Generalized hypergeometric functions XI

Nice hypergeometric series involving the silver ratio $\delta_S$

Today we show the proof of this result involving hypergeometric series and the silver ratio $\delta_S= 1+\sqrt{2}$ proposed by @mamekebi $${}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2}$$ The proof relies on some properties of the rising factorial and some identities of the dilogarithm function.

Proof

Recall the recursion formula for the argument and the degree of the rising factorial: $$(x)_{n+1} = (n+x)(x)_{n} = x(x+1)_{n}\quad \tag{1}$$ and the duplication formula $$(x)_{2n} = 2^{2n} \left(\frac{x}{2}\right)_n\left(\frac{x+1}{2}\right)_n \tag{2}$$ Hence $$\begin{align*} {}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = &\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}\left(\frac{1}{2}\right)_{j}}{\left(\frac{3}{2}\right)_{j}\left(\frac{3}{2}\right)_{j}} \frac{(-1)^j}{j!}\\ =&\frac{1}{2}\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}\left(j+\frac{1}{2}\right)} \frac{(-1)^j}{j!} \quad \textrm{from (1)}\\ =&\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}\left(2j+1\right)} \frac{(-1)^j}{j!}\\ \end{align*}$$ We previously showed the proof of this series expansion: $$\begin{align*}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} =& \sum_{j=0}^{\infty} \frac{2^{2j}}{\binom{2j}{j}(2j+1)}(-1)^jx^{2j} = \sum_{j=0}^{\infty} \frac{2^{2j}j!^2}{(2j)!(2j+1)}(-1)^jx^{2j} \\ =& \sum_{j=0}^{\infty} \frac{2^{2j}(1)_{j}(1)_{j}}{(1)_{2j}(2j+1)}(-1)^jx^{2j} \quad \left( (1)_{j} = j!\right)\\ =& \sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}}\frac{(-1)^jx^{2j}}{j!} \quad \textrm{ from (1) and (2)} \end{align*}$$ Integrating from $0$ to $1$ $$\int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}(2j+1)}\frac{(-1)^j}{j!} ={}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right]$$ $$\begin{align*} I = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = &\int_{0}^{1}\frac{\ln\left(\sqrt{x^2+1}+x\right)}{x\sqrt{1+x^2}} dx\\ =& 2\int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{(w+1)(w-1)}dw \quad \left(w \mapsto \sqrt{x^2+1}+x\right) \\ =& -\int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{1-w}dw - \int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{w+1}dw\\ =& \int_{0}^{-\sqrt{2}}\frac{\ln(1-t)}{t}dw - \int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{w+1}dw \quad \left(t\mapsto 1-w\right) \\ \stackrel{IBP}{=} & \int_{0}^{-\sqrt{2}}\frac{\ln(1-t)}{t}dw - \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\int_{-1-\sqrt{2}}^{-1}\frac{\ln(1-w)}{w}dw\\ =& -\int_{-\sqrt{2}}^{0}\frac{\ln(1-t)}{t}dw - \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\int_{-1-\sqrt{2}}^{0}\frac{\ln(1-w)}{w}dw+\int_{-1}^{0}\frac{\ln(1-w)}{w}dw\\ =& -\operatorname{Li}_{2}(-\sqrt{2})- \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\operatorname{Li}_{2}(-1-\sqrt{2})+\operatorname{Li}_{2}(-1) \end{align*}$$ From the identity $$\operatorname{Li}_2(-z)-\operatorname{Li}_2(1-z)+\frac{1}{2}\operatorname{Li}_2(1-z^2)=-\frac{{\pi}^2}{12}-\ln z \cdot \ln(z+1)$$ If we put $z = \sqrt{2}$ $$\operatorname{Li}_2(-\sqrt{2})=\operatorname{Li}_2(1-\sqrt{2})-\frac{1}{2}\operatorname{Li}_2(-1)-\frac{{\pi}^2}{12}-\ln \sqrt{2} \cdot \ln(\sqrt{2}+1) \quad \quad (3)$$ From the identity $$\operatorname{Li}_2(z) +\operatorname{Li}_2\left(\frac{1}{z}\right) = - \frac{\pi^2}{6} - \frac{\ln^2(-z)}{2}$$ If we put $z = 1-\sqrt{2}$

Therefore $$\operatorname{Li}_2\left(-1-\sqrt{2}\right) = -\operatorname{Li}_2(1-\sqrt{2}) - \frac{\pi^2}{6} - \frac{\ln^2(\sqrt{2}-1)}{2} \quad \quad (4)$$ Using the fact that $$\operatorname{Li}_{2}(-1)= -\frac{\pi^2}{12}$$ and from $(3)$ and $(4)$ we have: $$\begin{align*} I =& \frac{\pi^2}{8} +\ln \sqrt{2} \cdot \ln(\sqrt{2}+1)- \ln(1+\sqrt{2})\ln(2+\sqrt{2}) + \frac{\ln^2(\sqrt{2}-1)}{2}\\ =& \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2} \end{align*}$$ $$\boxed{ {}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2}}$$

Generalized hypergeometric functions X

Series involving the binomial coefficient

Today we show the proof of this series proposed by @qq3023625451 $$\sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 = \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)}$$ In the proof we use some properties of the generalized hypergeometric function ${}_pF_{q}$

Proof:

