{"id":532,"date":"2020-08-06T10:27:24","date_gmt":"2020-08-06T10:27:24","guid":{"rendered":"https:\/\/integralsandseries.in\/?p=532"},"modified":"2025-11-15T06:27:32","modified_gmt":"2025-11-15T06:27:32","slug":"evaluating-very-nasty-logarithmic-integrals-part-ii","status":"publish","type":"post","link":"https:\/\/integralsandseries.in\/?p=532","title":{"rendered":"Evaluating very nasty logarithmic integrals: Part II"},"content":{"rendered":"\n

\nIn this post, we’ll evaluate some more nasty logarithmic integrals. Please read part 1<\/a> of this series if you haven’t done so already. <\/p>\n

Integral #3<\/h2>\n

\nWe’ll start by finding a closed form for the integral:\n$$\nI_1 = \\int_0^1 \\frac{\\log^2(1+x^2)}{1+x^2}dx\n$$\nThis integral can be reduced to Euler sums like our previous problem. But this time, the resulting Euler sums cannot be evaluated using the method of residues. Therefore, we’ll have to use a different approach.\n<\/p>\n

\nLet us first consider the following integral:\n$$ I_2 = \\int_0^1 \\frac{\\log^2(1+ix)}{1+x^2}dx $$\nThroughout this post, $\\log $ denotes the principal branch of the logarithmic function defined by $\\log z = \\log|z| + i\\text{arg}(z)$, with $-\\pi < |\\text{arg}(z)| \\leq \\pi$. We have\n$$\n\\begin{aligned}\nI_2 &= \\frac{1}{2}\\int_0^1\\log^2(1+ix)\\left(\\frac{1}{1+ix}+\\frac{1}{1-ix} \\right)dx \\\\\n&= \\frac{\\log^3(1+ix)}{6i}\\Big|_0^1 + \\frac{1}{2}\\int_0^1 \\frac{\\log^2(1+ix)}{1-ix}dx \\\\\n&= \\frac{\\log^3(1+i)}{6i} + \\frac{i}{2}\\int_{\\frac{1}{2}}^{\\frac{1-i}{2}}\\frac{\\log^2(2(1-x))}{x}dx \\\\\n&= \\frac{\\log^3(1+i)}{6i} + \\frac{i}{2}\\int_{\\frac{1}{2}}^{\\frac{1-i}{2}}\\frac{\\log^2(1-x)+\\log^2(2)+2\\log(2)\\log(1-x)}{x}dx \\\\\n&= \\frac{\\log^3(1+i)}{6i} + \\frac{i\\log^2(2)\\log\\left(1-i\\right)}{2} + i\\log(2) \\left[\\text{Li}_2\\left(\\frac{1}{2}\\right)-\\text{Li}_2\\left(\\frac{1-i}{2}\\right) \\right] \\\\ &\\quad + \\frac{i}{2}\\int_{\\frac{1}{2}}^{\\frac{1-i}{2}}\\frac{\\log^2(1-x)}{x}dx \\quad \\color{blue}{\\cdots (1)}\n\\end{aligned}\n$$\nWe can use equation (2) from (B)<\/a> to evaluate $\\int_{\\frac{1}{2}}^{\\frac{1+i}{2}}\\frac{\\log^2(1-x)}{x}dx$. \n$$\n\\begin{aligned}\n\\int_{\\frac{1}{2}}^{\\frac{1-i}{2}}\\frac{\\log^2(1-x)}{x}dx &= \\log^2\\left( \\frac{1+i}{2}\\right)\\log\\left(\\frac{1-i}{2}\\right) + 2\\log\\left(\\frac{1+i}{2} \\right)\\text{Li}_2\\left(\\frac{1+i}{2}\\right) \\\\ &\\quad -2\\text{Li}_3\\left(\\frac{1+i}{2}\\right) + \\log^3(2) + 2\\log(2)\\text{Li}_2\\left(\\frac{1}{2}\\right) + 2\\text{Li}_3\\left(\\frac{1}{2} \\right) \\quad \\color{blue}{\\cdots (2)}\n\\end{aligned}\n$$\n<\/p>\n\n\n\n

