# Evaluate $\int^1_0 \log^2(1-x) \log^2(x) \, dx$

I have no idea where to even start. WolframAlpha cant compute it either.

$$\int^1_0 \log^2(1-x) \log^2(x) \, dx$$

I think it can be done with series, but I am not sure, can someone help a little so I can get a start??

Thanks!

• i only know the result $-8 \zeta (3)+24-\frac{4 \pi ^2}{3}-\frac{\pi ^4}{90}$ – Dr. Sonnhard Graubner Oct 5 '14 at 19:51
• Differentiate the integral representation of the beta function twice w.r.t. $p$ and twice w.r.t. $q$, and then set $p=q=0$. This results into polygamma values which are known to be related to zeta values. The final result simplifies to what is written in the previous comment. – Start wearing purple Oct 5 '14 at 19:53
• @O.L., I may need you to be more explicit. What do you mean by "w.r.t p" which variable is $p$ and which variable is $q$ I know the beta function integral, but I am not sure which variables represent what. Can you let me know? – Amad27 Oct 5 '14 at 19:56
• Another tactic is repeated integration by parts. – David H Oct 5 '14 at 20:24

Using the series of $\ln^2(1-x)$, \begin{align} \int^1_0\ln^2{x}\ln^2(1-x) \ {\rm d}x &=\sum^\infty_{n=1}\frac{2H_n}{n+1}\int^1_0x^{n+1}\ln^2{x}\ {\rm d}x\\ &=\sum^\infty_{n=1}\frac{4H_n}{(n+1)(n+2)^3} \end{align} Then integrate $f(z)=\dfrac{(\gamma+\psi_0(-z))^2}{(z+1)(z+2)^3}$ along an infinitely large square. The integral vanishes which implies the sum of its residues is zero. At the positive integers, \begin{align} \sum^\infty_{n=1}{\rm Res}(f,n) &=\sum^\infty_{n=1}\operatorname*{Res}_{z=n}\left[\frac{1}{(z+1)(z+2)^3(z-n)^2}+\frac{2H_n}{(z+1)(z+2)^3(z-n)}\right]\\ &=\sum^\infty_{n=1}\frac{2H_n}{(n+1)(n+2)^3}-\sum^\infty_{n=1}\frac{4n+5}{(n+1)^2(n+2)^4}\\ &=\sum^\infty_{n=1}\frac{2H_n}{(n+1)(n+2)^3}+\frac{\pi^4}{30}+2\zeta(3)-\frac{91}{16} \end{align} At $z=0$, \begin{align} {\rm Res}(f,0) &=\operatorname*{Res}_{z=0}\frac{1}{z^2(z+1)(z+2)^3}\\ &=-\frac{5}{16} \end{align} At $z=-1$, \begin{align} {\rm Res}(f,-1) &=0 \end{align} At $z=-2$, \begin{align} {\rm Res}(f,-2) &=\frac{1}{2}\lim_{z\to-2}\frac{{\rm d}^2}{{\rm d}z^2}\frac{(\gamma+\psi_0(-z))^2}{z+1}\\ &=-\frac{\pi^4}{36}+2\zeta(3)+\frac{2\pi^2}{3}-6 \end{align} Therefore, \begin{align} \int^1_0\ln^2{x}\ln^2(1-x)\ {\rm d}x &=2\left[-\frac{\pi^4}{30}-2\zeta(3)+\frac{91}{16}+\frac{5}{16}+\frac{\pi^4}{36}-2\zeta(3)-\frac{2\pi^2}{3}+6\right]\\ &=-\frac{\pi^4}{90}-8\zeta(3)-\frac{4\pi^2}{3}+24 \end{align}