The Pocchhammer symbol (rising factorial) is defined as the product: $$(x)_{n} = x(x+1)\cdots(x+n-1) = \prod_{j=0}^{n-1}(x+j)$$ while the generalized hypergeometric function is defined $${}_pF_{q} \left[{a_{1},a_2,...,a_{p}\atop b_{1},b_{2},...,b_{q}}; z \right] = \sum_{n=0}^{\infty} \frac{(a_{1})_{n}(a_{2})_{n}\cdots (a_{p})_n}{(b_{1})_{n}(b_{2})_{n}\cdots (b_{q})_{n} } \frac{z^n}{n!}$$

Note that $(1)_{n} = n!$

The rising factorial obey the duplication formula: $$(x)_{2n} = 2^{2n} \left(\frac{x}{2}\right)_{n} \left(\frac{1+x}{2}\right)_{n} \tag{1}$$ and the recursion formula for the argument $$(n+x)(x)_{n} = x(x+1)_{n} \tag{2}$$ Hence $$\begin{align*} S = \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)(2n)!^5}{ 2^{10n}n!^{10}}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)(1)_{2n}^5}{ 2^{10n}(1)_{n}^{10}}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)2^{10n}\left(\frac{1}{2}\right)_{n}^{5}(1)_{n}^{5}}{ 2^{10n}(1)_{n}^{10}}\quad \textrm{from (1)}\\ =& \sum_{n=0}^{\infty} \frac{(-1)^n4\left(n+\frac{1}{4}\right)\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n}{ (1)_n(1)_n(1)_n(1)_nn!}\\ =& \sum_{n=0}^{\infty} \frac{\left(\frac{5}{4}\right)_{n}\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n\left(\frac{1}{2}\right)_n}{\left(\frac{1}{4}\right)_{n} (1)_n(1)_n(1)_n(1)_n} \frac{(-1)^n}{n!} \quad \textrm{from (2)}\\ =& {}_{6}F_{5} \left[ {\frac{1}{2}, \frac{5}{4}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \atop \frac{1}{4},1,1,1,1};-1 \right] \end{align*}$$ One of the Whipple's identities states that $${}_{6}F_{5} \left[{a, 1+\frac{1}{2}a , b,c,d,e \atop \frac{1}{2}a, 1+a-b, 1+a-c, 1+a-d, 1+a-e }; -1\right] = \frac{\Gamma(1-a-d)\Gamma(1+a-e)}{\Gamma(1+a)\Gamma(1+a-d-e)} {}_{3}F_{2}\left[ {1+a-b-c,d,e \atop 1+a-b,1+a-c}; 1\right]$$ Hence $$S = {}_{6}F_{5} \left[ {\frac{1}{2}, \frac{5}{4}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \atop \frac{1}{4},1,1,1,1};-1 \right]= \frac{1}{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)} {}_{3}F_{2} \left[{\frac{1}{2},\frac{1}{2},\frac{1}{2} \atop 1, 1}; 1\right]$$ ${}_{3}F_{2}$ obey the Dixon's well poised sum: $${{}_{3}F_{2}}\left[{a,b,c\atop a-b+1,a-c+1};1\right]=\frac{\Gamma\left(\frac{1% }{2}a+1\right)\Gamma\left(a-b+1\right)\Gamma\left(a-c+1\right)\Gamma\left(% \frac{1}{2}a-b-c+1\right)}{\Gamma\left(a+1\right)\Gamma\left(\frac{1}{2}a-b+1% \right)\Gamma\left(\frac{1}{2}a-c+1\right)\Gamma\left(a-b-c+1\right)},$$ Hence, using $\displaystyle \Gamma\left(\frac{3}{2}\right) = \frac{\sqrt{\pi}}{2}$ and $\displaystyle \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$ $$\begin{align*} S = \frac{1}{\Gamma\left(\frac{3}{2}\right) \Gamma\left(\frac{1}{2}\right)} {}_{3}F_{2} \left[{\frac{1}{2},\frac{1}{2},\frac{1}{2} \atop 1, 1}; 1\right] =& \frac{\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)} \\ =& \frac{4\Gamma\left(\frac{5}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\pi^2 \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \\ =& \frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{1}{4}\right)}{\pi^2 \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \quad \left(\Gamma(z+1) = z\Gamma(z) \right) \\ =& \frac{2}{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{3}{4}\right)} \quad \left( \Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin\left(\pi z\right)}\right)\\ =& \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)} \end{align*}$$ $$\boxed{\sum_{n=0}^{\infty} \frac{(-1)^n(1+4n)}{2^{10n}}{\binom{2n}{n}}^5 = \frac{2}{\Gamma^{4}\left(\frac{3}{4}\right)}}$$

Integral of the day XXXIV

Integral involving the golden mean $\phi$

Today we show the proof of the following integral posted by @integralsbot $$\int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx = 2\pi\ln^2(\phi) + \frac{\pi^3}{15}$$ The proof relies on a "discrete" Laplace transform a technique that we have used in the past here and here

Proof:

Consider the following discrete Laplace transform (proof in Appendix 1): $$1+ 2\sum_{k=1}^{\infty} e^{-kt}\cos(kx) = \frac{\sinh t}{\cosh t -\cos x } \quad t>0$$ If $\cosh t = \frac{\sqrt{5}}{2}$ then $$t = \ln\left(\frac{1+\sqrt{5}}{2} \right) = \ln(\phi)$$ $$\sinh\left(\frac{1+\sqrt{5}}{2} \right) = \frac{1}{2}$$ therefore $$\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}-\cos x} = 1+ 2\sum_{k=1}^{\infty} \phi^{-k}\cos(kx)$$ where $\displaystyle \phi = \frac{1+\sqrt{5}}{2}$

