# Integral $\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx$

Here is an integral I derived while evaluating another. It appears to be rather tough, but some here may not be so challenged :)

Show that: $$\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx=\frac{11}{9}\zeta(3)-\frac{\pi}{72\sqrt{3}}\left(5\psi_{1}\left(\frac13\right)+4\psi_{1}\left(\frac23\right)-3\psi_{1}\left(\frac56\right)\right)$$ $$=\frac{11}{9}\zeta(3)+\frac{4\pi^{3}}{27\sqrt{3}}-\frac{2\pi}{9\sqrt{3}}\psi_{1}\left(\frac13\right)=\frac{11}{9}\zeta(3)-\frac{4\pi}{9}\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)$$ $$=\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)-2\operatorname{Cl}_{2}\left(\frac{2\pi}{3}\right)-\frac{4\pi}{9}\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)$$

I attempted all kinds of 'starts' to no satisfactory end, but things look promising. There are some mighty sharp folks here that may be better at deriving the solution.

I thought perhaps the identity:

$$\frac{\log^{2}(1-(x-x^{2}))}{x}=2\sum_{n=1}^{\infty}\frac{H_{n}}{n+1}x^{n}(1-x)^{n+1}$$

or the Beta function could be used if given enough ingenuity.

This led me to the no-less-imposing Euler/reciprocal of central binomial coefficients sum below. It would be great to just show the middle sum is equivalent to the right sum:

$$1/4\sum_{n=1}^{\infty}\frac{H_{n}n\Gamma^{2}(n)}{(n+1)(2n+1)\Gamma(2n)}=1/2\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)(2n+1)\binom{2n}{n}}=1/3\zeta(3)-2/3\sum_{n=1}^{\infty}\frac{1}{n^{3}\binom{2n}{n}}$$

Is there a general form for $$\sum_{n=1}^{\infty}\frac{H_{n}}{\binom{2n}{n}}x^{n}?$$

I tried starting with the identity: $$\sum_{n=1}^{\infty}\frac{\Gamma^{2}(n)}{\Gamma(2n)}x^{n-1}=\frac{4\sin^{-1}\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}}$$ and using various manipulations to hammer into the needed form. This, too, turned monstrous.

There appears to be a relation to Clausen functions (as with other log integrals such as $$\int_{0}^{1}\frac{\log(x)}{x^{2}-x+1}dx$$), to wit:

I use Cl for sin and CL for cos Clausen functions

$$\operatorname{Cl}_{2}\left(\frac{\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\sin(\frac{\pi k}{3})}{k^{2}}=\frac{\sqrt{3}}{72}\left(\psi_{1}(1/6)+\psi_{1}(1/3)-\psi_{1}(2/3)-\psi_{1}(5/6)\right)$$

$$=\frac{\sqrt{3}}{6}\psi_{1}(1/3)-\frac{\pi^{2}\sqrt{3}}{9}$$

and

$$\operatorname{Cl}_{3}\left(\frac{\pi}{3}\right)-\operatorname{Cl}_{3}\left(\frac{2\pi}{3}\right)=\sum_{k=1}^{\infty}\frac{\cos(\frac{\pi k}{3})}{k^{3}}-2\sum_{k=1}^{\infty}\frac{\cos(\frac{2\pi k}{3})}{k^{3}}=\frac{11}{9}\zeta(3)$$

Another approach. I also broke the integral up as such:

$$\int_{0}^{1}\frac{\log^{2}(x^{2}-x+1)}{x}dx=\int_{0}^{1}\frac{\log^{2}(1-xe^{\frac{\pi i}{3}})}{x}dx+2\int_{0}^{1}\frac{\log(1-xe^{\pi i/3})\log(1-xe^{-\pi i/3})}{x}dx+\int_{0}^{1}\frac{\log^{2}(1-xe^{-\pi i/3})}{x}dx$$

The middle integral right of the equal sign is the one that has given me the fit.

I think this is a fun and head-scratchin' integral that has led me to other discoveries. Maybe a generalization could be obtained with other powers of log such as n = 3, 4, etc.

I wonder if they can also be evaluated in terms of Clausens and then into closed forms involving $$\zeta(n+1)$$ and derivatives of digamma, $$\psi_{n-1}(z)?$$.