• where can I learn about this method? I have never learned this, can you tell me a starting point? – Amad27 Nov 10 '14 at 17:00
• @Amad27 You may want to learn some complex analysis and familarise yourself with the residue theorem first. After which, you may refer to this paper: algo.inria.fr/flajolet/Publications/FlSa98.pdf. – M.N.C.E. Nov 11 '14 at 6:36
• I read that page. It does not show anything about how you integrated along the square? Please tell me; how do you integrate along a square. And what do you mean the integral "vanishes," making the residue 0? – Amad27 Nov 11 '14 at 8:16
• @Amad27 I am integrating $f(z)$ over a square contour with vertices $\pm(N+\frac{1}{2})\pm i(N+\frac{1}{2})$ such that we avoid the poles at the integers on the real axis. To check that the contour integral $\to 0$ as $N\to\infty$, we parameterise and note that $\psi_0(z)\sim\ln{z}$ for large $z$. One can then apply the ML-inequality. But by the residue theorem, $\displaystyle\oint_{C_N}f(z)\ {\rm d}z=0=2\pi i\sum\text{(residues in contour)}$. The paper above explains how one can determine $f(z)$ such that the unknown sum can be computed. – M.N.C.E. Nov 12 '14 at 16:34

Starting with the Beta function \begin{align} B(x, y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, dt \end{align} differentiate with respect to $x$ and $y$ twice. This leads to \begin{align} I &= \int_{0}^{1} t^{x-1} (1-t)^{y-1} \, \ln^{2}(t) \, \ln^{2}(1-t) \, dt \\ &= \partial_{x}^{2} \partial_{y}^{2} B(x,y). \end{align} Now the integral in question is the case for $x=y=1$. For this, it is seen that \begin{align} \partial_{x}^{2} \left. B(x,y) \right|_{x=1} = \frac{1}{y} \left[ \psi'(1) - \psi'(y+1) + \left( \psi(1) - \psi(y+1) \right)^{2} \right]. \end{align} Differential with respect to $y$ yields \begin{align} \partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=1} &= \frac{1}{y} \left[ - \psi'''(y+1) - 2 \psi''(y+1) \left( \psi(1) - \psi(y+1) \right) + 2 \left( \psi'(y+1) \right)^{2} \right] \\ & \hspace{5mm} - \frac{2}{y^{2}} \left[ - \psi''(y+1) - 2 \psi'(y+1) \left( \psi(1) - \psi(y+1) \right) \right] \\ & \hspace{10mm} + \frac{2}{y^{3}} \left[ \psi'(1) - \psi'(y+1) + \left( \psi(1) - \psi(y+1) \right)^{2} \right]. \end{align} Now \begin{align} \partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=y=1} &= 4 - \psi'''(2) + 4 \psi''(2) - 4 \psi'(2) + 2 \left( \psi'(2) \right)^{2} . \end{align} Since \begin{align} \psi'(2) &= \frac{\pi^{2}}{6} -1 \\ \psi''(2) &= 2 - 2\zeta(3) \\ \psi'''(2) &= \frac{\pi^{4}}{45} - 6 \end{align} then \begin{align} \partial_{y}^{2} \partial_{x}^{2} \left. B(x,y) \right|_{x=y=1} = 24 - \frac{4 \pi^{2}}{3} - \frac{\pi^{4}}{90} - 8 \zeta(3). \end{align}