Hence $$\begin{align*} I = \int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx =& \frac{1}{2} \int_0^\pi \frac{x^2}{\frac{\sqrt{5}}{2}-\cos x} dx \\ =& \int_0^\pi \left(\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}-\cos x}\right) x^2 dx\\ =& \int_0^\pi \left(1+ 2\sum_{k=1}^{\infty} e^{-k\ln(\phi)}\cos(kx)\right) x^2 dx\\ =& \int_{0}^{\pi }\left[ x^2 + 2\sum_{k=1}^{\infty} \phi^{-k} \cos(kx)x^2 \right] dx \\ =& \int_{0}^{\pi } \left[x^2 dx + 2\sum_{k=1}^{\infty} \phi^{-k} \int _{0}^{\pi} \cos(kx)x^2 \right] dx \\ =& \int_{0}^{\pi } \left[x^2 dx + 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3} \int _{0}^{k\pi} \cos(w)w^2\right] dx \\ \stackrel{IBP}{=} &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ w^2\sin(w)\Big|_{0}^{k\pi} -2\int_{0}^{k\pi} \sin(w)w dw \right] \\ \stackrel{IBP}{=} &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ (k\pi)^2\sin(k \pi ) +2\cos(w)w\Big|_{0}^{k\pi} +2 \int_{0}^{\pi k} \cos(w) dw \right] \\ = &\frac{\pi^3}{3}+ 2\sum_{k=1}^{\infty} \frac{\phi^{-k}}{k^3}\left[ (k\pi)^2\sin(k \pi ) +2\cos(k \pi )k\pi -2\sin(\pi k)\right] \\ =& \frac{\pi^3}{3} + \underbrace{2\sum_{k=1}^{\infty} \phi^{-k}\frac{(k^2\pi^2-2)\sin(k\pi)}{k^3}}_{\sin(\pi k) = 0 } + \underbrace{4\sum_{k=1}^{\infty} \phi^{-k}\pi \frac{\cos(k \pi)}{k^2} }_{\cos(\pi k) =(-1)^k}\\ =& \frac{\pi^3}{3} + 4\pi \sum_{k=1}^{\infty} \frac{(-\phi^{-1})^k}{k^2} \\ =& \frac{\pi^3}{3} + 4\pi \operatorname{Li}_{2}(-\phi^{-1}) \\ \end{align*}$$ where we used the series expansion for the dilogarithm function $\operatorname{Li}_{2}(z)$: $$\operatorname{Li}_{2}(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2}$$ Landen and Euler proved (proof in the Appendix 2): $$\operatorname{Li}_{2}\left(-\phi^{-1} \right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2(\phi)$$ Therefore $$\boxed{ \int_0^\pi \frac{x^2}{\sqrt{5}-2\cos x} dx = 2\pi\ln^2(\phi) + \frac{\pi^3}{15} }$$

Appendix 1

$$1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx = \frac{\sinh t}{\cosh t - \cos x} \quad { t>0 }$$ Proof: $$\begin{align*} 1+2\sum_{k=1}^{\infty} e^{-kt}\cos kx =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{-kt}e^{ikx} \right)\\ =& 1+2\Re\left(\sum_{k=1}^{\infty}e^{k(ix-t)} \right)\\ =& 1+2\Re\left(\frac{1}{e^{t-ix}-1} \right)\\ =& 1+2\Re\left(\frac{e^{-t}}{e^{-ix}-e^{-t}} \right)\\ =& 1+2e^{-t}\Re\left(\frac{1}{\cos x-i\sin x- e^{-t}}\right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{(\cos x-e^{-t})^2 + \sin^2 x} \right)\\ =& 1+2e^{-t}\Re\left(\frac{\cos x-e^{-t} - i\sin x}{e^{-2t}-2e^{t}\cos x + 1} \right)\\ =& 1+\Re\left(\frac{\cos x -e^{-t}+ i\sin x}{\cosh t-\cos x} \right)\\ =& \frac{\cosh t- \cos x}{\cosh t- \cos x}+\frac{\cos x-e^{-t}}{(\cosh t- \cos x )} \\ =& \frac{\sinh t}{\cosh t - \cos x} \end{align*}$$

Appendix 2

$$\operatorname{Li}_{2}\left(-\phi^{-1} \right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2(\phi)$$

Proof:

