# Can you compute $\int_0^1\frac{\log(x)\log(1-x)}{x}dx$ more precisely than $1.20206$ and do a comparision with $\zeta(3)$?

I know from Wolfram Alpha that $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx=1.20206$$ and in the other hand, too from this online tool that $$\int\frac{\log(x)\log(1-x)}{x}dx=\mathrm{Li}_3(x)-\mathrm{Li}_2(x)\log(x)+constant.$$

Question. I would like made a comparision, and need obtain $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx,$$ more precisely than $1.20206$. I believe that could be $\zeta(3)$, but now I don't sure, and I don't know if holding this claim could be deduce easily.

Can you compute $\int_0^1\frac{\log(x)\log(1-x)}{x}dx$ more precisely than $1.20206$ to discard that this value is $\zeta(3)$, Apéry constant, or claim that the equality $$\int_0^1\frac{\log(x)\log(1-x)}{x}dx=\zeta(3)$$ holds and is known/easily deduced (perhaps from some of its known formulas involving integrals)?

This definite integral was computed as a summand, when I made some changes of variable in Beuker's integral (see [1]), and now i don't know if too I could be wrong in my computations.

I've searched in this site about this integral $\int\frac{\log(x)\log(1-x)}{x}dx$, and in Wikipedia about a possible identity between $\zeta(3)$ and particular values of logarithmic integrals $\mathrm{Li}_2(x)$ and $\mathrm{Li}_3(x)$.

References:

• My changes from Beukers integral (see Refereces) to obtain the integral as a summand (there were two summands) were first $x=e^{z/y}$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense, and there are no mistakes. Too I am assuming that this way is know. Very thanks much all users. – user243301 Dec 18 '15 at 18:55
• Did you try expanding $\ln(1-x)$? – Clayton Apr 26 at 19:37
• Taylor series ... expand $\ln(1-x)$ in terms of powers of $x$. – GEdgar Apr 26 at 19:41
• Seems like taylor series then by parts is the best way forward here – George Dewhirst Apr 26 at 20:07

$$\operatorname{Li}_s(z) = \sum_{k=1}^\infty {z^k \over k^s} = z + {z^2 \over 2^s} + {z^3 \over 3^s} + \cdots \,.$$

Therefore, since $\log(1)=0$, we have:

$$\operatorname{Li}_3(1) = \sum_{k=1}^\infty {1 \over k^3} = \zeta(3)$$

$$\operatorname{Li}_3(1)-\log(1)\operatorname{Li}_2(1) = \zeta(3)$$

It remains to show that $$\lim_{x\to0} \operatorname{Li}_3(x)-\log(x)\operatorname{Li}_2(x)=0$$

Note that if $z<1$,

$$\operatorname{Li}_2(z) = \sum_{k=1}^\infty {z^k \over k^2} = z + {z^2 \over 2^2} + {z^3 \over 3^2} + \cdots \\< z + {z \over 2^2} + {z \over 2^2}+ {z \over 4^2} + {z \over 4^2} + {z \over 4^2} + {z \over 4^2} + {z \over 8^2}+\cdots \leq z\sum_{k=0}^\infty {1 \over 2^k} = 2z$$

Since $\log(z)<z$, for $z>1$, we also have $\log(z^2)<2z$ thus $\log(x)<2\sqrt{x}$. Thus $\log(\frac1x)>-2\sqrt{x}$, thus $\log(u)>-2\sqrt{\frac1u}$, thus $0>\log(u)\operatorname{Li}_2(u)>-2\sqrt{u}$.

We may use the Squeeze Theorem to finish the result.

• Then I understand is a bad question, since is easy to compute the limit (I will try it) too I will try don't make more bad questions. Very thanks much @wythagoras. Now I don't know how comute the limit, but I will try it. – user243301 Dec 18 '15 at 18:33
• Very thanks for details @wythagoras – user243301 Dec 18 '15 at 18:43

There is a variety of possibilities how to show that this integral indeed equals $$\zeta(3)$$, i.e. Apéry's Constant. I would like to show some of them

I: Taylor Series Expansion of $$\log(1-x)$$

As it was first suggested within the comments (and done by FDP) we may expand the $$\log(1-x)$$ term as Taylor Series. Specifically, by using the MacLaurin Series of the aforementioned logarithm we obtain

\begin{align*} \int_0^1\frac{\log(1-x)\log(x)}x\mathrm dx&=\int_0^1\frac{\log(x)}x\left[-\sum_{n=1}^\infty\frac{x^n}n\right]\mathrm dx\\ &=-\sum_{n=1}^\infty\frac1n\int_0^1x^{n-1}\log(x)\mathrm dx\\ &=-\sum_{n=1}^\infty\frac1n\left[-\frac1{n^2}\right]\\ &=\sum_{n=1}^\infty\frac1{n^3}\\ &=\zeta(3) \end{align*}

This might be the most straightforward approach possible.

