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I am looking for a closed form for

$$I=\int_0^1\frac{(x^3-3x^2+x)\log(x-x^2)}{(x^2-x+1)^3}\mathrm dx\approx 0.851035604949$$

Wolfram does not evaluate $I$.

I suspect $I$ has a closed form, because if $I$ has a closed form then the answer to my question In Pascal's triangle without the $1$s, what is the sum of squares of reciprocals? has a closed form. In that question, I give reasons for why I suspect a closed form answer. But of course I could be wrong.

My attempt

Here is the graph of $y=\frac{(x^3-3x^2+x)\log(x-x^2)}{(x^2-x+1)^3}$.

enter image description here

I thought about translating the function to an odd function in order to take advantage of rotational symmetry, but the curve clearly does not have rotational symmetry.

I tried substitution and Maclaurin series, but got nowhere.

Dan
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    Another possible path with contour integration, although it's quite overkill and really only simplifies the log term. Using a slight variant of this, we can recover from the imaginary parts of the residue sum the relation$$I=\frac{2\pi}{3\sqrt3}-1+2\int_0^\infty\frac{x^3+3x^2+x}{(x^2+x+1)^3}\log(x+x^2),dx$$and notice that substituting $x\to\frac1x$ further reduces this to$$I=\frac{2\pi}{3\sqrt3}-1+2\int_0^\infty\frac{x^3+3x^2+x}{(x^2+x+1)^3}\color{red}{\log(1+x)},dx$$ – user170231 Jan 23 '24 at 21:21

3 Answers3

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$$I=\int_0^1\frac{x(1-3x+x^2)\ln(x-x^2)}{(1-x+x^2)^3}dx=\frac{2\pi}{9\sqrt 3}-\frac13+\frac{4}{3\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$ Where $\operatorname{Cl}_2(x)$ is the Clausen function.

$$I=\int_0^1\frac{x(1-3x+x^2)\ln(x-x^2)}{(1-x+x^2)^3}dx$$ $$\overset{x\to 1-x}=\int_0^1 \frac{(1-x)(-1+x+x^2)\ln(x-x^2)}{(1-x-x^2)^3}dx$$ $$\Rightarrow 2I=\int_0^1 \frac{(3x-3x^2-1)\ln(x-x^2)}{(1-x-x^2)^3}dx$$

$$\int_0^1 \frac{(3x-3x^2-1)\ln x}{(1-x+x^2)^3}dx\overset{x\to 1-x}=\int_0^1 \frac{(3x-3x^2-1)\ln (1-x)}{(1-x+x^2)^3}dx$$ $$\Rightarrow I=\int_0^1 \frac{(3x-3x^2-1)\ln x}{(1-x+x^2)^3}dx$$

$$=2\int_0^1 \frac{\ln x}{(1-x+x^2)^3}dx-3\int_0^1 \frac{\ln x}{(1-x+x^2)^2}dx$$

$$\left(\frac{x(1-x)(2x-1)}{3(1-x+x^2)^2}\right)'=\frac{2}{(1-x+x^2)^3}-\frac{3}{(1-x+x^2)^2}+\frac{\frac23}{1-x+x^2}$$ $$\Rightarrow I=\int_0^1 \ln x\left(\frac{x(1-x)(2x-1)}{3(1-x+x^2)^2}\right)'dx-\frac23\int_0^1 \frac{\ln x}{1-x+x^2}dx$$

$$\int_0^1 \ln x\left(\frac{x(1-x)(2x-1)}{3(1-x+x^2)^2}\right)'dx\overset{IBP}=\frac13\int_0^1 \frac{(1-x)(1-2x)}{(1-x+x^2)^2}dx=\frac{2\pi}{9\sqrt 3}-\frac13$$

$$\int_0^1 \frac{\ln x}{1-x+x^2}dx=\int_0^1 \frac{\ln x}{1-2\cos\left(\frac{\pi}{3}\right)x+x^2}dx=$$ $$=\frac{1}{\sin\left(\frac{\pi}{3}\right)}\sum_{n=1}^\infty \sin\left(\frac{n\pi}{3}\right)\int_0^1 x^{n-1} \ln x\, dx=-\frac{2}{\sqrt 3}\sum_{n=1}^\infty \frac{\sin\left(\frac{n\pi}{3}\right)}{n^2}=-\frac{2}{\sqrt 3}\operatorname{Cl}_2\left(\frac{\pi}{3}\right)$$