Consider the following derivative: $$\frac{d}{dx} \operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) = = -\left[ \ln\left(1+ \frac{x}{1-x}\right)\right]\left(\frac{1}{x} + \frac{1}{1-x} \right)= \ln(1-x)\left(\frac{1}{x} + \frac{1}{1-x} \right)$$ Integrating $$\operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) = -\operatorname{Li}_{2}(x) - \frac{1}{2} \ln^2(1-x)$$ $$\Longrightarrow \operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) + \operatorname{Li}_{2}(x) = - \frac{1}{2} \ln^2(1-x) \tag{1}$$ If we put $\displaystyle x = \frac{3-\sqrt{5}}{2}$ $$\Longrightarrow \operatorname{Li}_{2}\left(\frac{1-\sqrt{5}}{2} \right) + \operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) = - \frac{1}{2} \ln^2\left(\frac{1+\sqrt{5}}{2}\right) \tag{2}$$ Now we will show that $$\frac{1}{2} \operatorname{Li}_{2}(x^2) = \operatorname{Li}_{2}(x) +\operatorname{Li}_{2}(-x) \tag{3}$$ From the factorization $$(1-x^r) = (1-x)(1-wx)(1-w^2x)\cdots(1-w^{r-1}x)$$ where $\displaystyle w = e^{\frac{i2\pi}{r}}$ Taking logarithms and dividing by $x$ both sides $$\frac{\ln(1-x^r)}{x} = \frac{\ln(1-x)}{x} + \frac{\ln(1-wx)}{x} + \cdots +\frac{\ln(1-w^{r-1}x)}{x}$$ Integrating $$\frac{1}{r}\operatorname{Li}_{2}(x^r) = \operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(xw) + \cdots + \operatorname{Li}_{2}(w^{r-1}x) + C$$ If $x \to 0$ then $C\to 0$ $$\frac{1}{r}\operatorname{Li}_{2}(x^r) = \operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(xw) + \cdots + \operatorname{Li}_{2}(w^{r-1}x)$$ If we put $r=2$ $$\frac{1}{2} \operatorname{Li}_{2}(x^2) = \operatorname{Li}_{2}(x) +\operatorname{Li}_{2}(-x) \tag{3}$$ From (1) and (3) we have $$\operatorname{Li}_{2}\left(\frac{-x}{1-x}\right) + \frac{1}{2}\operatorname{Li}_{2}(x) - \operatorname{Li}_2(-x) = -\frac{1}{2}\ln^2(1-x) \tag{4}$$ If we solve the equation $$\frac{-x}{1-x} = x^2 \Longrightarrow x^2-x+1=0 \Longrightarrow x = \frac{1-\sqrt{5}}{2}$$ Hence from (4) we have $$\frac{3}{2}\operatorname{Li}_{2} \left(\frac{3-\sqrt{5}}{2}\right)- \operatorname{Li}_{2}\left(\frac{\sqrt{5}-1}{2}\right) = -\frac{1}{2}\ln^2\left(\frac{1+\sqrt{5}}{2}\right) \tag{5}$$ If we put $\displaystyle x = \frac{3-\sqrt{5}}{2}$ in the relation (Proof in the Appendix 2 here) $$\operatorname{Li}_{2}(x) + \operatorname{Li}_{2}(1-x) = \frac{\pi^2}{6} -\ln(x)\ln(1-x) \tag{6}$$ we have $$\operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) + \operatorname{Li}_{2}\left(\frac{\sqrt{5}-1}{2}\right) = \frac{\pi^2}{6} -\ln\left(\frac{3-\sqrt{5}}{2}\right)\ln\left(\frac{\sqrt{5}-1}{2}\right) \tag{7}$$ Solving the system given by (5) and (7) we have $$\operatorname{Li}_{2}\left(\frac{3-\sqrt{5}}{2}\right) = \frac{\pi^2}{15} - \frac{1}{4}\ln^2\left(\frac{3-\sqrt{5}}{2}\right) \tag{8}$$ Then from (2) and (8) we get $$\operatorname{Li}_{2}\left(\frac{1-\sqrt{5}}{2}\right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2\left(\frac{\sqrt{5}-1}{2}\right)$$ which is basically the same as $$\operatorname{Li}_{2}\left(-\phi^{-1}\right) = -\frac{\pi^2}{15} +\frac{1}{2}\ln^2\left(\phi\right)$$

Integral of the day XXXIII

An integral representation of the dilogarithm function

Today we show the proof of this integral posted by @Ali39342137 $$\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \frac{\pi^3}{12} -\frac{\pi}{2}\ln^2\left(2\right)$$ In general we found $$\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \quad |a|\geq1$$ In the proof we make use of Fourier series and some properties of the dilogarithm function $\operatorname{Li}_{2}(z)$.

Proof:

Consider the following Fourier series (Proof in the Appendix 1): $$\sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} = \arctan\left(\frac{p\sin x}{1-p\cos x}\right) \quad 0\lt x\lt 2\pi, \quad p^2\leq 1$$ Then, if $p = -q$ $$\sum_{k=1}^{\infty } \frac{(-1)^kq^{k}\sin(kx)}{k} = \arctan\left(\frac{-q\sin x}{1+q\cos x}\right) = - \arctan\left(\frac{q\sin x}{1+q\cos x}\right)$$ Therefore $$\arctan\left(\frac{\sin x}{\frac{1}{q}+\cos x}\right) = \sum_{k=1}^{\infty } \frac{(-1)^{k+1}q^{k}\sin(kx)}{k} \quad \frac{1}{q}\geq 1$$ Hence $$\begin{align*} I = \int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx =& \int_{0}^{\pi} x \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}\sin(kx)}{k}dx\\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k} \int_{0}^{\pi} x\sin(kx)dx\\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \int_{0}^{k\pi} w\sin(w)dx \quad (w \mapsto kx)\\ \stackrel{IBP}{=}& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \left[-\cos(x)x\Big|_{0}^{k\pi}+ \int_{0}^{k\pi}\cos(w)dx \right] \\ =& \sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}}{k^3} \left[-k\pi\cos(\pi k) + \sin(\pi k) \right] \\ =& \pi \underbrace{\sum_{k=1}^{\infty } \frac{(-1)^{k}\left(\frac{1}{a}\right)^{k}\cos(k\pi) }{k^2}}_{\cos(k \pi) = (-1)^k} + \underbrace{\sum_{k=1}^{\infty } \frac{(-1)^{k+1}\left(\frac{1}{a}\right)^{k}\sin(\pi k) }{k^3}}_{\sin(k\pi)=0} \\ =& \pi \sum_{k=1}^{\infty } \frac{\left(\frac{1}{a}\right)^{k} }{k^2} \\ =& \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \end{align*}$$ If $a = 2$ $$\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{2}\right)$$ From the relation (proof in the Appendix 2) $$\operatorname{Li}_{2}(z) +\operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6} -\ln(z)\ln(1-z)$$ if we put $\displaystyle z = \frac{1}{2}$ $$\operatorname{Li}_{2}\left(\frac{1}{2} \right) = \frac{\pi^2}{12} -\frac{1}{2}\ln^2\left(2\right)$$ Therefore $$\boxed{\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{2+\cos x}\right) dx = \frac{\pi^3}{12} -\frac{\pi}{2}\ln^2\left(2\right)}$$ $$\boxed{\int_{0}^{\pi} x\arctan\left(\frac{\sin x}{a+\cos x}\right) dx = \pi \operatorname{Li}_{2}\left(\frac{1}{a}\right) \quad |a|\geq1 }$$