\nTo simplify $\\text{Li}_2\\left(\\frac{1+i}{2} \\right)$ and $\\text{Li}_2\\left(\\frac{1-i}{2} \\right)$, we can use the following Dilogarithm identity:\n$$\n\\text{Li}_2(1-z) + \\text{Li}_2\\left(1-z^{-1} \\right) = -\\frac{1}{2}\\log^2(z)\n$$\nThis is easy to verify by differentiating both sides of the above equation with respect to $z$. Plugging in $z=\\frac{1+i}{2}$ gives\n$$\n\\begin{aligned}\n\\text{Li}_2\\left(\\frac{1-i}{2}\\right) &= -\\text{Li}_2(i) – \\frac{1}{2}\\log^2\\left(\\frac{1+i}{2}\\right) = \\frac{5\\pi^2}{96}-\\frac{\\log^2(2)}{8}+i\\left(-G + \\frac{\\pi}{8}\\log(2)\\right) \\\\\n\\text{Li}_2\\left(\\frac{1+i}{2}\\right) &= \\overline{\\text{Li}_2\\left(\\frac{1-i}{2}\\right)} = \\frac{5\\pi^2}{96}-\\frac{\\log^2(2)}{8}-i\\left(-G + \\frac{\\pi}{8}\\log(2)\\right)\n\\end{aligned}\n$$\n<\/p>\n\n\n\n

\nNow, we have everything needed to simplify equation (1). This is a cumbersome task so I will only present the final result:\n$$\n\\begin{aligned}\nI_2 &= -\\frac{3\\pi^3}{128}- \\frac{G \\log(2) }{2} + \\frac{7\\pi \\log^2(2)}{32} + i\\Bigg( -\\frac{ G \\pi }{4} + \\frac{7\\pi^2 \\log(2)}{192} + \\frac{\\log^3(2)}{48}+\\frac{7}{8}\\zeta(3) \\\\ &\\quad – \\text{Li}_3\\left(\\frac{1+i}{2} \\right)\\Bigg) \\quad \\color{blue}{\\cdots (3)}\n\\end{aligned}\n$$\nTo the best of my knowledge, $\\text{Li}_3\\left(\\frac{1+i}{2} \\right)$ can not be simplified further. So, we’ll leave it as it is. We can now extract $I_1$ from the real part of $I_2$.\n$$\n\\begin{aligned}\n\\text{Re }I_2 &= \\int_0^1 \\frac{\\frac{1}{4}\\log^2(1+x^2) – \\arctan^2(x)}{1+x^2}dx \\\\\n&= \\frac{1}{4}\\int_0^1 \\frac{\\log^2(1+x^2)}{1+x^2}dx – \\frac{\\arctan^3(x)}{3}\\Big|_0^1 \\\\\n&= \\frac{1}{4}I_1 – \\frac{\\pi^3}{192}\n\\end{aligned}\n$$\nTherefore,\n$$\n\\boxed{I_1 = -2G\\log(2) – \\frac{7\\pi^3}{96} + \\frac{7\\pi \\log^2(2)}{8} + 4\\text{ Im }\\text{Li}_3\\left(\\frac{1+i}{2} \\right)}\\quad \\color{blue}{\\cdots (4)}\n$$\nNow, let’s turn our attention to another integral:\n$$ I_3 = \\int_0^1 \\frac{\\log(x)\\log(1+x^2)}{1+x^2}dx $$\nNotice that with the help of some algebra, we can write:\n$$\nI_3 = -\\frac{1}{2}\\int_0^1 \\frac{\\log^2\\left(\\frac{x}{1+x^2} \\right)}{1+x^2}dx + \\frac{1}{2}\\int_0^1 \\frac{\\log^2(1+x^2)}{1+x^2}dx + \\frac{1}{2}\\int_0^1 \\frac{\\log^2(x)}{1+x^2}dx\n$$\nThe leftmost integral can be dealt with the trigonometric substitution $x=\\tan \\theta$:\n$$\n\\begin{aligned}\n\\int_0^1 \\frac{\\log^2\\left(\\frac{x}{1+x^2} \\right)}{1+x^2}dx &= \\int_0^{\\frac{\\pi}{4}} \\log^2\\left(\\sin \\theta \\cos\\theta \\right) \\; d\\theta \\\\\n&= \\int_0^{\\frac{\\pi}{4}} \\log^2\\left(\\frac{\\sin(2\\theta)}{2} \\right)\\; d\\theta \\\\\n&= \\frac{1}{2}\\int_0^{\\frac{\\pi}{2}}\\log^2\\left(\\frac{\\sin \\theta}{2} \\right)\\; d\\theta \\\\\n&= \\frac{1}{2}\\lim_{s\\to 1}\\frac{d^2}{ds^2}\\int_0^{\\frac{\\pi}{2}}\\left(\\frac{\\sin\\theta}{2} \\right)^{s-1}d\\theta \\\\\n&= \\frac{1}{2}\\lim_{s\\to 1}\\frac{d^2}{ds^2}\\left[\\frac{2^{-s}\\sqrt{\\pi}\\Gamma\\left(\\frac{s}{2}\\right)}{\\Gamma\\left(\\frac{1+s}{2} \\right)} \\right] \\\\\n&= \\frac{\\pi^3}{48}+ \\pi \\log^2(2)\n\\end{aligned}\n$$\nThe middle integral has already been evaluated. As for the rightmost integral, we have:\n$$\n\\begin{aligned}\n\\int_0^1 \\frac{\\log^2(x)}{1+x^2}dx &= \\sum_{n=0}^\\infty (-1)^{n}\\int_0^1 x^{2n}\\log^2(x)\\; dx \\\\\n&= 2\\sum_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)^3} \\\\\n&= \\frac{\\pi^3}{16}\n\\end{aligned}\n$$\nThis gives us\n$$\n\\boxed{I_3 =-\\frac{\\pi^3}{64} -G \\log(2) – \\frac{\\pi \\log^2(2)}{16} +2 \\text{ Im }\\text{Li}_3\\left(\\frac{1+i}{2}\\right)} \\quad \\color{blue}{\\cdots (5)}\n$$\n<\/p>\n\n\n\n