First, letting $x=e^{-t}$ gives \begin{align} \int_0^1x^k\log(x)^2\,\mathrm{d}x &=\int_0^\infty e^{-(k+1)t}\,t^2\,\mathrm{d}t\\ &=\frac2{(k+1)^3}\tag{1} \end{align} Also, squaring the power series yields $$\log(1-x)^2=\sum_{k=1}^\infty\frac{2H_{k-1}}{k}x^k\tag{2}$$ Therefore, \begin{align} &\int_0^1\log(1-x)^2\log(x)^2\,\mathrm{d}x\\ &=\int_0^1\sum_{k=1}^\infty\frac{2H_{k-1}}{k}x^k\log(x)^2\,\mathrm{d}x\tag{3a}\\ &=\sum_{k=1}^\infty\frac{4H_{k-1}}{k(k+1)^3}\tag{3b}\\ &=\sum_{k=1}^\infty\sum_{j=1}^{k-1}\frac4{jk(k+1)^3}\tag{3c}\\ &=\sum_{k=1}^\infty\sum_{j=1}^k\frac4{jk(k+1)^3}-\sum_{k=1}^\infty\frac4{k^2(k+1)^3}\tag{3d}\\ &=\sum_{j=1}^\infty\frac4j\sum_{k=j}^\infty\left(\frac1k-\frac1{k+1}-\frac1{(k+1)^2}-\frac1{(k+1)^3}\right)\tag{3e}\\ &-4\sum_{k=1}^\infty\left(-\frac3k+\frac3{k+1}+\frac1{k^2}+\frac2{(k+1)^2}+\frac1{(k+1)^3}\right)\tag{3f}\\ &=4\zeta(2)-\sum_{j=1}^\infty\frac4j\left[\sum_{k=j}^\infty\left(\frac1{k^2}+\frac1{k^3}\right)-\frac1{j^2}-\frac1{j^3}\right]\tag{3g}\\[6pt] &-4(-6+3\zeta(2)+\zeta(3))\tag{3h}\\[15pt] &=24-8\zeta(2)-4\zeta(3)\tag{3i}\\[4pt] &-\sum_{k=1}^\infty\sum_{j=1}^k\frac4j\left(\frac1{k^2}+\frac1{k^3}\right)+4\zeta(3)+4\zeta(4)\tag{3j}\\[2pt] &=24-8\zeta(2)+4\zeta(4)\tag{3k}\\[6pt] &-\sum_{k=1}^\infty\frac{4H_k}{k^2}-\sum_{k=1}^\infty\frac{4H_k}{k^3}\tag{3l}\\ &=24-8\zeta(2)-6\zeta(4)-8\zeta(3)+2\zeta(2)^2\tag{3m}\\[6pt] &=24-\frac{4\pi^2}3-\frac{\pi^4}{90}-8\zeta(3)\tag{3n} \end{align} Explanation:
$\text{(3a)}$: apply $(2)$
$\text{(3b)}$: apply $(1)$
$\text{(3c)}$: expand $H_{k-1}$
$\text{(3d)}$: add and subtract the $j=k$ term
$\text{(3e)}$: partial fractions: $\frac1{k(k+1)^3}=\frac1k-\frac1{k+1}-\frac1{(k+1)^2}-\frac1{(k+1)^3}$
$\text{(3f)}$: partial fractions: $\frac1{k^2(k+1)^3}=-\frac3k+\frac3{k+1}+\frac1{k^2}+\frac2{(k+1)^2}+\frac1{(k+1)^3}$
$\text{(3g)}$: apply $\sum\limits_{j=1}^\infty\frac4j\sum\limits_{k=j}^\infty\left(\frac1k-\frac1{k+1}\right)=\sum\limits_{j=1}^\infty\frac4{j^2}=4\zeta(2)$ and
$\hphantom{\text{(3g):}}$ $\sum\limits_{k=j}^\infty\left(\frac1{(k+1)^2}+\frac1{(k+1)^3}\right)=\sum\limits_{k=j}^\infty\left(\frac1{k^2}+\frac1{k^3}\right)-\frac1{j^2}-\frac1{j^3}$ to $\text{(3e)}$
$\text{(3h)}$: evaluate $\text{(3f)}$
$\text{(3i)}$: collect the evaluated terms in $\text{(3g)}$ and $\text{(3h)}$
$\text{(3j)}$: change order of summation in $\text{(3g)}$ and evaluate sums
$\text{(3k)}$: collect the evaluated terms in $\text{(3i)}$ and $\text{(3j)}$
$\text{(3l)}$: collect $H_k$ in $\text{(3j)}$
$\text{(3m)}$: use this answer to evaluate the sums in $\text{(3l)}$
$\text{(3n)}$: use $\zeta(2)=\frac{\pi^2}6$ and $\zeta(4)=\frac{\pi^4}{90}$

(Too long for comment)

For what it's worth, I'd like to expand on my comment above about attacking this integral via integration by parts. My main reason for wanting to solve the problem this way is not because it is the easiest way and certainly not because it is the quickest way, but rather because it seems to be the most elementary way, relying for the most part on methods accessible to high school students. (My apologies to the OP for just now getting around to writing this, but partially due to the length of the derivation I simply didn't have the time until now.)

