\nIn this post, we will evaluate the famous Euler sum:\n$$\\sum_{n=1}^\\infty (-1)^{n+1}\\frac{H_n}{n^3}\\quad \\color{blue}{\\cdots (*)} $$\nwhere $H_n = \\sum_{k=1}^n \\frac{1}{k}=\\int_0^1 \\frac{1-t^n}{1-t}dt$ is the $n$-th harmonic number. This series resists contour integration techniques which makes it’s computation quite challenging. We will start with the following well known generating function identity:\n$$\\sum_{n=1}^\\infty H_n t^n = -\\frac{\\log(1-t)}{1-t} ,\\quad -1 \\leq t < 1 $$\nDividing the above equation by $t$ and integrating both sides, gives us:\n$$\n\\begin{aligned}\n\\sum_{n=1}^\\infty H_n \\frac{x^n}{n} &= -\\int_0^x\\frac{\\log(1-t)}{t(1-t)}dt \\\\\n&= -\\int_0^x \\left(\\frac{1}{t} + \\frac{1}{1-t} \\right)\\log(1-t) \\; dt \\\\\n&= \\text{Li}_2(x) + \\frac{1}{2}\\log^2(1-x) \n\\end{aligned}\n$$\nwhere $-1\\leq x < 1$.<\/p>\n
\nNow, we will use the same process to evaluate $\\sum_{n=1}^\\infty \\frac{H_n}{n^2}x^n$ but we’ll have to divide the evaluation into two parts.\n<\/p>\n\n\n\n
\n(I)<\/b> First consider the case when $0 \\leq x < 1$. We don’t need this for evaluating (*) but I’ll do it anyway for completeness. We have:\n$$\n\\begin{aligned}\n\\sum_{n=1}^\\infty \\frac{H_n}{n^2} x^n &= \\int_0^x\\left(\\frac{\\text{Li}_2(t)}{t} + \\frac{1}{2}\\frac{\\log^2(1-t)}{t}\\right)dt \\\\\n&= \\text{Li}_3(x) + \\frac{1}{2}\\int_0^x \\frac{\\log^2(1-t)}{t} dt \\\\\n&= \\text{Li}_3(x) + \\frac{1}{2}\\int_{\\log(1-x)}^0 \\frac{t^2 e^t}{1-e^t}dt \\quad (t\\mapsto 1-e^t) \\\\\n&= \\text{Li}_3(x) + \\frac{1}{2}\\sum_{n=1}^\\infty \\int_{\\log(1-x)}^0 t^2 e^{nt} \\; dt \\\\\n&= \\text{Li}_3(x) + \\frac{1}{2}\\sum_{n=1}^\\infty \\left(\\frac{2}{n^3} -\\frac{\\log^2(1-x)}{n}(1-x)^n+\\frac{2\\log(1-x)}{n^2}(1-x)^n-2\\frac{(1-x)^n}{n^3}\\right) \\\\\n&= \\text{Li}_3(x) + \\frac{\\log^2(1-x)\\log(x)}{2} + \\log(1-x)\\text{Li}_2(1-x) – \\text{Li}_3(1-x) + \\zeta(3) \\quad\\color{blue}{\\cdots (1)}\n\\end{aligned}\n$$\n(II)<\/b> Similarly, for the case $-1\\leq x < 0$, we obtain:\n<\/p>\n
\n$$\n\\begin{aligned}\n\\sum_{n=1}^\\infty \\frac{H_n}{n^2}x^n &= \\text{Li}_3(x) -\\frac{1}{2}\\int_x^0 \\frac{\\log^2(1-t)}{t}dt \\\\\n&= \\text{Li}_3(x) +\\frac{1}{2}\\int_{0}^{-x} \\frac{\\log^2(1+t)}{t}dt \\quad (t\\mapsto -t) \\\\\n&= \\text{Li}_3(x) +\\frac{1}{2}\\int_{0}^{\\log(1-x)} \\frac{t^2 e^t}{e^t-1}dt \\quad (t\\mapsto e^t-1) \\\\\n&= \\text{Li}_3(x) +\\frac{1}{2} \\sum_{n=0}^\\infty \\int_0^{\\log(1-x)} t^2 e^{-nt} dt \\\\\n&= \\text{Li}_3(x) +\\frac{\\log^3(1-x)}{6} + \\frac{1}{2}\\sum_{n=1}^\\infty \\int_0^{\\log(1-x)} t^2 e^{-nt} dt \\\\\n&= \\text{Li}_3(x) +\\frac{\\log^3(1-x)}{6} + \\frac{1}{2}\\sum_{n=1}^\\infty \\left(\\frac{2}{n^3} -\\log^2(1-x)\\frac{(1-x)^{-n}}{n}-2\\log(1-x)\\frac{(1-x)^{-n}}{n^2}-2\\frac{(1-x)^{-n}}{n^3}\\right) \\\\\n&= \\text{Li}_3(x) -\\frac{\\log^3(1-x)}{3} +\\frac{\\log^2(1-x)\\log(-x)}{2} -\\log(1-x)\\text{Li}_2\\left( \\frac{1}{1-x}\\right)-\\text{Li}_3\\left(\\frac{1}{1-x}\\right)+\\zeta(3) \\\\\n&\\quad\\color{blue}{\\cdots (2)}\n\\end{aligned}\n$$\n<\/p>\n\n\n\n
\nThis is the hardest step in the evaluation. Once again, we will divide the evaluation into two parts. <\/p>\n
\n(I)<\/b> Similar to section 1, we’ll first consider the case when $0\\leq x < 1$. \n<\/p>\n
Using integration by parts, we obtain:\n$$\n\\begin{aligned}\n\\int_0^x \\frac{\\zeta(3)-\\text{Li}_3(1-t)}{t}dt &= \\log(t)\\left(\\zeta(3)-\\text{Li}_3(1-t) \\right) \\Big|_{0}^x – \\int_0^x \\frac{\\log(t)\\text{Li}_2(1-t)}{1-t}dt \\\\\n&= \\log(x)\\left(\\zeta(3)-\\text{Li}_3(1-x) \\right) – \\frac{1}{2}\\text{Li}_2^2 (1-x) + \\frac{\\zeta^2 (2)}{2} \\quad \\color{blue}{\\cdots (3)}\n\\end{aligned}\n$$\n<\/p>\n\n
\n$$\n\\begin{aligned}\n\\int_0^x \\frac{\\log(1-t)\\text{Li}_2(1-t)}{t}dt &= \\log(t)\\log(1-t)\\text{Li}_2(1-t) \\Big|_0^x – \\int_0^x \\log(t) \\left(\\frac{-\\text{Li}_2(1-t)}{1-t} + \\frac{\\log(1-t)\\log(t)}{1-t}\\right)dt \\\\\n&= \\log(x)\\log(1-x)\\text{Li}_2(1-x) + \\frac{1}{2}\\text{Li}_2^2(1-x)-\\frac{\\zeta^2(2)}{2} – \\int_0^x \\frac{\\log^2(t) \\log(1-t)}{1-t}dt \\\\\n&\\quad \\color{blue}{\\cdots (4)}\n\\end{aligned}\n$$\n<\/p>\n
\n$$\n\\begin{aligned}\n\\int_0^x \\frac{\\log(t)\\log^2(1-t)}{2t}dt &= \\frac{1}{4}\\log^2(t)\\log^2(1-t)\\Big|_0^x +\\frac{1}{2}\\int_0^x \\frac{\\log^2(t)\\log(1-t)}{1-t}dt \\\\\n&= \\frac{1}{4}\\log^2(x)\\log^2(1-x) +\\frac{1}{2}\\int_0^x \\frac{\\log^2(t)\\log(1-t)}{1-t}dt \\quad \\color{blue}{\\cdots (5)}\n\\end{aligned}\n$$\n<\/p>\n
\nPutting together the results of equations (3), (4) and (5) gives us:\n$$\n\\begin{aligned}\n\\sum_{n=1}^\\infty \\frac{H_n}{n^3}x^n &= \\text{Li}_4(x) +\\frac{1}{4}\\log^2(x)\\log^2(1-x) + \\log(x)\\log(1-x)\\text{Li}_2(1-x)\\\\ &\\quad + \\log(x)\\left(\\zeta(3)-\\text{Li}_3(1-x) \\right) -\\frac{1}{2}\\int_0^x \\frac{\\log^2(t)\\log(1-t)}{1-t}dt \\quad \\color{blue}{\\cdots (6)}\n\\end{aligned}\n$$\n<\/p>\n
