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 \[ \boxed{\int_{0}^{1} \ln^2(x)\ln(x^2+x+1) dx = \frac{2}{3}\psi^{(1)}\left(\frac{4}{3}\right)+\frac{8}{9}\zeta(3) +\frac{4\pi^3}{27\sqrt{3}} +2\ln\sqrt{27} -6 +\frac{\pi}{\sqrt{3}}-\frac{\pi^2}{3}} \]