{"id":444,"date":"2020-07-23T17:33:33","date_gmt":"2020-07-23T17:33:33","guid":{"rendered":"https:\/\/integralsandseries.in\/?p=444"},"modified":"2025-11-13T15:21:22","modified_gmt":"2025-11-13T15:21:22","slug":"euler-sums-involving-square-of-harmonic-numbers","status":"publish","type":"post","link":"https:\/\/integralsandseries.in\/?p=444","title":{"rendered":"Euler Sums involving square of Harmonic numbers"},"content":{"rendered":"\n

\nIn my previous post on Euler sums<\/a>, we evaluated sums containing $H_n$ and $H_n^{(2)}$. In this post, we’ll derive some further results using the integral $\\int_0^x \\frac{\\log^3(1-t)}{t}dt$. Our starting point is the following generating function identity:\n$$\n\\sum_{n=1}^\\infty (H_n)^2 x^n = \\frac{\\log^2(1-x)+\\text{Li}_2(x)}{1-x} ,\\quad -1\\leq x < 1\n$$\nThis can derived by plugging $H_n = \\int_0^1\\frac{1-t^n}{1-t}dt$ and interchanging the sum and the integral. We can rewrite the above equation using the fact that $\\sum_{n=1}^\\infty H_n^{(2)} x^n = \\frac{\\text{Li}_2(x)}{1-x}$.\n$$\n\\sum_{n=1}^\\infty (H_n)^2 x^n = \\frac{\\log^2(1-x)}{1-x} + \\sum_{n=1}^\\infty H_n^{(2)} x^n , \\quad -1\\leq x < 1 \\quad \\color{blue}{\\cdots (1)}\n$$\n<\/p>\n\n\n\n

1. Evaluation of $\\int_0^x \\frac{\\log^n(1-t)}{t}dt$<\/h2>\n

\nIn this section, we will derive a formula for the integral $\\int_0^x \\frac{\\log^n(1-t)}{t}dt$ where $n$ is a positive integer. First, we’ll consider the case when $0\\leq x < 1$. We have:\n$$\n\\begin{aligned}\n\\int_0^x \\frac{\\log^n (1-t)}{t}dt &= \\int_0^{-\\log(1-x)} \\frac{t^n e^{-t}}{1-e^{-t}}dt \\quad (t\\mapsto 1-e^{-t}) \\\\\n&= \\sum_{j=1}^\\infty \\int_0^{-\\log(1-x)}t^n e^{-jt} dt \\\\\n&= -\\sum_{j=1}^\\infty \\left[e^{-jt}\\sum_{i=0}^n \\frac{(-1)^{n-i}\\log^{n-i}(1-x)}{j^{i+1}}n^{\\underline{i}} \\right]_0^{-\\log(1-x)} \\\\\n&= – \\sum_{j=1}^\\infty \\left( (1-x)^j \\sum_{i=0}^n \\frac{(-1)^{n-i}\\log^{n-i}(1-x)}{j^{i+1}}n^{\\underline{i}}-\\frac{n!}{j^{n+1}} \\right) \\\\\n&= n! \\zeta(n+1) + \\sum_{i=0}^n (-1)^{n-i+1} n^{\\underline{i}}\\log^{n-i}(1-x) \\text{Li}_{i+1}(1-x) \\\\\n&\\quad \\color{blue}{\\cdots (2)}\n\\end{aligned}\n$$\n<\/p>\n

\nA similar calculation shows that for $-1\\leq x < 0$, we have:\n$$\n\\begin{aligned}\n\\int_x^0 \\frac{\\log^n(1-t)}{t}dt &= -\\frac{\\log^{n+1}(1-x)}{n+1} – n! \\zeta(n+1) + \\sum_{i=0}^{n} n^{\\underline{i}}\\log^{n-i}(1-x)\\text{Li}_{i+1}\\left(\\frac{1}{1-x}\\right) \\\\\n&\\quad \\color{blue}{\\cdots (3)}\n\\end{aligned}\n$$\n<\/p>\n\n\n\n

2. Sums with $(H_n)^2$<\/h2>\n

\nWe can divide equation (1) by $x$ and integrate both sides to get some interesting results. \n$$\n\\begin{aligned}\n\\sum_{n=1}^\\infty \\frac{(H_n)^2}{n}x^n &= -\\frac{\\log^3(1-x)}{3}-\\log(1-x)\\text{Li}_2(x)+\\text{Li}_3(x), \\quad -1\\leq x < 1\\quad \\color{blue}{\\cdots (4)} \\\\\n\\sum_{n=1}^\\infty \\frac{(H_n)^2}{n^2}x^n &= \\text{Li}_4(x) + \\frac{\\text{Li}_2^2(x)}{2}-\\frac{1}{3}\\int_0^x \\frac{\\log^3(1-t)}{t}dt, \\quad -1\\leq x \\leq 1\\quad \\color{blue}{\\cdots (5)}\n\\end{aligned}\n$$\nEquation (4) was derived using results from section 4 of this post<\/a>.\n<\/p>\n

\nOne can now plug in $x=-1$ in equation (5) to get:\n$$ \\boxed{\\sum_{n=1}^\\infty \\frac{(H_n)^2}{n^2}(-1)^{n+1} = \\frac{41\\pi^4}{1440} + \\frac{\\pi^2 \\log^2(2)}{12}-\\frac{\\log^4(2)}{12}-\\frac{7}{4}\\log(3)\\zeta(3)-2\\text{Li}_4\\left(\\frac{1}{2}\\right)} $$\nOf course, equation (3) was used to evaluate $\\int_{-1}^0 \\frac{\\log^3(1-t)}{t}dt$.\n<\/p>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"

In my previous post on Euler sums, we evaluated sums containing $H_n$ and $H_n^{(2)}$. In this post, we’ll derive some further results using the integral $\\int_0^x \\frac{\\log^3(1-t)}{t}dt$. Our starting point is the following generating function identity: $$ \\sum_{n=1}^\\infty (H_n)^2 x^n … 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":[1],"tags":[4,5],"class_list":["post-444","post","type-post","status-publish","format-standard","hentry","category-uncategorized","tag-euler-sums","tag-polylogarithm"],"_links":{"self":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/444","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=444"}],"version-history":[{"count":32,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/444\/revisions"}],"predecessor-version":[{"id":1164,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=\/wp\/v2\/posts\/444\/revisions\/1164"}],"wp:attachment":[{"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=444"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=444"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/integralsandseries.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=444"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}