\nOne can also evaluate $I_4 = \\int_0^1\\frac{\\log(1+x^2)\\arctan(x)}{x}dx$ by noting that\n$$ I_4 = \\text{Im}\\int_0^1 \\frac{\\log^2(1+ix)}{x}dx = \\text{Im}\\int_0^{-i}\\frac{\\log^2(1-x)}{x}dx $$\nand using equation (3) from (B)<\/a>.The end result is:\n$$\n\\boxed{I_4 =-\\frac{3\\pi^3}{64}+G\\log(2)-\\frac{\\pi \\log^2(2)}{16}+2\\text{ Im }\\text{Li}_3\\left(\\frac{1+i}{2}\\right) } \\quad \\color{blue}{\\cdots (6)}\n$$\nUsing $I_3$ and $I_4$, we can evaluate $I_5 = \\int_0^1 \\frac{\\log(x)\\arctan(x)}{x(1+x^2)}dx$ as follows:\n$$\n\\begin{aligned}\nI_5 &= \\int_0^1 \\log(x)\\arctan(x)\\left(\\frac{1}{x}-\\frac{x}{1+x^2} \\right) dx \\\\\n&= \\int_0^1 \\frac{\\log(x)\\arctan(x)}{x}dx – \\int_0^1 \\frac{x\\log(x)\\arctan(x)}{1+x^2}dx \\\\\n&= -\\frac{1}{2}\\int_0^1 \\frac{\\log^2(x)}{1+x^2}dx + \\frac{1}{2} \\int_0^1 \\frac{\\log(x)\\log(1+x^2)}{1+x^2} dx + \\frac{1}{2}\\int_0^1 \\frac{\\log(1+x^2)\\arctan(x)}{x}dx \\quad (\\text{IBP}) \\\\\n&= -\\frac{\\pi^3}{32} + \\frac{I_3 + I_4}{2} \\\\\n&= -\\frac{\\pi^3}{16} – \\frac{\\pi \\log^2(2)}{16} + 2\\text{ Im }\\text{Li}_3\\left(\\frac{1+i}{2}\\right)\n\\end{aligned}\n$$\nOn the other hand, we have:\n$$\n\\begin{aligned}\nI_5 &= \\int_0^1 \\log(x) \\left(\\sum_{n=0}^\\infty (-1)^n \\tilde{H}_n x^{2n} \\right) dx\\\\ &= \\sum_{n=0}^\\infty (-1)^n \\tilde{H}_n \\int_0^1 x^{2n}\\log(x) \\; dx \\\\\n&= -\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{(2n+1)^2} \n\\end{aligned}\n$$\nwhere $\\tilde{H}_n = \\sum_{j=0}^n \\frac{1}{2j+1}$. This gives us an interesting Euler sum:\n$$\n\\boxed{\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{(2n+1)^2} = \\frac{\\pi^3}{16} + \\frac{\\pi \\log^2(2)}{16} – 2\\text{ Im }\\text{Li}_3\\left(\\frac{1+i}{2}\\right)}\\quad \\color{blue}{\\cdots (7)}\n$$\nOf course, one can proceed in a similar manner to create more crazy integrals. The following problem is left as an exercise for the reader.\n<\/p>\n