Appendix 1 $$\sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} = \arctan\left(\frac{p\sin x}{1-p\cos x}\right) \quad 0\lt x\lt 2\pi, \quad p^2\leq 1$$

Proof

Take the principal branch of the $\ln(z)$ function. Hence $$\begin{align*} \sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} =& \Im\left( \sum_{k=1}^{\infty } \frac{(pe^{ix})^k}{k} \right) = \Im\left( -\ln(1-pe^{ix})\right)\\ =& -\arg\left(1-pe^{ix}\right)\\ =& -\arg\left(1-p\cos(x)-pi\sin(x)\right)\\ \end{align*}$$ Since $|p|\lt 1$ then $1-p\cos(x)>0$ therefore $$\begin{align*} \sum_{k=1}^{\infty } \frac{p^{k}\sin(kx)}{k} =& -\arg\left(1-p\cos(x)-pi\sin(x)\right)\\ =& -\arctan\left( \frac{-p\sin x}{1-p\cos(x)}\right) \\ =& \arctan\left( \frac{p\sin x}{1-p\cos(x)}\right) \quad \left(\arctan \textrm{ is odd } \right) \end{align*}$$

Appendix 2 $$\operatorname{Li}_{2}(z) +\operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6} -\ln(z)\ln(1-z)$$

Proof $$\begin{align*} \operatorname{Li}_{2}(z) = &\int_{z}^{0} \frac{\ln(1-t)}{t} dt \stackrel{IBP}{=} -\ln(z)\ln(1-z) -\int_{0}^{z}\frac{\ln (t)}{1-t}dt\\ =& -\ln(z)\ln(1-z) + \int_{1}^{1-z}\frac{\ln (1-w)}{w}dw \quad \left( w\mapsto 1-t \right)\\ =& -\ln(z)\ln(1-z) + \int_{1}^{0}\frac{\ln (1-w)}{w}dw + \int_{0}^{1-z}\frac{\ln (1-w)}{w}dw\\ =& -\ln(z)\ln(1-z) + \int_{1}^{0}\frac{\ln (1-w)}{w}dw - \int_{1-z}^{0}\frac{\ln (1-w)}{w}dw\\ =& -\ln(z)\ln(1-z) + \operatorname{Li}_{2}(1) - \operatorname{Li}_{2}(1-z)\\ \end{align*}$$ Therefore $$\operatorname{Li}_{2}(z) + \operatorname{Li}_{2}(1-z) = -\ln(z)\ln(1-z) + \zeta(2) = -\ln(z)\ln(1-z) + \frac{\pi^2}{6}$$

Application of Ramanujan's master theorem II

A useful Mellin transform related to the Bernoulli and zeta numbers

Today we show the proof of this Mellin transform posted by @integralsbot $$\int_{0}^{\infty} x^{s-n-2} \left(\frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\right) dx =\Gamma(s-n)\zeta(s-n) \quad 0\lt \Re(s)\lt1$$ The proof relies on the Ramanujan's master theorem, some properties of the Bernoulli numbers and the binomial coefficient.

We also show the proof of this nice application: an integral representation for the Apery's constant (also posted by @integralsbot ) : $$\int_0^{\infty} \left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)\frac{1}{x^4} dx = \frac {\zeta (3)} { 8\pi^2}$$

Proof:

We will apply the Ramanujan master theorem: If a complex-valued function $f(x)$ has an expansion of the form $$f(x)=\sum_{k=0}^\infty \frac{\,\varphi(k)\,}{k!}(-x)^k$$ then the Mellin transform of $f(x)$ is given by $$\int_0^\infty x^{s-1} f(x) dx = \Gamma(s)\,\varphi(-s)$$ where $\Gamma(s)$ is the gamma function.

