A NICE PAIR OF LOGARITHMIC INTEGRALS-PART II

In a previous post we proved the following pair of Logarithmic integrals


\int_0^1 \frac{\ln^2\left(x\right)}{x^2+x +1}\,dx=\frac{8\pi^3}{81 \sqrt{3}}


\int_0^1 \frac{\ln^2\left(x\right)}{x^2-x +1}\,dx=\frac{10\pi^3}{81 \sqrt{3}}


Today we will prove another apparently similar pair of Logarithmic integrals, but with a very different type of answer, namely:


\int_0^1 \frac{\ln(x)}{x^2+x+1}\,dx=\frac19\psi^\prime\left(\frac23 \right)-\frac19\psi^\prime\left(\frac13 \right)


\int_0^1 \frac{\ln(x)}{x^2-x+1}\,dx=\frac14\psi^\prime\left(\frac{2}{3}\right)-\frac{1}{12}\psi^\prime\left(\frac{1}{3}\right)- \frac{\pi^2}{9}}


During the calculations I found a way to express \psi^\prime\left(\frac{1}{6}\right) and \psi^\prime\left(\frac{5}{6}\right) in terms of \psi^\prime\left(\frac{1}{3}\right) and \psi^\prime\left(\frac{2}{3}\right), a result that I didn´t know previously and could not find in the literature, but could I ws able to confirm it numerically using WolframAlpha special function calculator.


\begin{aligned} I&=\int_0^1 \frac{\ln(x)}{x^2+x+1}\,dx\\ &=\int_0^1 \frac{(x-1)\ln(x)}{x^3-1}\,dx\\ &=\frac13\int_0^1 \frac{(x^{1/3}-1)\ln(x^{1/3})}{x-1}\left(x^{1/3-1}\right)\,dx\\ &=\frac19\int_0^1 \frac{\left(x^{2/3-1}-x^{1/3-1}\right)\ln(x)}{x-1}\,dx\\ &=\frac19\int_0^1 \frac{\left(x^{2/3-1}-x^{1/3-1}\right)\ln(x)}{x-1}\,dx\\ &=\frac19\int_0^1 \frac{x^{\frac13-1}\ln(x)}{x-1}\,dx-\frac19\int_0^1 \frac{x^{\frac23-1}\ln(x)}{x-1}\,dx\\ &=\frac19\psi^\prime\left(\frac23 \right)-\frac19\psi^\prime\left(\frac13 \right) ​\qquad \blacksquare \end{aligned}


Where We used the integral representation of the trigamma function


\psi^{\prime}(x) &=-\int_{0}^{1} \frac{t^{x-1} \ln t}{1-t} d t


\begin{aligned} I&=\int_0^1 \frac{\ln(x)}{x^2-x+1}\,dx\\ &=\int_0^1 \frac{(x+1)\ln(x)}{1+x^3}\,dx\\ &=\int_0^1 \frac{x\ln(x)}{1+x^3}\,dx+\int_0^1 \frac{\ln(x)}{1+x^3}\,dx\\ &=\frac{1}{36}\left(\psi^\prime\left(\frac{5}{6}\right)-\psi^\prime\left(\frac{1}{3}\right)\right)+\frac{1}{36}\left(\psi^\prime\left(\frac{2}{3}\right)-\psi^\prime\left(\frac{1}{6}\right)\right)\\ &=\frac{1}{36}\left(\psi^\prime\left(\frac{5}{6}\right)-\psi^\prime\left(\frac{1}{6}\right)+\psi^\prime\left(\frac{2}{3}\right)-\psi^\prime\left(\frac{1}{3}\right)\right)\\ &=\frac{1}{36}\left(8\psi^\prime\left(\frac{2}{3}\right)-2\psi^\prime\left(\frac{1}{3}\right)-4 \pi^2}+\psi^\prime\left(\frac{2}{3}\right)-\psi^\prime\left(\frac{1}{3}\right) \right)\\ &=\frac{1}{36}\left(9\psi^\prime\left(\frac{2}{3}\right)-3\psi^\prime\left(\frac{1}{3}\right)-4 \pi^2} \right)\\ &=\frac14\psi^\prime\left(\frac{2}{3}\right)-\frac{1}{12}\psi^\prime\left(\frac{1}{3}\right)- \frac{\pi^2}{9}} \qquad \blacksquare \end{aligned}


Where We used (see appendix)


\int_{0}^{1} \frac{x^{m}\ln(x)}{1+x^{n}} d x=\frac{1}{4 n^2}\left(\psi^\prime\left(\frac{m+n+1}{2 n}\right)-\psi^\prime\left(\frac{m+1}{2 n}\right)\right)

Appendix

Recall the integral proved here:


\int_{0}^{1} \frac{x^{m}}{1+x^{n}} d x=\frac{1}{2 n}\left(\psi\left(\frac{m+n+1}{2 n}\right)-\psi\left(\frac{m+1}{2 n}\right)\right)(A.1)


Differentiating (A.1) w.r. to m we obtain


\int_{0}^{1} \frac{x^{m}\ln(x)}{1+x^{n}} d x=\frac{1}{4 n^2}\left(\psi^\prime\left(\frac{m+n+1}{2 n}\right)-\psi^\prime\left(\frac{m+1}{2 n}\right)\right)(A.2)

Recall the duplication formula of the Trigamma function proved here:


4\psi^{\prime} \left(2x \right)= \psi^{\prime}(x)+\psi^{\prime} \left( x+\frac{1}{2}\right)(A.3)

Letting x=\frac13 in (A.3) we obtain


\psi^\prime\left(\frac{5}{6}\right)=4\psi^\prime\left(\frac{2}{3}\right)-\psi^\prime\left(\frac{1}{3}\right)(A.4)


Recall the reflection formula of the Trigamma function


\psi^{\prime}(x)+\psi^{\prime}(1-x)=\frac{\pi^2} {\sin^2(\pi x)}(A.5)

Letting x=\frac16 in (A.5) we obtain


\begin{aligned} \psi^\prime\left(\frac{1}{6}\right)&= 4\pi^2-\psi^\prime\left(\frac{5}{6}\right)\\ &=4\pi^2-4\psi^\prime\left(\frac{2}{3}\right)+\psi^\prime\left(\frac{1}{3}\right) \end{aligned}(A.6)


Subtracting (A.6) from (A.4) we conclude that


\boxed{\psi^\prime\left(\frac{5}{6}\right)-\psi^\prime\left(\frac{1}{6}\right)=8\psi^\prime\left(\frac{2}{3}\right)-2\psi^\prime\left(\frac{1}{3}\right)-4 \pi^2}(A.7)

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