\nExercise 1: <\/b> Using the method of residues, show that\n$$\\sum_{n=0}^\\infty \\frac{(-1)^n\\tilde{H}_n}{2n+1} = \\frac{G}{2}+\\frac{\\pi \\log(2)}{8} \\quad \\color{blue}{\\cdots (8)}$$\n<\/p>\n\n\n\n

Integral #4<\/h2>\n

\nMany years ago, I encountered the following integral:\n$$I_6 = \\int_0^1 \\frac{x \\arctan(x)\\log(1-x^2)}{1+x^2}dx$$\nAt that time, I couldn’t find a solution to this problem. Hence, I ended up asking it on math.stackexchange.com. The answers that I received there involved evaluating complex logarithmic integrals by brute force. Recently, I discovered a much simpler way to solve it using the method of residues. \n<\/p>\n

\nLet’s start by breaking down $I_6$ into Euler sums.\n$$\n\\begin{aligned}\nI_6 &= \\int_0^1 x\\log(1-x^2) \\left(\\sum_{n=0}^\\infty (-1)^n \\tilde{H}_n x^{2n+1} \\right)dx \\\\\n&= \\sum_{n=0}^\\infty (-1)^n \\tilde{H}_n \\int_0^1 x^{2n+2}\\log(1-x^2) dx \\\\\n&= \\sum_{n=0}^\\infty (-1)^{n+1} \\tilde{H}_n \\left(\\frac{\\psi_0\\left(n+\\frac{5}{2} \\right)+\\gamma}{2n+3} \\right) \\\\\n&= \\sum_{n=0}^\\infty (-1)^{n+1}\\left(\\tilde{H}_{n+1}-\\frac{1}{2n+3} \\right)\\left(\\frac{-2\\log(2)+2\\tilde{H}_{n+1}}{2n+3} \\right) \\\\\n&= \\sum_{n=0}^\\infty (-1)^n \\left(\\tilde{H}_{n}-\\frac{1}{2n+1} \\right)\\left(\\frac{-2\\log(2)+2\\tilde{H}_{n}}{2n+1} \\right) \\\\\n&= -2\\log(2)\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{2n+1}+2G\\log(2) + 2\\sum_{n=0}^\\infty \\frac{(-1)^n (\\tilde{H}_n)^2}{2n+1} -2\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{(2n+1)^2} \\\\\n&= G\\log(2) – \\frac{\\pi \\log^2(2)}{4} + 2\\sum_{n=0}^\\infty \\frac{(-1)^n (\\tilde{H}_n)^2}{2n+1} -2\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{(2n+1)^2} \\quad \\color{blue}{\\cdots (9)}\n\\end{aligned}\n$$\nThe result of exercise 1 was used in the last step.\n<\/p>\n

\nNow, integrate the function $f(z) = \\pi\\csc(\\pi z) \\frac{\\left( \\gamma + \\psi_0 \\left(-z+\\frac{3}{2} \\right)\\right)^2}{-2z+1}$ over the positively oriented square, $C_N$, with vertices $\\pm \\left(N+\\frac{1}{4}\\right)\\pm i\\left(N+\\frac{1}{4} \\right)$. It takes a bit of effort to see that\n$$\\lim_{N\\to \\infty}\\int_{C_N} f(z)\\; dz = 0$$\nThis implies that the sum of residues of $f(z)$ at it’s poles is equal to $0$. We have\n<\/p>\n