II: Integration By Parts

Choosing $$u=\log(1-x)$$ and $$\mathrm dv=\frac{\log(x)}x$$ we can apply Integration By Parts which gives

\begin{align*} \int_0^1\frac{\log(1-x)\log(x)}x&=\underbrace{\left[\log(1-x)\frac{\log^2(x)}2\right]_0^1}_{\to0}+\frac12\int_0^1\frac{\log^2(x)}{1-x}\mathrm dx\\ &=\frac12\int_0^1\log^2(x)\left[\sum_{n=0}^\infty x^n\right]\mathrm dx\\ &=\frac12\sum_{n=0}^\infty\int_0^1x^n\log^2(x)\mathrm dx\\ &=\frac12\sum_{n=0}^\infty\left[\frac2{(n+1)^3}\right]\\ &=\sum_{n=1}^\infty\frac1{n^3}\\ &=\zeta(3) \end{align*}

Again, we utilized a series expansion, this time the one of the geometric series.

III: Integral Representation of the Zeta Function

To use the Integral Representation of the Zeta Function here we need to reshape the integral a little bit. Starting with substitution $$\log(x)\mapsto -x$$ followed by Integration By Parts again we find

\begin{align*} \int_0^1\frac{\log(1-x)\log(x)}x\mathrm dx&=-\int_\infty^0(-x)\log(1-e^{-x})\mathrm dx\\ &=-\int_0^\infty x\log(1-e^{-x})\mathrm dx\\ &=\underbrace{\left[\frac{x^2}2\log(1-e^{-x})\right]_0^\infty}_{\to0}+\frac12\int_0^\infty\frac{x^2}{1-e^{-x}}e^{-x}\mathrm dx\\ &=\frac1{\Gamma(3)}\int_0^\infty\frac{x^{3-1}}{e^x-1}\mathrm dx\\ &=\zeta(3) \end{align*}

Overall this is more or less the same as the second approach, but I wanted to bring the integral representation into play. While this approach seems to omit the usage of a series representation we need it actually in order to prove the here used representation for the Zeta Function.

IV: The Trilogarithm $$\operatorname{Li}_3(1)$$

Similiar to the second approach we may chose Integration By Parts as suitable technique but instead we will apply it with $$u=\log(x)$$ and $$\mathrm dv=\frac{\log(1-x)}x$$ to get

\begin{align*} \int_0^1\frac{\log(1-x)\log(x)}x\mathrm dx&=\underbrace{\left[\log(x)(-\operatorname{Li}_2(x))\right]_0^1}_{\to0}+\int_0^1\frac{\operatorname{Li}_2(x)}x\mathrm dx\\ &=[\operatorname{Li}_3(x)]_0^1\\ &=\zeta(3) \end{align*}

A quick look at the series representation of the Trilogarithm verifies the last line.

• Hey, that's nicely done, but you took them all 😒. – LeBlanc Apr 27 at 6:29
• @Zacky Thank you! I'm sorry, but I couldn't resist ^^ – mrtaurho Apr 27 at 7:37

\begin{align}J&=\int_0^1 \frac{\ln(1-x)\ln x}{x} dx\\ &=-\int_0^1 \left(\sum_{n=1}^\infty \frac{x^{n-1}}{n}\right)\ln x\,dx\\ &=-\sum_{n=1}^\infty \frac{1}{n}\int_0^1 x^{n-1}\ln x\,dx\\ &=\sum_{n=1}^\infty \frac{1}{n^3}\\ &=\zeta(3) \end{align}

An informal argument: $$\log (1-x) = - \sum_{k=0}^\infty \frac{x^{k+1}}{k+1}, \quad |x| < 1.$$ Then $$\frac{\log x \log(1-x)}{x} = -\sum_{k=0}^\infty \frac{x^k \log x}{k+1},$$ and integrating term by term gives $$\int_{x=0}^1 x^k \log x \, dx = \left[ \frac{x^{k+1} \log x}{k+1} \right]_{x=0}^1 - \int_{x=0}^1 \frac{x^k}{k+1} \, dx = - \frac{1}{(k+1)^2}.$$ Therefore, $$\int_{x=0}^1 \frac{\log x \log(1-x)}{x} \, dx = \sum_{k=1}^\infty \frac{1}{k^3} = \zeta(3).$$

