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

Question

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

Answer 1

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*}

Answer 2

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}

Answer 3

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 $$ 0.1041096792619493789449118629712286069593 $$

Answer 4

http://mathworld.wolfram.com/PolygammaFunction.html $\displaystyle{9}$ $\displaystyle{10}$)

$$\displaystyle{{\psi _1}\left( z \right) + {\psi _1}\left( {1 - z} \right) = \frac{{{\pi ^2}}}{{{{\sin }^2}\left( {\pi z} \right)}} \Rightarrow {\psi _1}\left( {\frac{2}{3}} \right) = \frac{{4{\pi ^2}}}{3} - {\psi _1}\left( {\frac{1}{3}} \right)}$$$$\displaystyle{{\psi _1}\left( {\frac{5}{6}} \right) = 4{\pi ^2} - {\psi _1}\left( {\frac{1}{6}} \right)}$$ $$\displaystyle{{\psi _1}\left( {2z} \right) = \frac{1}{4}\left( {{\psi _1}\left( z \right) + {\psi _1}\left( {z + \frac{1}{2}} \right)} \right) \Rightarrow {\psi _1}\left( {2 \cdot \frac{1}{6}} \right) = \frac{1}{4}\left( {{\psi _1}\left( {\frac{1}{6}} \right) + {\psi _1}\left( {\frac{2}{3}} \right)} \right)}$$$$\displaystyle{{\psi _1}\left( {\frac{1}{6}} \right) = 5{\psi _1}\left( {\frac{1}{3}} \right) - \frac{{4{\pi ^2}}}{3}}$$. $$\displaystyle{{\psi _1}\left( z \right) = \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + z} \right)}^2}}}} }$$Lemma 1 $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\dfrac{{\pi n}}{3}} \right)}}{{{n^2}}}} = \frac{1}{{2\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) - \frac{1}{{3\sqrt 3 }}{\pi ^2}}$$ $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\dfrac{{\pi n}}{3}} \right)}}{{{n^2}}}} = \sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\dfrac{{\pi \left( {6n - 1} \right)}}{3}} \right)}}{{{{\left( {6n - 1} \right)}^2}}}} + \sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\dfrac{{\pi \left( {6n - 2} \right)}}{3}} \right)}}{{{{\left( {6n - 2} \right)}^2}}}} +}$$$$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\dfrac{{\pi \left( {6n - 4} \right)}}{3}} \right)}}{{{{\left( {6n - 4} \right)}^2}}}} + \sum\limits_{n = 1}^\infty {\frac{{\sin \left( {\dfrac{{\pi \left( {6n - 5} \right)}}{3}} \right)}}{{{{\left( {6n - 5} \right)}^2}}}} = }$$ $$\displaystyle{ = \sum\limits_{n = 1}^\infty {\frac{{\sin \left( { - \dfrac{\pi }{3}} \right)}}{{{{\left( {6n - 1} \right)}^2}}} + \sum\limits_{n = 1}^\infty {\frac{{\sin \left( { - \dfrac{{2\pi }}{3}} \right)}}{{{{\left( {6n - 2} \right)}^2}}}} + \sum\limits_{n = 1}^\infty {\frac{{\sin \left( { - \dfrac{{4\pi }}{3}} \right)}}{{{{\left( {6n - 4} \right)}^2}}}} } + \sum\limits_{n = 1}^\infty {\frac{{\sin \left( { - \dfrac{{5\pi }}{3}} \right)}}{{{{\left( {6n - 5} \right)}^2}}}} = }$$ $$\displaystyle{ = \frac{{\sqrt 3 }}{{72}}\left( { - \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 5/6} \right)}^2}}}} - \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 4/6} \right)}^2}}}} + \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 2/6} \right)}^2}}}} + \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( {n + 1/6} \right)}^2}}}} } \right) = }$$ $$\displaystyle{ = \frac{1}{{24\sqrt 3 }}\left( {{\psi _1}\left( {\frac{1}{6}} \right) - {\psi _1}\left( {\frac{5}{6}} \right) + {\psi _1}\left( {\frac{1}{3}} \right) - {\psi _1}\left( {\frac{2}{3}} \right)} \right) = .. = \frac{1}{{2\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) - \frac{1}{{3\sqrt 3 }}{\pi ^2}}$$ Lemmma 2: $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n\pi /3} \right)}}{{{n^3}}}} = \frac{1}{3}\zeta \left( 3 \right)}$$ $$\displaystyle{I = \int\limits_0^1 {\frac{{{{\log }^2}\left( {{x^2} - x + 1} \right)}}{x}dx} = \int\limits_0^1 {\frac{{{{\log }^2}\left( {{x^2} - x + 1} \right)}}{{1 - x}}dx} \Rightarrow }$$$$\displaystyle{I = \frac{1}{2}\int\limits_0^1 {\frac{{{{\log }^2}\left( {{x^2} - x + 1} \right)}}{{x\left( {1 - x} \right)}}dx} = \frac{1}{2}\int\limits_0^1 {\frac{{{{\log }^2}\left( {1 - x\left( {1 - x} \right)} \right)}}{{x\left( {1 - x} \right)}}dx} = }$$ $$\displaystyle{ = \int\limits_0^1 {\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n + 1}}{x^n}{{\left( {1 - x} \right)}^n}} dx} = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n + 1}}B\left( {n + 1,n + 1} \right)} = }$$$$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{n + 1}} \cdot \frac{{\Gamma \left( {n + 1} \right)\Gamma \left( {n + 1} \right)}}{{\Gamma \left( {2n + 2} \right)}}} = 2\sum\limits_{n = 1}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} }$$ $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} }$$ $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^2}\left( {\begin{array}{*{20}{c}} {2n}\\ n \end{array}} \right)}}} = - \frac{{\zeta \left( 3 \right)}}{9} + \frac{\pi }{{18\sqrt 3 }}\left( {{\psi _1}\left( {\frac{1}{3}} \right) - {\psi _1}\left( {\frac{2}{3}} \right)} \right)}$$ $\displaystyle{{\psi _1}}$ $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{{n^2}\left( {2n} \right)!}}} = \frac{\pi }{{9\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) - \frac{{2{\pi ^3}}}{{27\sqrt 3 }} - \frac{{\zeta \left( 3 \right)}}{9}}$$ $$\displaystyle{\sum\limits_{n = 1}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{{n^2}\left( {2n} \right)!}}} = \frac{1}{2} + \sum\limits_{n = 2}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{{n^2}\left( {2n} \right)!}}} = \frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{{{H_{n + 1}}{{\left( {\left( {n + 1} \right)!} \right)}^2}}}{{{{\left( {n + 1} \right)}^2}\left( {2n + 2} \right)!}}} = \frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{{{H_{n + 1}}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} = }$$ $$\displaystyle{ = \frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{{\left( {{H_n} + \frac{1}{{n + 1}}} \right){{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} = \frac{1}{2} + \sum\limits_{n = 1}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} + \sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}}}{{\left( {n + 1} \right)\left( {2n + 2} \right)!}}} \Rightarrow }$$ $$\displaystyle{ \Rightarrow \sum\limits_{n = 1}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} = \sum\limits_{n = 1}^\infty {\frac{{{H_n}}}{{{n^2}\left( {\begin{array}{*{20}{c}} {2n}\\ n \end{array}} \right)}}} - \frac{1}{2} - \sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}}}{{\left( {n + 1} \right)\left( {2n + 2} \right)!}}} }$$ $$\displaystyle{I = \int\limits_0^1 {\frac{{{{\log }^2}\left( {{x^2} - x + 1} \right)}}{x}dx} = 2\sum\limits_{n = 1}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} = 2\sum\limits_{n = 1}^\infty {\frac{{{H_n}{{\left( {n!} \right)}^2}}}{{{n^2} \cdot \left( {2n} \right)!}}} - 1 - 2\sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}}}{{\left( {n + 1} \right)\left( {2n + 2} \right)!}}} \Rightarrow }$$ $$\displaystyle{ \Rightarrow I = \frac{{2\pi }}{{9\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) - \frac{{4{\pi ^3}}}{{27\sqrt 3 }} - \frac{{2\zeta \left( 3 \right)}}{9} - 1 - 2\underbrace {\sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}}}{{\left( {n + 1} \right)\left( {2n + 2} \right)!}}} }_{{S_1}}}$$$$\displaystyle{{S_1} = \sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}}}{{\left( {n + 1} \right)\left( {2n + 2} \right)!}}} = \int\limits_0^1 {\underbrace {\sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}{x^n}}}{{\left( {2n + 2} \right)!}}} }_{{S_2}}\;dx} }$$ $$\displaystyle{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}{z^{2n + 2}}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} = \frac{1}{2}{\arcsin ^2}z}$$https://en.wikipedia.org/wiki/List_of_mathematical_series