\n$$\n\\begin{aligned}\n\\mathop{\\text{Res}}\\limits_{z=-n} f(z) &= (-1)^n \\frac{\\left(\\gamma +\\psi_0\\left(n+\\frac{3}{2}\\right) \\right)^2}{2n+1} = (-1)^n \\frac{\\left(-2\\log(2)+2\\tilde{H}_n \\right)^2}{2n+1}, \\quad n\\in\\{0,1,2,\\cdots\\} \\\\\n\\mathop{\\text{Res}}\\limits_{z=n} f(z) &= (-1)^{n-1} \\frac{\\left(\\gamma+\\psi_0\\left(-n+\\frac{3}{2} \\right) \\right)^2}{2n-1} = (-1)^{n-1} \\frac{\\left(-2\\log(2)+2\\tilde{H}_{n-1} -\\frac{2}{2n-1}\\right)^2}{2n-1}, \\quad n\\in\\{1,2,3,\\cdots\\} \\\\\n\\mathop{\\text{Res}}\\limits_{z=\\frac{2n+1}{2}} f(z) &= (-1)^{n-1} \\frac{\\pi H_n}{n} – (-1)^{n-1} \\frac{3\\pi}{2n^2}, \\quad n\\in\\{1,2,3,\\cdots\\}\n\\end{aligned}\n$$\n<\/p>\n

\nSumming up the residues and performing some algebraic simplifications gives:\n$$\n\\begin{aligned}\n&\\; 8\\sum_{n=0}^\\infty \\frac{(-1)^n (\\tilde{H}_n)^2}{2n+1}-8\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{(2n+1)^2}-16\\log(2)\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{2n+1}+ 4\\sum_{n=0}^\\infty \\frac{(-1)^n}{(2n+1)^3} \\\\ &\\quad + 8\\log(2)\\sum_{n=0}^\\infty \\frac{(-1)^n }{(2n+1)^2} + \\pi\\sum_{n=1}^\\infty \\frac{(-1)^{n-1}H_n}{n} – \\frac{3\\pi}{2}\\sum_{n=1}^\\infty \\frac{(-1)^{n-1}}{n^2} = 0 \\\\\n&\\implies 8\\left(\\sum_{n=0}^\\infty \\frac{(-1)^n (\\tilde{H}_n)^2}{2n+1}-\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{(2n+1)^2} \\right) + \\frac{\\pi^3}{8} + \\pi \\left(\\frac{\\pi^2}{12}-\\frac{\\log^2(2)}{2}\\right)-\\frac{3\\pi}{2}\\left(\\frac{\\pi^2}{12} \\right) = 0 \\\\\n&\\implies \\sum_{n=0}^\\infty \\frac{(-1)^n (\\tilde{H}_n)^2}{2n+1}-\\sum_{n=0}^\\infty \\frac{(-1)^n \\tilde{H}_n}{(2n+1)^2} = -\\frac{\\pi^3}{96}+\\frac{\\pi \\log^2(2)}{16} \\quad \\color{blue}{\\cdots (10)}\n\\end{aligned}\n$$\nIn the above calculation, we used the result of exercise 1 and that\n$$\\sum_{n=1}^\\infty \\frac{(-1)^{n-1}H_n}{n} = \\frac{\\pi^2}{12}-\\frac{\\log^2(2)}{2}$$\nThis follows from (A)<\/a>.\nFinally, plugging equation (10) into (9), gives\n$$\n\\boxed{I_6 = -\\frac{\\pi^3}{48}-\\frac{\\pi}{8}\\log^2 (2) +G\\log (2)} \\quad \\color{blue}{\\cdots (11)}\n$$\n<\/p>\n","protected":false},"excerpt":{"rendered":"

In this post, we’ll evaluate some more nasty logarithmic integrals. Please read part 1 of this series if you haven’t done so already. Integral #3 We’ll start by finding a closed form for the integral: $$ I_1 = \\int_0^1 \\frac{\\log^2(1+x^2)}{1+x^2}dx … 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-532","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\/532","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=532"}],"version-history":[{"count":128,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/532\/revisions"}],"predecessor-version":[{"id":1169,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/532\/revisions\/1169"}],"wp:attachment":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=532"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=532"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=532"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}