Evaluate the integral,

$$ \int_{0}^{1} \ln(x)\ln(1-x)\,dx$$

I solved this problem, by writing power series and then calculating the series and found the answer to be $ 2 -\zeta(2) $, but I don't think that it is best solution to this problem. I want to know if it can be solved by any other nice/elegant method.

up vote 9 down vote accepted

Integrating by parts,

$$ \int \ln(x) \ln(1-x) \, dx = x \ln(x) \ln(1-x) - x \ln(1-x)+ \int \frac{x \ln (x)}{1-x} \, dx - \int \frac{x}{1-x} \, dx$$

where

$$ \int \frac{x}{1-x} \, dx = - \int \ dx + \int \frac{1}{1-x} \, dx = -x - \ln(1-x) + C_{1}$$

and $$ \begin{align} \int \frac{x \ln (x)}{1-x} \, dx &= -x \ln (x) - \ln(x) \ln(1-x) + \int dx + \int \frac{\ln (1-x)}{x} \, dx \\ &= -x \ln (x) - \ln(x) \ln(1-x) + x - \text{Li}_{2}(x) + C_{2}. \end{align}$$

$\text{Li}_{2}(x)$ is the dilogarithm function.

So we have $$ \begin{align} \int \ln(x) \ln(1-x) \, dx &= x \ln(x) \ln(1-x) - x \ln(1-x) - x \ln(x) - \ln(x) \ln(1-x) + 2x \\ &- \text{Li}_{2}(x) + \ln(1-x) + C . \end{align} $$

Therefore,

$$ \int_{0}^{1} \ln(x) \ln(1-x) \ dx = \lim_{x \to 1} \left[-x \ln(1-x)+\ln(1-x) \right] + 2 - \text{Li}_{2}(1) = 2 - \zeta(2) .$$

  • I also tried this, but it turns out too lengthy. +1 as usual. – Tunk-Fey Sep 3 '14 at 19:39
  • Thanks for your compliment but I think that's too much, rather your answers and sos440 are very impressive either at MSE or I&S, you both never ceases to amaze me. $\ddot\smile$ – Tunk-Fey Sep 3 '14 at 20:18
  • @Tunk-Fey, so you are a user in I&S! – Zaid Alyafeai Sep 3 '14 at 20:25
  • @ZaidAlyafeai No, I am not, maybe one day I'll join there. For the time being, I only participate in here and Brilliant.org. I only visit and observe I&S as a non-member. Anyway, I also enjoy your posts in I&S especially for the series and integral involving polylog. – Tunk-Fey Sep 3 '14 at 20:40
  • @Tunk-Fey What is I&S? – user940 Sep 3 '14 at 21:53

\begin{align} \int_0^1\ln(1-x)\ln x\ dx&=\int_0^1\sum_{n=1}^\infty\frac{x^n}n\ln x\ dx\\ &=\sum_{n=1}^\infty\frac{1}n\int_0^1 x^n\ln x\ dx\\ &=-\sum_{n=1}^\infty\frac{1}n\cdot\frac1{(n+1)^2}\\ &=\sum_{n=1}^\infty\left[\frac{1}n-\frac1{n+1}-\frac1{(n+1)^2}\right]\\ &=1-\left[\sum_{n=1}^\infty\frac1{n^2}-1\right]\\ &=\large\color{blue}{2-\zeta(2)=2-\frac{\pi^2}6}. \end{align}


Note :

$\displaystyle\ \ \int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots$

  • Ups, I didn't see the OP. I'll edit it if an idea comes to mind. Sorry. – Tunk-Fey Sep 3 '14 at 6:18
  • Thanks @JackD'Aurizio. Unfortunately, this is not what OP wants, he asked for other methods. Anyway, have you found a way to obtain the problem asked by V-Moy: a closed-form for $$\int_0^{\Large\frac{\pi}{4}}\ln^2\cos x\ dx\ ?$$I tried to answer but without success so far. – Tunk-Fey Sep 4 '14 at 4:20
  • 1
    The line below (608) in pi314.net/eng/hypergse13.php#x15-12200013 was the key: Landen identities give that the value of such an integral depends on $\operatorname{Li}_3\left(\frac{1+i}{2}\right)$ as "claimed" by Mathematica. – Jack D'Aurizio Sep 4 '14 at 4:24
  • 1
    $\zeta(2)$ might not look to significant to the OP. Add the fact that it is $\frac{\pi^2}6$ – Ali Caglayan Sep 7 '14 at 22:14

Using the reflection formula

$$\log(x)\log(1-x) =\zeta(2)-\mathrm{Li}_2(x)-\mathrm{Li}_2(1-x) $$

\begin{align} \int^1_0\log(x)\log(1-x) &=\zeta(2)-\int^1_0\mathrm{Li}_2(x)\,dx-\int^1_0\mathrm{Li}_2(1-x)\,dx\\ &=\zeta(2)-2\int^1_0\mathrm{Li}_2(x)\,dx\\ &=\zeta(2)-2\zeta(2)-2\int^1_0\log(1-x)\,dx\\ &=2-\zeta(2) \end{align}

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x:\ {\large ?}}$.

\begin{align} &\color{#66f}{\large\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x} =\int_{x\ =\ 0}^{x\ =\ 1}\ln\pars{1 - x}\dd\bracks{x\ln\pars{x} - x + 1} \\[3mm]&=\left.\bracks{x\ln\pars{x} - x + 1}\ln\pars{1 - x}\right\vert_{0}^{1} -\int_{0}^{1}\bracks{x\ln\pars{x} - x + 1}\,{-1 \over 1 - x}\,\dd x =\int_{0}^{1}{x\ln\pars{x} \over 1 - x}\,\dd x + 1 \\[3mm]&=-\lim_{\mu\ \to\ 1}\partiald{}{\mu} \int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x + 1 =-\lim_{\mu\ \to\ 1}\partiald{\Psi\pars{\mu + 1}}{\mu} + 1 \end{align} where $\ds{\Psi\pars{z}}$ is the Digamma Function $\ds{\bf 6.3.1}$ and we used the identity $\ds{\bf 6.3.22}$.