The Bernoulli numbers are defined through the generating function $$\frac{x}{e^x -1} = \sum_{k=0}^{\infty} \frac{ B_{k}x^k}{k!}$$ Therefore $$\begin{align*} \frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!} = & \sum_{k=0}^{\infty} \frac{B_{k} x^k}{k!} - \sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\\ =& \sum_{k=n+1}^{\infty} \frac{B_{k} x^k}{k!}\\ =& \sum_{j=0}^{\infty} \frac{B_{j+n+1}x^{j+n+1}}{(j+n+1)!} \end{align*}$$ The zeta function are related to the Bernoulli numbers through the following expression $$B_n = -\zeta(1-n)n \quad n=2,3,4,..$$ Hence $$\begin{align*} \frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!} = &\sum_{j=0}^{\infty} \frac{B_{j+n+1}x^{j+n+1}}{(j+n+1)!} \\ =& \sum_{j=0}^{\infty} \frac{\zeta(-j-n)(j+n+1)x^{j+n+1}}{(j+n+1)!}\\ =& \sum_{j=0}^{\infty} \frac{\zeta(-j-n)x^{j+n+1}}{(j+n)!}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n) j!(-1)^j}{(j+n)!} \frac{(-x)^{j}}{j!}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n) j!n!(-1)^j}{n! (j+n)!} \frac{(-x)^{j}}{j!}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n)(-1)^j}{n! \binom{n+j}{n} } \frac{(-x)^{j}}{j!} \end{align*}$$ The formula $$\binom{v}{m} = \frac{(v-m+1)_{m}}{m!} \tag{1}$$ relate the binomial coefficient to the Pochhammer polynomial (rising factorial) while the Pochhammer polynomial obey the reflection formula $$(-x)_{n} = (-1)^n(x-n+1)_{n} \tag{2}$$ Hence $$\begin{align*} \frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!} =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n)(-1)^j}{n! \binom{n+j}{n} } \frac{(-x)^{j}}{j!}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n)(-1)^j}{(j+1)_{n}} \frac{(-x)^{j}}{j!} \quad \textrm{from (1)}\\ =& x^{n+1}\sum_{j=0}^{\infty} \frac{\zeta(-j-n)}{(-j-n)_{n}} \frac{(-x)^{j}}{j!} \quad \textrm{from (2)}\\ \end{align*}$$ Back to the integral: $$\int_{0}^{\infty} x^{s-n-2} \left(\frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\right) dx = \int_{0}^{\infty} x^{s-1} \left(\sum_{j=0}^{\infty} \frac{\zeta(-j-n)}{(-j-n)_{n}} \frac{(-x)^{j}}{j!}\right) dx$$ Denote $$f(x) = \sum_{j=0}^{\infty} \frac{\zeta(-j-n)}{(-j-n)_{n}} \frac{(-x)^{j}}{j!}$$ $$\varphi(j) =\frac{\zeta(-j-n)}{(-j-n)_{n}}$$ Applying the Ramanujan's master theorem $$\begin{align*} I = \int_{0}^{\infty} x^{s-n-2} \left(\frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\right) dx =& \Gamma(s)\varphi(-s)\\ =& \Gamma(s)\frac{\zeta(s-n)}{(s-n)_{n}} \end{align*}$$ Finally, recall that the rising facotiral can be expressed as the quotient of two gamma functions: $$(x)_{n} = \frac{\Gamma(n+x)}{\Gamma(x)}$$ Hence $$I = \Gamma(s)\frac{\zeta(s-n)}{(s-n)_{n}} = \Gamma(s)\frac{\zeta(s-n)\Gamma(s-n)}{\Gamma(s)} = \Gamma(s-n)\zeta(s-n)$$ Therefore, we can conclude $$\boxed{\int_{0}^{\infty} x^{s-n-2} \left(\frac{x}{e^x -1} -\sum_{k=0}^{n} \frac{B_{k}x^k}{k!}\right) dx =\Gamma(s-n)\zeta(s-n) \quad 0<\Re(s)<1 }$$

Application: Integral representation of the Apery's constant

$$\int_0^{\infty} \left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)\frac{1}{x^4} dx = \frac {\zeta (3)} { 8\pi^2}$$

Proof

Put $n=2$ in the Mellin transform, then $$\int_0^{\infty} x^{s-4}\left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)dx = -\Gamma(s-2)\zeta(s-2)$$ Under some technical conditions the Mellin transform converges uniformly and we can take the limit as $s \to 0$ $$\lim_{s \to 0} \int_0^{\infty} x^{s-4}\left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)dx = \int_0^{\infty} \left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)\frac{1}{x^4} dx = \lim_{s \to 0} -\Gamma(s-2)\zeta(s-2)$$ What is the limit on the right hand side? $$\begin{align*} \lim_{s \to 0} -\Gamma(s-2)\zeta(s-2) =& \lim_{s \to 0} -\Gamma(s-2)\zeta(s-2)\\ =& \lim_{t \to -2} -\Gamma(t)\zeta(t)\\ =& \lim_{t \to -2} -\frac{\zeta(t)}{\frac{1}{\Gamma(t)}}\\ =& \lim_{t \to -2} -\frac{\zeta(t)}{\frac{t(t+1)(t+2)}{\Gamma(t+3)}} \quad \left(\Gamma(t+1) = t\Gamma(t) \right)\\ =& -\frac{\lim_{t \to -2} \zeta'(t)}{\lim_{t \to -2} \frac{3t^2+6t+2-(t^2+3t+2)\psi(t+3)}{\Gamma(t+3)}} \quad \textrm{(L'Hôpital's rule )}\\ =& -\frac{\zeta'(-2)}{2}\\ \end{align*}$$ The derivative of the zeta function at the negative even integers is given by $$\zeta^{\prime}(-2n) = (-1)^n \frac {(2n)!} {2 (2\pi)^{2n}} \zeta (2n+1)$$ If we put $n=1$ $$\zeta^{\prime}(-2) = - \frac {\zeta (3)} { 4\pi^2}$$ Hence $$\lim_{s \to 0} -\Gamma(s-2)\zeta(s-2) = \frac {\zeta (3)} { 8\pi^2}$$ Therefore we can conclude $$\boxed{\int_0^{\infty} \left(1-\frac{x}{2}+\frac{x^2}{12} -\frac{x}{e^x-1}\right)\frac{1}{x^4} dx = \frac {\zeta (3)} { 8\pi^2} }$$