\n(II)<\/b> Now, let $-1\\leq x < 0$. We have:\n$$\n\\begin{aligned}\n-\\int_x^0 \\frac{\\zeta(3)-\\text{Li}_3\\left(\\frac{1}{1-t}\\right)}{t}dt &= \\int_0^{\\frac{-x}{1-x}}\\frac{\\zeta(3)-\\text{Li}_3(1-t)}{t(1-t)}dt \\quad \\left(t\\mapsto\\frac{-t}{1-t} \\right)\\\\\n&= \\int_0^{\\frac{-x}{1-x}}\\frac{\\zeta(3)-\\text{Li}_3(1-t)}{t}dt + \\int_0^{\\frac{-x}{1-x}}\\frac{\\zeta(3)-\\text{Li}_3(1-t)}{1-t}dt \\\\\n&= \\log\\left(\\frac{-x}{1-x}\\right)\\left(\\zeta(3) – \\text{Li}_3\\left(\\frac{1}{1-x}\\right) \\right) – \\frac{1}{2}\\text{Li}_2^2\\left(\\frac{1}{1-x}\\right) + \\frac{\\zeta^2(2)}{2} \\\\\n&\\quad + \\zeta(3)\\log(1-x) + \\text{Li}_4\\left(\\frac{1}{1-x} \\right) – \\zeta(4) \\quad \\color{blue}{\\cdots (7)}\n\\end{aligned}\n$$\nHere, we used equation (3) to simplify the integral after the transformation. Similarly, we obtain:\n$$\n\\begin{aligned}\n\\int_x^0 \\frac{\\log(1-t)\\text{Li}_2\\left(\\frac{1}{1-t}\\right)}{t}dt &= \\int_0^{\\frac{-x}{1-x}}\\frac{\\log(1-t)\\text{Li}_2(1-t)}{t(1-t)}dt \\\\\n&= \\int_0^{\\frac{-x}{1-x}}\\frac{\\log(1-t)\\text{Li}_2(1-t)}{t}dt + \\int_0^{\\frac{-x}{1-x}}\\frac{\\log(1-t)\\text{Li}_2(1-t)}{1-t}dt \\\\\n&= -\\log\\left(\\frac{-x}{1-x} \\right)\\log(1-x)\\text{Li}_2\\left(\\frac{1}{1-x}\\right)+\\frac{1}{2}\\text{Li}_2^2\\left(\\frac{1}{1-x}\\right)-\\frac{\\zeta^2(2)}{2} \\\\\n&\\quad +\\text{Li}_4\\left(\\frac{1}{1-x}\\right)+\\log(1-x)\\text{Li}_3\\left(\\frac{1}{1-x}\\right) -\\zeta(4) \\\\\n&\\quad -\\int_0^{\\frac{-x}{1-x}}\\frac{\\log^2(t)\\log(1-t)}{1-t}dt \n \\quad \\color{blue}{\\cdots (8)}\n\\end{aligned}\n$$\n<\/p>\n
\n$$\n\\begin{aligned}\n\\int_x^0 \\left( \\frac{\\log^3(1-t)}{3t} – \\frac{\\log^2(1-t)\\log(-t)}{2t}\\right) dt &= \\frac{1}{6}\\log^3(1-x)\\log\\left(\\frac{-x}{1-x}\\right) + \\frac{1}{24}\\log^4(1-x) \\\\\n&\\quad +\\frac{1}{4}\\log^2\\left(\\frac{-x}{1-x}\\right)\\log^2(1-x) + \\frac{1}{2}\\int_0^{\\frac{-x}{1-x}}\\frac{\\log^2(t)\\log(1-t)}{1-t}dt \\\\ &\\quad \\color{blue}{\\cdots (9)}\n\\end{aligned}\n$$\n<\/p>\n
Finally, putting these results together gives:<\/p>\n