We may reduce the evaluation of the definite integral $\mathcal{I}:=\int_{0}^{1}\ln^2{(x)}\ln^2{(1-x)}\,\mathrm{d}x$ to the problem of evaluating a sum of relatively less difficult integrals via integration by parts (IBPs). We'll use $u$ and $u^\prime$ to denote the following functions:

$$\begin{cases} u{(x)}:=\ln^2{(x)}\ln{(1-x)},\\ v^{\prime}{(x)}:=\ln{(1-x)}.\\ \end{cases}$$

The derivative of $\ln^2{(x)}\ln{(1-x)}$ is:

$$\frac{d}{dx}\left(\ln^2{(x)}\ln{(1-x)}\right)=\frac{2\ln{(x)}\ln{(1-x)}}{x}-\frac{\ln^2{(x)}}{1-x};$$

the indefinite integral of $\ln{(1-x)}$ is:

\begin{align} \int\ln{(1-x)}\,\mathrm{d}x &=x\ln{(1-x)}-\int\frac{x}{x-1}\,\mathrm{d}x\\ &=x\ln{(1-x)}-x-\ln{(1-x)}+\color{grey}{constant}. \end{align}

Then, choosing $v{(x)}$ by selecting a null constant term in the last line above, we have:

$$\begin{cases} u^{\prime}{(x)}=\frac{2\ln{(x)}\ln{(1-x)}}{x}-\frac{\ln^2{(x)}}{1-x},\\ v{(x)}=x\ln{(1-x)}-x-\ln{(1-x)}.\\ \end{cases}$$

IBPs in this way yields:

\begin{align} \mathcal{I} &=\int_{0}^{1}\ln^2{(x)}\ln^2{(1-x)}\,\mathrm{d}x\\ &=\int_{0}^{1}u{(x)}\,v^{\prime}{(x)}\,\mathrm{d}x\\ &=\left[u{(x)}\,v{(x)}\right]_{0}^{1}-\int_{0}^{1}u^{\prime}{(x)}\,v{(x)}\,\mathrm{d}x\\ &=\left[\ln^2{(x)}\ln{(1-x)}\left(x\ln{(1-x)}-x-\ln{(1-x)}\right)\right]_{0}^{1}\\ &~~~~~ -\int_{0}^{1}\left(\frac{2\ln{(x)}\ln{(1-x)}}{x}-\frac{\ln^2{(x)}}{1-x}\right)\left(x\ln{(1-x)}-x-\ln{(1-x)}\right)\,\mathrm{d}x.\\ \end{align}

Note that after integrating by parts both boundary terms vanish identically on the interval $[0,1]$. Thus,