$$ \color{#66f}{\large\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x} =-\Psi'\pars{2} + 1=-\Psi'\pars{1} + 2=-\zeta\pars{2} + 2 $$ Here we used the identities: $$ \Psi'\pars{z + 1} = \Psi'\pars{z} - {1 \over z^{2}}\,,\qquad \Psi^{\rm\pars{n}}\pars{1}=\pars{-1}^{n + 1}\,n!\,\zeta\pars{n + 1}\,,\quad n = 1,2,3,\ldots $$

Since $\ds{\zeta\pars{2} = {\pi^{2} \over 6}}$: $$ \color{#66f}{\large\int_{0}^{1}\ln\pars{x}\ln\pars{1 - x}\,\dd x} =\color{#66f}{\large 2 - {\pi^{2} \over 6}} \approx {\tt 0.3551} $$

  • i love your answers!+1 – RE60K Jan 19 '15 at 3:51
  • @ADG Thanks a lot. – Felix Marin Jan 19 '15 at 4:06

You could start from the Beta function $$ B(p+1,r+1) = \int_0^1 x^p (1-x)^r\; dx = \dfrac{\Gamma(p+1) \Gamma(r+1)}{\Gamma(p+r+2)}$$ take the derivatives with respect to $p$ and $r$, and evaluate at $p=r=0$.

I've found a solution that is interesting, but probably not elegant, and definitely not short.

$I = \displaystyle\int_0^1 \ln(x)\ln(1 - x) dx$

Basic results:

  • $\lim\limits_{n \to 0} \dfrac{x^n - 1}{n} = \log x$, or $\lim\limits_{n \to 1}\dfrac{x^{n-1} - 1}{n - 1} = \log x$.
  • $\dfrac{d}{dn}\beta(n, n) = 2\beta(n, n)(\psi_0(n) - \psi_0(2n))$ where $\psi_0(n)$ is the digamma function.
  • $\dfrac{d^2}{dn^2}\beta(n, n) = 4\beta(n, n)(\psi_0(n) - \psi_0(2n))^2 + 2\beta(n, n)(\psi_1(n) - 2\psi_1(2n))$, where $\psi(1)(n)$ is the polygamma function.
  • $\psi_0(1) - \psi_0(2) = -1$ according to the recurrence relation.
  • $\psi_1(2) = \psi_1(1) - 1$ according to the recurrence relation.
  • $\psi_1(1) = \zeta(2)$.

Solution:
$\begin{align} I & = \lim\limits_{n \to 1} \displaystyle\int_0^1 \dfrac{(x^{n - 1} - 1)((1 - x)^{n - 1} - 1)}{(n - 1)^2} dx\\ & = \lim\limits_{n \to 1}\displaystyle\int_0^1 \dfrac{x^{n-1}(1-x)^{n-1} - x^{n - 1} - (1-x)^{n-1} + 1}{(n-1)^2} dx\\ & = \lim\limits_{n \to 1} \dfrac{\beta(n,n) - \frac{1}{n} - \frac{1}{n} + 1}{(n-1)^2}\\ & = \lim\limits_{n \to 1} \dfrac{\beta(n,n)(\psi_0(n)-\psi_0(2n)) + \frac{2}{n^2}}{2(n-1)} \quad [\text{l'Hospital's rule}]\\ & = \lim\limits_{n \to 1} \dfrac{4\beta(n,n)(\psi_0(n)-\psi_0(2n))^2 + 2\beta(n,n)(\psi_1(n)-2\psi_1(2n))- \frac{4}{n^3}}{2}\quad [\text{l'Hospital's rule}]\\ & = 2\beta(1, 1)(\psi_0(1) - \psi_0(2))^2 + \beta(1, 1)(\psi_1(1) - 2\psi_1(2)) - 2\\ & = 2(-1)^2 + 1(\psi_1(1) - 2\psi_1(1) + 2) - 2\\ & = 2 - \psi_1(1)\\ & = 2-\zeta(2) \end{align}$

You could expand $\ln(1-x) =-\sum_{n=1}^{\infty} \frac{x^n}{n} $ and evaluate $\int_0^1 x^n \ln x\,dx$, probably by an induction via integration by parts.

From your description, you may have already done this.

It sure is easier to write this than to do it.

Noting $$ \frac{d}{dx}[x(1-\ln(1-x))+\ln(1-x)]=-\ln(1-x) $$ we have \begin{eqnarray} \int_0^1\ln x\ln(1-x)dx&=&-\int_0^1\ln xd[x(1-\ln(1-x))+\ln(1-x)]\\ &=&-[x(1-\ln(1-x))+\ln(1-x)]\ln x\bigg|_0^1+\int_0^1\frac{x(1-\ln(1-x))+\ln(1-x)}{x}dx\\ &=&\int_0^1(1-\ln(1-x)+\frac{\ln(1-x)}{x})dx\\ &=&\int_0^1(1-\ln(1-x))dx+\int_0^1\frac{\ln(1-x)}{x}dx\\ &=&2-\zeta(2). \end{eqnarray} Here we used the well-known result $$ \int_0^1\frac{\ln(1-x)}{x}dx=-\zeta(2). $$

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.