5

I need your help to evaluate the following integral:

$$I=\int_0^1 \frac{\log^3(1-x)}{x}dx$$

Using the fact that for $|x|<1$ I get

$$\log(1-x)=-\sum_{n=1}^{\infty}\frac{x^k}{k}$$

One can rewrite $I$ as

$$I=-\sum_{k=0}^{\infty}\frac{1}{k+1}\left(\int_0^1 x^k\log^2(1-x)dx\right)$$

I tried to rewrite $\log(1-x)$ as a sum but I get some "monstrous" summation to calculate.

Plexus
  • 530
techer
  • 149

3 Answers3

9

$$ \begin{align} \int_0^1\frac{\log^3(1-x)}{x}\,\mathrm{d}x &=-\int_0^\infty\frac{u^3}{1-e^{-u}}e^{-u}\,\mathrm{d}u\tag{1}\\ &=-\int_0^\infty u^3\sum_{k=1}^\infty e^{-ku}\,\mathrm{d}u\tag{2}\\ &=-\sum_{k=1}^\infty\frac1{k^4}\int_0^\infty u^3e^{-u}\,\mathrm{d}u\tag{3}\\[6pt] &=-\zeta(4)\,\Gamma(4)\tag{4}\\[9pt] &=-\frac{\pi^4}{90}\cdot6\tag{5}\\[6pt] &=-\frac{\pi^4}{15}\tag{6} \end{align} $$ Explanation:
$(1)$: $x=1-e^{-u}$
$(2)$: series $\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k$
$(3)$: substitute $u\mapsto\frac uk$
$(4)$: use a couple of special functions
$(5)$: evaluate special functions (see this answer for $\zeta(4)$)
$(6)$: simplification

Addendum

Note that just as $$ \int_0^\infty\frac{x^{a-1}}{e^x}\,\mathrm{d}x=\Gamma(a) $$ we have $$ \int_0^\infty\frac{x^{a-1}}{e^x-1}\,\mathrm{d}x=\Gamma(a)\zeta(a) $$

robjohn
  • 345,667
3

Since robjohn already provided a detailed solution, I will just outline a non-trivial consequence of the substitution $x\mapsto(1-x)$. Since $\int_{0}^{1} x^m\log^3(x)\,dx=-\frac{6}{(m+1)^3}$ for any $m\in\mathbb{N}$, the original integral equals $-6\,\zeta(4)=-\frac{\pi^4}{15}$. On the other hand: $$ \log(1-x) = \sum_{n\geq 1}\frac{x^n}{n},\qquad \log^2(1-x)=\sum_{n\geq 1}\frac{2 H_{n-1}}{n}x^{n} \tag{1} $$ lead, by summation by parts, to $$ \log^3(1-x) = -3\sum_{n\geq 1}\frac{H_{n-1}^2-H_{n-1}^{(2)}}{n}x^n \tag{2}$$ so: $$ \int_{0}^{1}\frac{\log^3(1-x)}{x}\,dx = -3\left[\sum_{n\geq 1}\frac{H_{n-1}^2}{n^2}-\sum_{n\geq 1}\frac{H_{n-1}^{(2)}}{n^2}\right]\tag{3} $$ and since $$ \sum_{n\geq 1}\frac{H_{n-1}^{(2)}}{n^2} = \sum_{1\leq m < n}\frac{1}{m^2 n^2}=\frac{\zeta(2)^2-\zeta(4)}{2} = \frac{\pi^4}{120}\tag{4} $$ surprising consequences of the substitution $x\mapsto(1-x)$ are: $$ \sum_{n\geq 1}\frac{H_{n-1}^2}{n^2}=\frac{11\pi^4}{360},\qquad \sum_{n\geq 1}\frac{H_n^2}{n^2}=\frac{17\pi^4}{360}.\tag{5}$$ As a further reference, please see pages 91-92 here.

Jack D'Aurizio
  • 353,855
2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & \equiv \int_{0}^{1}{\ln^{3}\pars{1 - x} \over x}\,\dd x \,\,\,\stackrel{x\ \mapsto\ 1 - x}{=}\,\,\, \int_{0}^{1}{\ln^{3}\pars{x} \over 1 - x}\,\dd x \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, \int_{0}^{1}\ln\pars{1 - x}\bracks{3\ln^{2}\pars{x} \,{1 \over x}}\dd x = -3\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln^{2}\pars{x} \,\dd x \\[5mm] &\ \pars{~\substack{\ds{\mrm{Li}_{s}:\ PolyLogarithm\ Function\,,\quad \mrm{Li}_{s}\pars{0} = 0}\\[2mm] \ds{\mrm{Li}_{s + 1}'\pars{z} = {\mrm{Li}_{s}\pars{z} \over z}\,,\quad\mrm{Li}_{1}\pars{z} = -\ln\pars{1 - z}}}~} \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, 3\int_{0}^{1}\mrm{Li}_{2}\pars{x}\bracks{2\ln\pars{x}\,{1 \over x}}\,\dd x = 6\int_{0}^{1}\mrm{Li}_{3}'\pars{x}\ln\pars{x}\,\dd x \\[5mm] & \stackrel{\mrm{IBP}}{=}\,\,\, -6\int_{0}^{1}\mrm{Li}_{3}\pars{x}\,{1 \over x}\,\dd x = -6\int_{0}^{1}\mrm{Li}_{4}'\pars{x}\,\dd x = -6\,\mrm{Li}_{4}\pars{1} \\[5mm] & = -6\ \underbrace{\zeta\pars{4}}_{\ds{\pi^{4} \over 90}} \qquad\qquad\qquad\pars{~\substack{\ds{\zeta:\ Riemann\ Zeta\ Function}\\[2mm] \ds{\mrm{Li}_{s}\pars{1} = \zeta\pars{s}}}~} \\[5mm] & = \bbx{-\,{\pi^{4} \over 15}} \\ & \end{align}

Felix Marin
  • 89,464