Zacky
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    "Clausen" function - just in time for Christmas :) – Dan Dec 17 '23 at 04:27
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    @Dan haha, that's quite funny. Santa Claus(en) makes an early appearance. – Zacky Dec 17 '23 at 04:31
  • Here is another integral that may also be related to the Clausen function, as discussed in the comments there. – Dan Jan 17 '24 at 07:46
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    @Dan Thanks for the ping, but I wouldn't say those 2 integrals are related (beside the appearance of the Clausen function). Anyway, I can post an answer there that finishes the integral, in case no one else does in the meantime. – Zacky Jan 17 '24 at 10:56
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Tedious work using algebra$$I=\int_0^1\frac{x(x^2-3x+1)}{(x^2-x+1)^3}\,\log(x(1-x))\, dx$$

Write $$F=\frac{x(x^2-3x+1)}{(x^2-x+1)^3}=\frac{x(x-a)(x-b)}{(x-c)^3\,(x-d)^3}$$ where $$a=\frac{3-\sqrt{5}}{2}\qquad b=\frac{3+\sqrt{5}}{2}\qquad c=\frac{1-i \sqrt{3}}{2}\qquad d=\frac{1+i \sqrt{3}}{2}$$

Partial fraction decomposition gives $$F=\frac {A_c}{(x-c)}+\frac {B_c}{(x-c)^2}+\frac {C_c}{(x-c)^3}+\frac {A_d}{(x-d)}+\frac {B_d}{(x-d)^2}+\frac {C_d}{(x-d)^3}$$ with

$$A_c=-\frac{i}{3 \sqrt{3}}\qquad B_c=\frac{3+i \sqrt{3}}{18} \qquad C_c=\frac{3-i \sqrt{3}}{9} $$ $$A_d=+\frac{i}{3 \sqrt{3}}\qquad B_d=\frac{3-i \sqrt{3}}{18} \qquad C_d=\frac{3+i \sqrt{3}}{9} $$

All antiderivarives $$J_n=\int \frac {\log(x(1-x)) } {(x-k)^n}\,dx$$ are "relatively" simple to compute.

For the definite integrals $$\int \frac {\log(x(1-x)) } {(x-c)}\,dx=\text{Li}_2\left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)-\text{Li}_2\left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)$$ $$\int \frac {\log(x(1-x)) } {(x-c)^2}\,dx=\frac{\pi }{\sqrt{3}}$$ $$\int \frac {\log(x(1-x)) } {(x-c)^3}\,dx=-\frac{i}{6} \left(3 \sqrt{3}+\pi \right)$$ $$\int \frac {\log(x(1-x)) } {(x-d)}\,dx=$$ $$\frac{i}{12 \sqrt{3}} \left(-\psi ^{(1)}\left(\frac{1}{6}\right)-\psi ^{(1)}\left(\frac{1}{3}\right)+\psi ^{(1)}\left(\frac{2}{3}\right)+\psi ^{(1)}\left(\frac{5}{6}\right)\right)$$ $$\int \frac {\log(x(1-x)) } {(x-d)^2}\,dx=\frac{\pi }{\sqrt{3}}$$ $$\int \frac {\log(x(1-x)) } {(x-d)^3}\,dx=\frac{i}{6} \left(3 \sqrt{3}+\pi \right)$$

Replacing the $(A_c,B_c,C_c,A_d,B_d,C_d)$ constants by their values, ths leads to $$I=\frac{2 \pi }{9 \sqrt{3}}-\frac{1}{3}+\frac 1 {54} \left(\psi ^{(1)}\left(\frac{1}{6}\right)+\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right)$$ already given by @David G. Stork

Claude Leibovici
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Mathematica gives:

$$\frac{1}{54} \left(-18+4 \sqrt{3} \pi +\psi ^{(1)}\left(\frac{1}{6}\right)+\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right)$$

using the PolyGamma function.

Indeed, the numerical value of this analytic form is $0.851036$.

David G. Stork
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