\begin{align} \mathcal{I} &=\left[\ln^2{(x)}\ln{(1-x)}\left(x\ln{(1-x)}-x-\ln{(1-x)}\right)\right]_{0}^{1}\\ &~~~~~ -\int_{0}^{1}\left(\frac{2\ln{(x)}\ln{(1-x)}}{x}-\frac{\ln^2{(x)}}{1-x}\right)\left(x\ln{(1-x)}-x-\ln{(1-x)}\right)\,\mathrm{d}x\\ &=0-0+\int_{0}^{1}\left(\frac{\ln^2{(x)}}{1-x}-\frac{2\ln{(x)}\ln{(1-x)}}{x}\right)\left(x\ln{(1-x)}-x-\ln{(1-x)}\right)\,\mathrm{d}x\\ &=\int_{0}^{1}\left(\frac{\ln^2{(x)}}{1-x}-\frac{2\ln{(x)}\ln{(1-x)}}{x}\right)\left(x\ln{(1-x)}-x-\ln{(1-x)}\right)\,\mathrm{d}x\\ &=\color{green}{\int_{0}^{1}\frac{x\ln^2{(x)}\ln{(1-x)}}{1-x}\,\mathrm{d}x}-\int_{0}^{1}\frac{x\ln^2{(x)}}{1-x}\,\mathrm{d}x\color{green}{-\int_{0}^{1}\frac{\ln^2{(x)}\ln{(1-x)}}{1-x}\,\mathrm{d}x}\\ &~~~~~ \color{brown}{-\int_{0}^{1}2\ln{(x)}\ln^2{(1-x)}\,\mathrm{d}x}+\int_{0}^{1}2\ln{(x)}\ln{(1-x)}\,\mathrm{d}x\color{brown}{+\int_{0}^{1}\frac{2\ln{(x)}\ln^2{(1-x)}}{x}\,\mathrm{d}x}\\ &=\color{green}{-\int_{0}^{1}\ln^2{(x)}\ln{(1-x)}\,\mathrm{d}x}\color{orange}{-\int_{0}^{1}\frac{x\ln^2{(x)}}{1-x}\,\mathrm{d}x}+2\int_{0}^{1}\ln{(x)}\ln{(1-x)}\,\mathrm{d}x\\ &~~~~~ \color{brown}{+2\int_{0}^{1}\frac{(1-x)\ln{(x)}\ln^2{(1-x)}}{x}\,\mathrm{d}x}\\ &=2\int_{0}^{1}\ln{(x)}\ln{(1-x)}\,\mathrm{d}x\color{orange}{+\int_{0}^{1}\mathrm{d}x-\int_{0}^{1}\frac{\ln^2{(x)}}{1-x}\,\mathrm{d}x}-\int_{0}^{1}\ln^2{(x)}\ln{(1-x)}\,\mathrm{d}x\\ &~~~~~ +2\int_{0}^{1}\frac{(1-x)\ln{(x)}\ln^2{(1-x)}}{x}\,\mathrm{d}x\\ &=1+2\int_{0}^{1}\ln{(x)}\ln{(1-x)}\,\mathrm{d}x-\int_{0}^{1}\frac{\ln^2{(x)}}{1-x}\,\mathrm{d}x-\int_{0}^{1}\ln^2{(x)}\ln{(1-x)}\,\mathrm{d}x\\ &~~~~~ +2\int_{0}^{1}\frac{(1-x)\ln{(x)}\ln^2{(1-x)}}{x}\,\mathrm{d}x\\ &=:1+2\,\mathcal{I}_{1}-\mathcal{I}_{2}-\mathcal{I}_{3}+2\,\mathcal{I}_{4}, \end{align}

where in the last line we've simply introduced the constant names $\mathcal{I}_{k},k\in\{1,2,3,4\}$, for convenience to denote the following integrals, respectively:

$$\begin{cases} \int_{0}^{1}\ln{(x)}\ln{(1-x)}\,\mathrm{d}x,\\ \int_{0}^{1}\frac{\ln^2{(x)}}{1-x}\,\mathrm{d}x,\\ \int_{0}^{1}\ln^2{(x)}\ln{(1-x)}\,\mathrm{d}x,\\ \int_{0}^{1}\frac{(1-x)\ln{(x)}\ln^2{(1-x)}}{x}\,\mathrm{d}x.\\ \end{cases}$$

Now, the OP noted in the question statement that WolframAlpha was unable to find a closed form for the definite integral $\mathcal{I}$. As one might expect, WA is also unable to find an anti-derivative for the corresponding indefinite integral. HOWEVER, it turns out that WolframAlpha does return anti-derivatives for each of the four integrals $\mathcal{I}_{k},k\in\{1,2,3,4\}$. WA is also capable of returning closed form values for each of the definite integrals except $\mathcal{I}_{4}$, which has the longest and most complicated anti-derivative of the bunch.

• this is a good answer. Just a question. Do you know about the M.N.C.E. method? How I can use that? It looks interesting – Amad27 Nov 10 '14 at 17:18
• @Amad27 I am aware of the method, but I don't understand it that well. It's based on residue theory, which is a standard component of any complex analysis course. – David H Nov 10 '14 at 17:41
• Do you where I can get a start or a textbook (that perhaps you read,introductory only) that I can start learning that. The method of Residues is of great interest to me! If you understand it even some bit, will you help me learn it? – Amad27 Nov 10 '14 at 18:01
• @Amad27 I believe the standard introductory text is Saff & Snider. Beyond that, I'm really not the best person to ask. – David H Nov 10 '14 at 18:18