\n$$\n\\begin{aligned}\n\\sum_{n=1}^\\infty \\frac{H_n}{n^3}x^n &= \\text{Li}_4(x) + 2\\text{Li}_4\\left( \\frac{1}{1-x}\\right) + \\log\\left(\\frac{-x}{1-x} \\right)\\left(-\\text{Li}_3\\left(\\frac{1}{1-x}\\right) +\\frac{\\log^3(1-x)}{6}-\\log(1-x)\\text{Li}_2\\left(\\frac{1}{1-x} \\right)\\right) \\\\\n&\\quad + \\log(1-x)\\text{Li}_3\\left(\\frac{1}{1-x}\\right) + \\frac{\\log^4(1-x)}{24} +\\frac{1}{4}\\log^2 \\left(\\frac{-x}{1-x}\\right)\\log^2(1-x) +\\zeta(3) \\log(-x)- 2\\zeta(4) \\\\\n&\\quad -\\frac{1}{2}\\int_0^{\\frac{-x}{1-x}}\\frac{\\log^2(t)\\log(1-t)}{1-t}dt \\quad \\color{blue}{\\cdots (10)}\n\\end{aligned}\n$$\n<\/p>\n
\nThe integral $\\int \\frac{\\log^2(t)\\log(1-t)}{1-t}dt$ can be evaluated in terms of Polylogarithms (refer to section 7.6 of “Polylogarithms and Associated Functions” by Leonard Lewin). Luckily for us, there is a much simpler was to evaluate this integral between the limits $0$ and $\\frac{1}{2}$. \n<\/p>\n\n\n\n
\nLet $I=\\int_0^{1}\\frac{\\log^2(t)\\log(1-t)}{1-t}dt$ and $J=\\int_0^{\\frac{1}{2}}\\frac{\\log^2(t)\\log(1-t)}{1-t}dt$. The integral, $I$, can be evaluated by the differentiation under the integral technique.\n$$\n\\begin{aligned}\nI &= \\lim_{y\\to 0^+}\\lim_{x\\to 1}\\frac{\\partial^2 }{\\partial x^2} \\frac{\\partial }{\\partial y} \\int_0^1 t^{x-1} (1-t)^{y-1} dt \\\\\n&= \\lim_{y\\to 0^+}\\lim_{x\\to 1}\\frac{\\partial^2 }{\\partial x^2} \\frac{\\partial }{\\partial y} \\left\\{\\frac{\\Gamma(x)\\Gamma(y)}{\\Gamma(x+y)} \\right\\}\n\\end{aligned}\n$$\nIt takes a bit of effort to compute the derivatives in terms of Polygamma functions so I won’t write the details here. One can easily verify that the end result is:\n$$I = -\\frac{\\pi^4}{180}$$\nNext, apply integration by parts to the integral $J$:\n$$\n\\begin{aligned}\nJ &= \\int_0^{\\frac{1}{2}}\\frac{\\log^2(t)\\log(1-t)}{1-t}dt \\\\\n&= -\\frac{\\log^2(t)\\log^2(1-t)}{2}\\Big|_0^{\\frac{1}{2}} + \\int_0^{\\frac{1}{2}} \\frac{\\log(t)\\log^2(1-t)}{t}dt \\\\\n&= -\\frac{\\log^4(2)}{2} + \\int_{\\frac{1}{2}}^1 \\frac{\\log^2(t)\\log(1-t)}{1-t}dt \\quad (t\\mapsto 1-t)\\\\\n&= -\\frac{\\log^4(4)}{2} + I-J\n\\end{aligned}\n$$\nNow, one can solve the above equation for $J$ to get:\n$$\n\\begin{aligned}\nJ &= -\\frac{\\log^4(2)}{4}+\\frac{I}{2} \\\\\n&= -\\frac{\\log^4(2)}{4}-\\frac{\\pi^4}{360}\n\\end{aligned}\n$$\n<\/p>\n\n\n\n