Integral of the day XXXII

Integral involving the polylogamma function $\psi^{(n)}(z)$

Today we show the proof of this nice result posted by @Ali39342137 $$\int_{0}^{1} \ln^2(x)\ln(x^2+x+1) dx = \frac{2}{3}\psi^{(1)}\left(\frac{4}{3}\right)+\frac{8}{9}\zeta(3) +\frac{4\pi^3}{27\sqrt{3}} +2\ln\sqrt{27} -6 +\frac{\pi}{\sqrt{3}}-\frac{\pi^2}{3}$$ The proof relies heavily on properties of the digamma, the trigamma and the tetragamma functions $\psi(z)$,$\psi^{(1)}(z)$, $\psi^{(2)}(z)$ and the polygamma function $\psi^{(n)}(z)$ in general.

Proof:

$$\begin{align*} I = \int_{0}^{1} \ln(x)\ln(1+x+x^2)dx =& \int_{0}^{1} \ln(x)\ln\left(\frac{1-x^3}{1-x}\right)dx\\ =& \underbrace{\int_{0}^{1} \ln(x)\ln\left(1-x^3\right)dx}_{J}- \underbrace{\int_{0}^{1} \ln(x)\ln\left(1-x\right)dx}_{K} \end{align*}$$ $$\begin{align*} J = \int_{0}^{1} \ln^2(x)\ln\left(1-x^3\right)dx =& -\int_{0}^{1} \sum_{j=1}^{\infty} \frac{x^{3j}}{j}\ln^2(x) dx \\ =& -\sum_{j=1}^{\infty}\frac{1}{j} \int_{0}^{1} x^{3j}\left(\frac{d^2}{dt^2}\Big|_{t=0+} x^t \right) dx\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} -\sum_{j=1}^{\infty} \frac{1}{j}\int_{0}^{1} x^{3j+t} dx\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} -\sum_{j=1}^{\infty}\frac{1}{j(3j+t+1)} \\ =& -\sum_{j=1}^{\infty}\left[ \frac{d^2}{dt^2}\Big|_{t=0+} \frac{1}{j(3j+t+1)} \right]\\ =& -2\sum_{j=1}^{\infty}\frac{1}{j(3j+1)^3} \\ =& 6\sum_{j=1}^{\infty} \frac{1}{(3j+1)^2} + 6\sum_{j=1}^{\infty} \frac{1}{(3j+1)^3} -2\sum_{j=1}^{\infty} \frac{1}{j(3j+1)} \quad \textrm{(partial fractions)}\\ =& 6\sum_{n=0}^{\infty} \frac{1}{(3n+4)^2} + 6\sum_{n=0}^{\infty} \frac{1}{(3n+4)^3} -2\sum_{n=0}^{\infty} \frac{1}{(n+1)(3n+4)} \quad (n=j-1)\\ =& \frac{2}{3}\sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{4}{3}\right)^2} + \frac{2}{9}\sum_{n=0}^{\infty} \frac{1}{\left(n+\frac{4}{3}\right)^3} -\frac{2}{3}\sum_{n=0}^{\infty} \frac{1}{(n+1)\left(n+\frac{4}{3}\right)} \end{align*}$$ Recall that the polygamma function has the series representation $$\psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}}$$ and that the difference between two digamma functions may be expressed as $$\psi(x)-\psi(y) = (x-y)\sum_{j=0}^{\infty} \frac{1}{(j+x)(j+y)} \tag{*}$$ Therefore $$J = \int_{0}^{1} \ln^2(x)\ln\left(1-x^3\right)dx = \frac{2}{3}\psi^{(1)}\left(\frac{4}{3}\right)-\frac{1}{9}\psi^{(2)}\left(\frac{4}{3}\right)-2\psi\left(\frac{4}{3}\right)+2\psi(1)$$ It is well known that $\displaystyle \psi(1) = -\gamma$