• Very thanks much @heropup – user243301 Dec 18 '15 at 18:34


You can add this "weird" answer to the excellent $$\texttt{@mrtaurho}$$ long list: \begin{align} &\bbox[10px,#ffd]{% \int_{0}^{1}{\ln\pars{1 - x}\ln\pars{x} \over x}\,\dd x} = \left.{\partial^{2} \over \partial\mu\,\partial\nu}\int_{0}^{1} {\bracks{\pars{1 - x}^{\mu} - 1}x^{\nu} \over x}\,\dd x\,\right\vert_{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}} \\[5mm] = &\ {\partial^{2} \over \partial\mu\,\partial\nu}\bracks{% \int_{0}^{1}x^{\nu - 1}\pars{1 - x}^{\mu}\,\dd x - \int_{0}^{1}x^{\nu - 1}\,\dd x} _{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}} \\[5mm] = &\ {\partial^{2} \over \partial\mu\,\partial\nu}\bracks{% {\Gamma\pars{\nu}\Gamma\pars{\mu + 1} \over \Gamma\pars{\nu + \mu + 1}} - {1 \over \nu}} _{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}} \\[5mm] = &\ {\partial^{2} \over \partial\mu\,\partial\nu}\braces{{1 \over \nu}\bracks{% {\Gamma\pars{\nu + 1}\Gamma\pars{\mu + 1} \over \Gamma\pars{\nu + \mu + 1}} - 1}} _{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}} \\[5mm] = &\ {1 \over 2}\,\partiald[2]{}{\nu}\braces{\Gamma\pars{\nu + 1} \partiald{}{\mu}\bracks{\Gamma\pars{\mu + 1} \over \Gamma\pars{\nu + \mu + 1}}} _{{\large\mu\ =\ 0} \atop {\large\,\,\nu\ =\ 0^{+}}} \\[5mm] = &\ {1 \over 2}\,\partiald[2]{}{\nu}\braces{\Gamma\pars{\nu + 1} \bracks{-\,{\gamma + \Psi\pars{\nu + 1} \over \Gamma\pars{\nu + 1}} }}_{\ \nu\ =\ 0^{+}} \\[5mm] = &\ -\,{1 \over 2}\,\Psi\,''\pars{1} = \bbx{\zeta\pars{3}} \end{align}

Let $x = e^{-y}$, we have $$\int_0^1 \frac{\log x\log(1-x)}{x} dx = \int_0^1 \frac{(-\log x)}{x} \sum_{n=1}^\infty \frac{x^n}{n} dx = \sum_{n=1}^\infty \frac{1}{n}\int_0^1 (-\log x) x^{n-1} dx\\ = \sum_{n=1}^\infty \frac{1}{n}\int_0^\infty y e^{-ny} dy = \sum_{n=1}^\infty \frac{1}{n^3} = \zeta(3)$$ Please note that we can switch the order of summation and integration because all the individual terms are non-negative.

• Very thanks much @achillehui incredible. I take notes from your solution. – user243301 Dec 18 '15 at 18:37
• @achillehui Are you exploiting Fubini-Tonelli with the counting measure on $\mathbb{N}$ and Lebesgue measure on $[0,1]$ to justify interchanging the series and integration? – Mark Viola Mar 20 '17 at 2:56
• @Dr.MV the last phrase "because all the individual terms are non-negative" is essentially Tonelli theorem. – achille hui Mar 20 '17 at 4:05
• @achillehui Yes, I know. So, you are exploiting it then? – Mark Viola Mar 20 '17 at 4:19

The Beta function and Feynman's trick are another way to go:

$$I=\int_{0}^{1}\frac{\log(x)\log(1-x)}{x}\,dx =\left.\frac{\partial^2}{\partial a \partial b}\int_{0}^{1}x^{a-1}(1-x)^{b}\,dx\,\right|_{\alpha,\beta=0^+}\tag{1}$$ hence: $$I = \left.\frac{\partial^2}{\partial a \partial b}\frac{\Gamma(a)\Gamma(b+1)}{\Gamma(a+b+1)}\,\right|_{\alpha,\beta=0^+}\tag{2}$$ and by exploiting $\Gamma'(z) = \Gamma(z)\cdot\psi(z)$ we get: $$I = -\frac{1}{2}\psi''(2)=\sum_{n\geq 0}\frac{1}{(n+1)^3}=\color{red}{\zeta(3)}\tag{3}$$ as wanted.

• I try understand your contribution too, now I am saturated. My changes to obtain the integral as a summand (there were two summands) were first $x=e^{z/y}$, second $u=z/y$ and finally $v=e^u$. I hope that it has sense. Very thanks much @JackD'Aurizio. – user243301 Dec 18 '15 at 18:54