Hay otros casos similares a los de Herglotz integral de $\int_0^1\frac{\ln\left(1+t^{4+\sqrt{15}}\right)}{1+t}\ \mathrm dt$?

Question

Este post de Boris Bukh menciona increíble Gustav Herglotz integral de $$\int_0^1\frac{\ln\left(1+t^{\,4\,+\,\sqrt{\vphantom{\large Un}\,15\,}\,}\right)}{1+t}\ \mathrm dt=-\frac{\pi^2}{12}\left(\sqrt{15}-2\right)+\ln2\cdot\ln\left(\sqrt3+\sqrt5\right)+\ln\frac{1+\sqrt5}{2}\cdot\ln\left(2+\sqrt3\right). $$ Me pregunto si hay otros irracional real algebraicas con exponentes $\alpha$ tal que la integral $$ \int_{0}^{1} \frac{\ln\left(1 + t^{\,{\large\alpha}}\right)}{1 + t}\,{\rm d}t $$ tiene una forma cerrada de la representación? Hay una fórmula general, dando resultados para este tipo de casos?

Están allí, algebraicas $\alpha$ de grado $> 2$ ?

Answer 1

Aquí está una lista de algunas de estas integrales: $$ \begin{align} \int_0^1\frac{\log\left(1+t^{2+\sqrt{3}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{12}\left(1-\sqrt{3}\right)+ \log 2 \log\left(1+\sqrt{3}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{3+\sqrt{8}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{24}\left(3-\sqrt{32}\right)+ \frac{1}{2}\log 2 \log\left(2\left(3+\sqrt{8}\right)^{3/2}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{4+\sqrt{15}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{12}\left(2-\sqrt{15}\right)+ \log\left(\frac{1+\sqrt{5}}{2}\right)\log\left(2+\sqrt{3}\right)+\\ &+\log 2 \log\left(\sqrt{3}+\sqrt{5}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{5+\sqrt{24}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{24}\left(5-\sqrt{96}\right)+ \frac{1}{2} \log\left(1+\sqrt{2}\right)\log\left(2+\sqrt{3}\right)+\\ &+\frac{1}{2}\log 2 \log\left(2\left(5+\sqrt{24}\right)^{3/2}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{6+\sqrt{35}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{12}\left(3-\sqrt{35}\right)+ \log\left(\frac{1+\sqrt{5}}{2}\right)\log\left(8+3\sqrt{7}\right)+\\ &+\log 2 \log\left(\sqrt{5}+\sqrt{7}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{8+\sqrt{63}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{12}\left(4-\sqrt{63}\right)+ \log\left(\frac{5+\sqrt{21}}{2}\right)\log\left(2+\sqrt{3}\right)+\\ &+\log 2 \log\left(3+\sqrt{7}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{11+\sqrt{120}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{24}\left(11-\sqrt{480}\right)+ \frac{1}{2} \log\left(1+\sqrt{2}\right)\log\left(4+\sqrt{15}\right)+\\ &+\frac{1}{2}\log\left(2+\sqrt{3}\right)\log\left(3+\sqrt{10}\right)+\\ &+ \frac{1}{2}\log\left(\frac{1+\sqrt{5}}{2}\right)\log\left(5+\sqrt{24}\right)+\\ &+ \frac{1}{2}\log 2 \log\left(2\left(11+\sqrt{120}\right)^{3/2}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{12+\sqrt{143}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{12}\left(6-\sqrt{143}\right)+ \log\left(\frac{3+\sqrt{13}}{2}\right)\log\left(10+3\sqrt{11}\right)+\\ &+\log 2 \log\left(\sqrt{11}+\sqrt{13}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{13+\sqrt{168}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{24}\left(13-\sqrt{672}\right)+ \frac{1}{2} \log\left(1+\sqrt{2}\right)\log\left(\frac{5+\sqrt{21}}{2}\right)+\\ &+\frac{1}{4}\log\left(2+\sqrt{3}\right)\log\left(15+\sqrt{224}\right)+\\ &+\frac{1}{4}\log\left(5+\sqrt{24}\right)\log\left(8+\sqrt{63}\right)+\\ &+ \frac{1}{2}\log 2 \log\left(2\left(13+\sqrt{168}\right)^{3/2}\right)\\ \\ \int_0^1\frac{\log\left(1+t^{14+\sqrt{195}}\right)}{1+t}\operatorname{d}t =& \frac{\pi^2}{12}\left(7-\sqrt{195}\right)+ \log\left(\frac{1+\sqrt{5}}{2}\right)\log\left(25+4\sqrt{39}\right)+\\ &+\log \left(\frac{3+\sqrt{13}}{2}\right) \log\left(4+\sqrt{15}\right)+\\ &+\log 2 \log\left(\sqrt{15}+\sqrt{13}\right)\\ \end{align} $$

Answer 2

Uno, que se desprende de la anterior, es $$ \int_0^1\frac{\log\left(1+t^{4-\sqrt{15}}\right)}{1+t}\,\mathrm{d}t =\log(2)^2-\int_0^1\frac{\log\left(1+t^{4+\sqrt{15}}\right)}{1+t}\,\mathrm{d}t $$ Voy a mirar por los demás.