Another easier one is $$\int_{0}^{1}\frac{\log(x^{2}-x+1)}{x}dx=\frac{-\pi^{2}}{18}=\frac{-1}{3}\zeta(2)?$$

• I don't see how Cauchy's theorem can help here. I have been able to find that the integral is equal to another integral: $$\int_0^1 dx \frac{\log{\left (1-\frac{x}{4} \right )}}{x \sqrt{1-x}}$$ But again, we can derive a sum from this and given what the result is, it is going to be a matter of manipulating the terms in the sum to match the trigammas. – Ron Gordon Sep 30 '15 at 10:58
• Since you have forgotten more about contour integration than I know, I thought perhaps you knew something I was not aware of . Anyway, may I ask how you arrived at that integrand?. I get a different numerical result. Maybe I messed up. If anyone can make headway, it'd be you RG :) I was going to try and manipulate the Clausen sums and see if I hit upon anything. I have to leave now, but I will probably play with it this evening. – Cody Sep 30 '15 at 12:52
• Hey Ron, I think that integral is equivalent to $\int_{0}^{1}\frac{\log(x^{2}-x+1)}{x}dx=\frac{-\pi^{2}}{18}$. I think you have a typo and need a square on your log as well. Then it is the same. That's a cool derivation. If you get time, please show how you arrived at that . :) – Cody Sep 30 '15 at 13:04

Asset at our disposal: $$\sum\limits_{n=0}^{\infty} \frac{x^{2n+2}}{(n+1)(2n+1)\binom{2n}{n}} = 4(\arcsin (x/2))^2$$

Differentiation followed by the substitution $x \to \sqrt{x}$ gives:

$\displaystyle \sum\limits_{n=0}^{\infty} \frac{x^{n}}{(2n+1)\binom{2n}{n}} = \frac{2\arcsin (\sqrt{x}/2)}{\sqrt{x}\sqrt{1-(\sqrt{x}/2)^2}}$

Thus, we split the series as: $$\sum\limits_{n=0}^{\infty} \frac{H_n}{(n+1)(2n+1)\binom{2n}{n}} \\= \sum\limits_{n=0}^{\infty} \frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}} - \sum\limits_{n=0}^{\infty} \frac{1}{(n+1)^2(2n+1)\binom{2n}{n}}$$

The first series can be dealt with using, $\displaystyle\frac{H_{n+1}}{n+1} = -\int_0^1 x^n\log(1-x)\,dx$

\begin{align*}\sum\limits_{n=0}^{\infty} \frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}}&= -\sum\limits_{n=0}^{\infty} \int_0^1 \frac{x^n\log(1-x)}{(2n+1)\binom{2n}{n}}\,dx\\ &= -2\int_0^1 \frac{\arcsin (\sqrt{x}/2)\log (1-x)}{\sqrt{x}\sqrt{1-(\sqrt{x}/2)^2}}\,dx\\ &= -8\int_0^{1/2} \frac{\arcsin x \cdot \log (1-4x^2)}{\sqrt{1-x^2}}\,dx\\ &= -8\int_0^{\pi/6} \theta \log (1-4\sin^2 \theta)\,d\theta\\ &= -8\int_0^{\pi/6} \theta \log \left(4\sin\left(\theta + \frac{\pi}{6}\right)\sin\left(\frac{\pi}{6}-\theta\right)\right) \end{align*}

Using the Fourier Series, $\displaystyle \log (2\sin \theta) = -\sum\limits_{n=1}^{\infty} \frac{\cos 2n\theta}{n}$ we get:

\begin{align*}&\int_0^{\pi/6} \theta\log \left(2\sin\left(\frac{\pi}{6}+\theta\right)\right)\,d\theta \\&= -\sum\limits_{n=1}^{\infty} \int_0^{\pi/6} \frac{\theta\cos \left(\dfrac{n\pi}{3}+2n\theta\right)}{n}\,d\theta\\&= -\frac{\pi}{12}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2}-\frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{\cos (2n\pi/3)}{n^3} +\frac{1}{4}\sum\limits_{n=1}^{\infty} \frac{\cos (n\pi/3)}{n^3}\end{align*}

and, \begin{align*}&\int_0^{\pi/6} \theta\log \left(2\sin\left(\frac{\pi}{6}-\theta\right)\right)\,d\theta \\&= -\sum\limits_{n=1}^{\infty} \int_0^{\pi/6} \frac{(\pi/6 - \theta)\cos \left(2n\theta\right)}{n}\,d\theta\\&= -\frac{1}{4}\zeta(3)+\frac{1}{4}\sum\limits_{n=1}^{\infty}\frac{\cos (n\pi/3)}{n^3}\end{align*}

