frac{\\pi}{16}\\left(\\frac{3\\zeta(3)}{4} \\right) + \\sum_{n=1}^\\infty \\frac{(-1)^n H_{2n}}{(2n+1)^3} -2 \\sum_{n=1}^\\infty \\frac{(-1)^{n+1}}{(2n-1)^4} &= 0 \\\\\n\\implies -2\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} + \\frac{35\\pi \\zeta(3)}{64} -2\\beta(4) = 0 \\\\\n\\implies \\boxed{\\sum_{n=1}^\\infty \\frac{(-1)^{n+1}H_{2n}}{(2n+1)^3} = -\\beta(4) + \\frac{35\\pi \\zeta(3)}{128}} \\color{blue}{\\cdots (5)}\n\\end{aligned}\n$$\nwhere $\\beta(s) = \\sum_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)^s}$ is the Dirichlet beta function. Finally, plugging equation (5) into (4) gives us:\n$$\n\\boxed{K = 2\\beta(4) – \\frac{35\\pi \\zeta(3)}{64}}\n$$\n<\/p>\n\n\n\n\n

\nFollowing are some more examples of Euler sums that can be evaluated using the contour integration method:\n$$\n\\begin{aligned}\n\\sum_{n=0}^\\infty\\frac{(-1)^n\\psi_2(n+1)}{2n+1} &= 8\\beta(4)+\\frac{\\pi^2}{3}G-\\frac{7\\pi}{2}\\zeta(3) \\\\\n\\sum_{n=0}^\\infty\\frac{(-1)^n\\psi_1(n+1)}{(2n+1)^2} &= 6\\beta(4)+\\frac{\\pi^2}{4}G-\\frac{7\\pi}{4}\\zeta(3) \\\\\n\\sum_{n=1}^\\infty \\frac{(-1)^{n+1} H_n}{(2n+1)^3} &= -3\\beta(4)+\\frac{7\\pi}{16}\\zeta(3)+\\frac{\\pi^3}{16}\\log(2)\n\\end{aligned}\n$$\nwhere $G=\\beta(2)$ denotes the Catalan’s constant. Some integrals that can be evaluated with the above Euler sums are:\n$$\n\\begin{aligned}\n\\int_0^1 \\frac{\\log^2(x)\\arctan(x)}{x\\left(1-x^2 \\right)}dx &=\\beta(4)+\\frac{7\\pi \\zeta(3)}{64}+\\frac{\\pi^3 \\log(2)}{32} \\\\\n\\int_0^1\\frac{\\log(x)\\arctan(x)\\text{arctanh}(x)}{x}dx &= \\frac{\\pi^2}{16}G-\\frac{7\\pi\\zeta(3)}{32} \\\\\n\\int_0^{\\frac{\\pi}{2}}\\frac{x}{\\sin x}\\log^2\\left(\\frac{1+\\cos x-\\sin x}{1+\\cos x+\\sin x}\\right)dx &= \\frac{\\pi^2}{6}G +4\\beta(4)\n\\end{aligned}\n$$\n<\/p>\n","protected":false},"excerpt":{"rendered":"

In this post, we’ll continue our exploration of logarithmic integrals and Euler sums. We’ll also discuss the contour integration method for evaluating Euler sums. It is recommended that the reader goes through the previous posts, (A) and (B), before reading … Continue reading →<\/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-476","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\/476","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=476"}],"version-history":[{"count":50,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/476\/revisions"}],"predecessor-version":[{"id":1157,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/476\/revisions\/1157"}],"wp:attachment":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=476"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=476"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=476"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}