For $\displaystyle \psi\left(\frac{4}{3}\right)$ we can use the following formula: $$\psi\left(\frac{1}{3}+J\right) +\frac{\pi}{2\sqrt{3}} = -\gamma -\ln\sqrt{27} + \sum_{j=1}^{J} \frac{3}{3j-2} \quad J=0,1,2,...$$ If we put $\displaystyle J=1$ $$\psi\left(\frac{4}{3}\right) = -\gamma -\ln\sqrt{27} +3 -\frac{\pi}{2\sqrt{3}}$$ For $\displaystyle \psi^{(2)}\left(\frac{4}{3}\right)$ Recall the recurrence formula for the polygamma function: $$\psi^{(n)}(x+1) = \psi^{(n)}(x) - \frac{n!}{(-x)^{n+1}} \quad n=1,2,3,...$$ If we put $n=2$ and $\displaystyle x = \frac{1}{3}$ $$\psi^{(2)}\left(\frac{4}{3}\right) = \psi^{(2)}\left(\frac{1}{3}\right)+54$$ From the reflection formula $$(-1)^n\psi^{(n)} (1-x) - \psi^{(n)} (x) = \pi \frac{\mathrm{d}^n}{\mathrm{d} z^n} \cot{\pi x}$$ if we put $n=2$ and $\displaystyle x=\frac{2}{3}$ $$\psi^{(2)}\left(\frac{1}{3}\right) - \psi^{(2)}\left(\frac{2}{3}\right) = -\frac{8\pi^3}{3\sqrt{3}}$$ while from the multiplication formula $$k^{m+1} \psi^{(m)}(kz) = \sum_{n=0}^{k-1} \psi^{(m)}\left(z+\frac{n}{k}\right)\qquad m \ge 1$$ If we put $m=2$, $k=3$ and $\displaystyle z=\frac{1}{3}$ $$27\psi^{(2)}\left(1\right) = \psi^{(2)}\left(\frac{1}{3}\right) + \psi^{(2)}\left(\frac{2}{3}\right) + \psi^{(2)}(1)$$ Hence $$\psi^{(2)}\left(\frac{1}{3}\right) = 13\psi^{(2)}(1) - \frac{4\pi^3}{3\sqrt{3}}$$ Finally, using the expression $$\psi^{(n)}(J) = (-1)^{n+1} n! \left[ \zeta(n+1)-\sum_{j=1}^{J-1} \frac{1}{j^{n+1}} \right]$$ $$\psi^{(2)}(1) = -2\zeta(3)$$ Therefore $$\psi^{(2)}\left(\frac{1}{3}\right) = -26\zeta(3) -\frac{4\pi^3}{3\sqrt{3}}$$ $$\psi^{(2)}\left(\frac{4}{3}\right) = -26\zeta(3) -\frac{4\pi^3}{3\sqrt{3}}+54$$ However, $\displaystyle \psi^{(1)}\left(\frac{4}{3}\right)$ do not have an expression in elementary functions. Therefore $$J = \int_{0}^{1} \ln^2(x)\ln\left(1-x^3\right)dx = \frac{2}{3}\psi^{(1)}\left(\frac{4}{3}\right)+\frac{26}{9}\zeta(3) +\frac{4\pi^3}{27\sqrt{3}} +2\ln\sqrt{27} -12 +\frac{\pi}{\sqrt{3}}$$ $$\begin{align*} K = \int_{0}^{1} \ln^2(x)\ln(1-x)dx =& -\int_{0}^{1}\sum_{j=1}^{\infty}\frac{x^j}{j}\ln^2(x)dx\\ =& -\sum_{j=1}^{\infty}\frac{1}{j} \int_{0}^{1}x^j \left(\frac{d^2}{dt^2}\Big|_{t=0} x^t\right)dx\\ =& \frac{d^2}{dt^2}\Big|_{t=0} -\sum_{j=1}^{\infty}\frac{1}{j}\int_{0}^{1}x^{j+t}dx\\ =& \frac{d^2}{dt2}\Big|_{t=0} -\sum_{j=1}^{\infty}\frac{1}{j(j+t+1)}\\ =& -\sum_{j=1}^{\infty}\left[\frac{d^2}{dt^2}\Big|_{t=0} \frac{1}{j(j+t+1)}\right]\\ =& -2\sum_{j=1}^{\infty} \frac{1}{j(j+1)^2}\\ =& 2\sum_{j=1}^{\infty} \frac{1}{(j+1)^2} + 2\sum_{j=1}^{\infty} \frac{1}{(j+1)^3} -2\sum_{j=0}^{\infty} \frac{1}{j(j+1)} \quad \textrm{(partial fractions)}\\ =& \underbrace{2\sum_{m=2}^{\infty} \frac{1}{m^2} + 2\sum_{m=2}^{\infty} \frac{1}{m^3}}_{m=j+1} -2\sum_{j=0}^{\infty} \frac{1}{j(j+1)}\\ =& 2\zeta(2)+2\zeta(3)-4 -2\sum_{j=0}^{\infty} \frac{1}{j(j+1)}\\ =& \frac{\pi^2}{3}+2\zeta(3)-4 -2\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)} \quad (n=j-1)\\ =& \frac{\pi^2}{3}+2\zeta(3)-4 -2\psi(2)+2\psi(1) \quad \textrm{ (from (*))} \\ =& \frac{\pi^2}{3}+2\zeta(3)-4 -2(1-\gamma)-2\gamma \\ =& \frac{\pi^2}{3}+2\zeta(3)-6 \\ \end{align*}$$ Therefore $$I = \int_{0}^{1} \ln^2(x)\ln(x^2+x+1) dx = K-J = \frac{2}{3}\psi^{(1)}\left(\frac{4}{3}\right)+\frac{8}{9}\zeta(3) +\frac{4\pi^3}{27\sqrt{3}} +2\ln\sqrt{27} -6 +\frac{\pi}{\sqrt{3}}-\frac{\pi^2}{3}$$ We can conclude $$\boxed{\int_{0}^{1} \ln^2(x)\ln(x^2+x+1) dx = \frac{2}{3}\psi^{(1)}\left(\frac{4}{3}\right)+\frac{8}{9}\zeta(3) +\frac{4\pi^3}{27\sqrt{3}} +2\ln\sqrt{27} -6 +\frac{\pi}{\sqrt{3}}-\frac{\pi^2}{3}}$$