Hence, $$\sum\limits_{n=0}^{\infty}\frac{H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}} = -\frac{2}{9}\zeta(3) + \frac{2\pi}{3}\sum\limits_{n=1}^{\infty}\frac{\sin (2n\pi/3)}{n^2}$$

Similarly we may deal with the second series:

\begin{align*}\sum\limits_{n=0}^{\infty} \frac{1}{(n+1)^2(2n+1)\binom{2n}{n}} &= 8\int_0^{1/2} \frac{\arcsin^2 (x)}{x}\,dx \\&= -4\zeta(3)+4\sum\limits_{n=1}^{\infty}\frac{\cos (n\pi/3)}{n^3}+\frac{4\pi}{3}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2}\end{align*}

Combining the results we get:

\begin{align*}\sum\limits_{n=1}^{\infty} \frac{H_n}{(n+1)(2n+1)\binom{2n}{n}} &= \frac{22}{9}\zeta(3) - \frac{2\pi}{3}\sum\limits_{n=1}^{\infty} \frac{\sin (2n\pi/3)}{n^2} \\&= \frac{22}{9}\zeta(3) - \frac{\pi}{9\sqrt{3}}\left(\psi'\left(\frac{1}{3}\right) - \psi'\left(\frac{2}{3}\right)\right)\end{align*}

• Very nice, r9m. Turning it into a log trig integral is the ticket. :) – Cody Oct 24 '15 at 7:08
• @Cody Thanks! couldn't have done it without your help! :-) – r9m Oct 24 '15 at 7:25
• Good job!:D (+1) – user 1357113 Oct 24 '15 at 8:08

The easier one is in fact very much easier: just write \begin{align} \int_0^1\frac{\ln\left(x^2-x+1\right)dx}{x}&=\int_0^1\frac{\ln\left(1+x^3\right)dx}{x}-\int_0^1\frac{\ln\left(1+x\right)dx}{x}=\\ &=\int_0^1\frac{\ln\left(1+x^3\right)d\left(x^3\right)}{3x^3}-\int_0^1\frac{\ln\left(1+x\right)dx}{x}=\\ &=-\frac23\int_0^1\frac{\ln\left(1+x\right)dx}{x}=-\frac23\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_0^1 x^{k-1}dx=\\ &=-\frac23\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}=-\frac{\zeta(2)}{3}=-\frac{\pi^2}{18}. \end{align}

• Nice swp. I derived it a different way by using $$Li_{2}(e^{\pm \pi i/3})$$, but your method is nice and straightforward. – Cody Oct 17 '15 at 8:03

Using the series $$2\sum_{k=1}^\infty\frac{H_kx^k}{k+1}=\frac{\log(1-x)^2}x$$ and the integral $$\int_0^{1/2}\left(\frac14-x^2\right)^k\,\mathrm{d}x=\frac{k!}{2^{k+1}(2k+1)!!}$$ we get \begin{align} \int_0^1\frac{\log\left(x^2-x+1\right)^2}x\,\mathrm{d}x &=\int_{-1/2}^{1/2}\frac{\log\left(x^2+\frac34\right)^2}{x+\frac12}\,\mathrm{d}x\\ &=\frac12\int_{-1/2}^{1/2}\frac{\log\left(x^2+\frac34\right)^2}{\frac14-x^2}\,\mathrm{d}x\\ &=\int_0^{1/2}\frac{\log\left(x^2+\frac34\right)^2}{\frac14-x^2}\,\mathrm{d}x\\ &=\int_0^{1/2}2\sum_{k=1}^\infty\frac{H_k\left(\frac14-x^2\right)^k}{k+1}\,\mathrm{d}x\\ &=\sum_{k=1}^\infty\frac{H_k}{k+1}\frac{k!}{2^k(2k+1)!!}\\ &=\sum_{k=1}^\infty\frac{2H_k}{(k+1)^2\binom{2k+2}{k+1}} \end{align} which converges at over $0.6$ digits per term to $$0.1041096792619493789449118629712286069593$$

• Thanks robjohn. Very insightful. I would not have thought of using that integral second from the top. Very clever, as usual. May I ask, is there some way we can now relate this to $$2/3\zeta(3)-4/3\sum_{n=1}^{\infty}\frac{1}{n^{3}\binom{2n}{n}}$$?. – Cody Oct 17 '15 at 4:35