$$\displaystyle{\sum\limits_{n = 0}^\infty {\frac{{{z^{2n}}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} = \frac{{2{{\arcsin }^2}\left( {\frac{z}{2}} \right)}}{{{z^2}}} \Rightarrow \sum\limits_{n = 0}^\infty {\frac{{{x^n}{{\left( {n!} \right)}^2}}}{{\left( {2n + 2} \right)!}}} = \frac{{2{{\arcsin }^2}\left( {\frac{{\sqrt x }}{2}} \right)}}{x}}$$, $$\displaystyle{{S_2} = \frac{{2{{\arcsin }^2}\left( {\frac{{\sqrt x }}{2}} \right)}}{x} - \frac{1}{2}}$$ $$\displaystyle{{S_1} = \int\limits_0^1 {\sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}{x^n}}}{{\left( {2n + 2} \right)!}}} dx} = - \frac{1}{2} + 2\int\limits_0^1 {\frac{{{{\arcsin }^2}\left( {\frac{{\sqrt x }}{2}} \right)}}{x}dx} = \mathop = \limits^{x = 4{u^2}} = - \frac{1}{2} + 4\int\limits_0^{1/2} {\frac{{{{\arcsin }^2}u}}{u}du} = \mathop = \limits^{\arcsin u = x} = }$$ $$\displaystyle{ = - \frac{1}{2} + 4\int\limits_0^{\pi /6} {\frac{{{x^2}}}{{\sin x}}\cos x\;dx} = - \frac{1}{2} + 4\int\limits_0^{\pi /6} {{x^2}\log \left( {\sin x} \right)'dx} = }$$$$\displaystyle{ - \frac{1}{2} + 4\left[ {{x^2}\log \left( {\sin x} \right)} \right]_0^{\pi /6} - 8\int\limits_0^{\pi /6} {x\log \left( {\sin x} \right)dx} = }$$ $$\displaystyle{ = - \frac{1}{2} - \frac{{{\pi ^2}\log 2}}{9} - 8\int\limits_0^{\pi /6} {x\log \left( {\sin x} \right)dx} = - \frac{1}{2} - \frac{{{\pi ^2}\log 2}}{9} + 8\int\limits_0^{\pi /6} {x\left( {\log 2 + \sum\limits_{n = 1}^\infty {\frac{{\cos 2nx}}{n}} } \right)dx} = }$$ $$\displaystyle{ = - \frac{1}{2} + 8\sum\limits_{n = 1}^\infty {\frac{1}{n}\int\limits_0^{\pi /6} {x\cos 2nx\;dx} } = - \frac{1}{2} + 2\sum\limits_{n = 1}^\infty {\left( {\frac{{ - 1}}{{{n^3}}} + \frac{{\cos \left( {n\pi /3} \right)}}{{{n^3}}} + \pi \frac{{\sin \left( {n\pi /3} \right)}}{{3{n^2}}}} \right)} = }$$ $$\displaystyle{ = - \frac{1}{2} - 2\zeta \left( 3 \right) + 2\sum\limits_{n = 1}^\infty {\frac{{\cos \left( {n\pi /3} \right)}}{{{n^3}}}} + \frac{{2\pi }}{3}\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\pi /3} \right)}}{{{n^2}}}} = - \frac{1}{2} - \frac{4}{3}\zeta \left( 3 \right) + \frac{{2\pi }}{3}\sum\limits_{n = 1}^\infty {\frac{{\sin \left( {n\pi /3} \right)}}{{{n^2}}}} }$$ $$\displaystyle{ = - \frac{1}{2} - \frac{4}{3}\zeta \left( 3 \right) + \frac{\pi }{{3\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) - \frac{{2{\pi ^3}}}{{9\sqrt 3 }}}$$ $$\displaystyle{I = \frac{{2\pi }}{{9\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) - \frac{{4{\pi ^3}}}{{27\sqrt 3 }} - \frac{{2\zeta \left( 3 \right)}}{9} - 1 - 2\underbrace {\sum\limits_{n = 1}^\infty {\frac{{{{\left( {n!} \right)}^2}}}{{\left( {n + 1} \right)\left( {2n + 2} \right)!}}} }_{{S_1}} = }$$ $$\displaystyle{ = \frac{{2\pi }}{{9\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) - \frac{{4{\pi ^3}}}{{27\sqrt 3 }} - \frac{{2\zeta \left( 3 \right)}}{9} - 1 - 2\left( { - \frac{1}{2} - \frac{4}{3}\zeta \left( 3 \right) + \frac{\pi }{{3\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) - \frac{{2{\pi ^3}}}{{9\sqrt 3 }}} \right) = }$$ $$\displaystyle{ = - \frac{{4\pi }}{{9\sqrt 3 }}{\psi _1}\left( {\frac{1}{3}} \right) + \frac{{22}}{9}\zeta \left( 3 \right) + \frac{{8{\pi ^3}}}{{27\sqrt 3 }}}$$$$\displaystyle{ = - \frac{{4\sqrt 3 \pi }}{{27}}{\psi _1}\left( {\frac{1}{3}} \right) + \frac{{22}}{9}\zeta \left( 3 \right) + \frac{{8\sqrt 3 {\pi ^3}}}{{81}}}$$