\nFinally, we have every thing needed to evaluate (*). Plugging in $x=-1$ in equation (10) and performing some simplifications gives us the beautiful result:\n$$\n\\boxed{\\sum_{n=1}^\\infty \\frac{H_n}{n^3}(-1)^{n+1} = \\frac{11\\pi^4}{360}+\\frac{\\pi^2}{12}\\log^2(2)-\\frac{\\log^4(2)}{12}-\\frac{7}{4}\\log(2)\\zeta(3) -2\\text{Li}_4\\left(\\frac{1}{2}\\right)}\n$$\nA similar result is obtained by plugging $x=\\frac{1}{2}$ into equation (6):\n$$\n\\boxed{\\sum_{n=1}^\\infty \\frac{H_n}{n^3 2^n}= \\frac{\\pi^4}{720}+\\frac{\\log^4(2)}{24}-\\frac{1}{8}\\log(2)\\zeta(3) +\\text{Li}_4\\left(\\frac{1}{2}\\right)}\n$$\n<\/p>\n\n\n\n
\nLet $H_n^{(2)} = \\zeta(2) – \\psi_1(n+1) = \\sum_{k=1}^n \\frac{1}{k^2}$. Then, it is easy to verify that:\n$$\n\\sum_{n=1}^\\infty t^n H_n^{(2)} = \\frac{\\text{Li}_2(t)}{1-t}, \\quad -1\\leq t < 1\n$$\nNow, dividing the above equation by $t$ and integrating both sides, gives us:\n$$\\begin{aligned}\n\\sum_{n=1}^\\infty \\frac{H_n^{(2)}}{n}x^n &= \\int_0^x \\text{Li}_2(t)\\left( \\frac{1}{t}+\\frac{1}{1-t}\\right) \\\\\n&= \\text{Li}_3(x) + \\int_0^x \\frac{\\text{Li}_2(t)}{1-t} dt \\\\\n&= \\text{Li}_3(x) -\\log(1-x)\\text{Li}_2(x) – \\int_0^x \\frac{\\log^2(1-t)}{t}dt \\end{aligned}\n$$\nFrom our previous calculation, we know that\n$$\n\\int_0^x \\frac{\\log^2(1-t)}{t}dt = 2\\sum_{n=1}^\\infty \\frac{H_n}{n^2}x^n -2\\text{Li}_3(x)\n$$\nHence, we obtain the following relation:\n$$\\sum_{n=1}^\\infty \\frac{H_n^{(2)}}{n}x^n =3\\text{Li}_3(x)-\\log(1-x)\\text{Li}_2(x) – 2\\sum_{n=1}^\\infty \\frac{H_n}{n^2}x^n \\quad \\color{blue}{\\cdots (11)}$$\nOnce again, divide the above equation by $x$ and integrate both sides to get:\n$$\n\\sum_{n=1}^\\infty \\frac{H_n^{(2)}}{n^2}x^n = 3\\text{Li}_4(x) + \\frac{1}{2}\\text{Li}_2^2(x) – 2\\sum_{n=1}^\\infty \\frac{H_n}{n^3}x^n \\quad \\color{blue}{\\cdots (12)}\n$$\nEquations (11) and (12) are valid for $-1\\leq x < 1$. Plugging in $x=\\frac{1}{2}$ in equation (12) and using the known closed form for $\\sum_{n=1}^\\infty \\frac{H_n}{n^3 2^n}$ gives us:\n$$\n\\boxed{ \\sum_{n=1}^\\infty \\frac{H_n^{(2)}}{n^2 2^n} = \\frac{\\pi^4}{1440}-\\frac{\\pi^2}{24}\\log^2(2) + \\frac{\\log^4(2)}{24}+ \\frac{1}{4}\\log(2)\\zeta(3) + \\text{Li}_4\\left(\\frac{1}{2}\\right) }\n$$\n<\/p>\n\n\n\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
In this post, we will evaluate the famous Euler sum: $$\\sum_{n=1}^\\infty (-1)^{n+1}\\frac{H_n}{n^3}\\quad \\color{blue}{\\cdots (*)} $$ where $H_n = \\sum_{k=1}^n \\frac{1}{k}=\\int_0^1 \\frac{1-t^n}{1-t}dt$ is the $n$-th harmonic number. This series resists contour integration techniques which makes it’s computation quite challenging. We will … Continue reading