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This post of Boris Bukh mentions amazing Gustav Herglotz's integral $$\int_0^1\frac{\ln\left(1+t^{\,4\,+\,\sqrt{\vphantom{\large A}\,15\,}\,}\right)}{1+t}\ \mathrm dt=-\frac{\pi^2}{12}\left(\sqrt{15}-2\right)+\ln2\cdot\ln\left(\sqrt3+\sqrt5\right)+\ln\frac{1+\sqrt5}{2}\cdot\ln\left(2+\sqrt3\right). $$ I wonder if there are other irrational real algebraic exponents $\alpha$ such that the integral $$ \int_{0}^{1} \frac{\ln\left(1 + t^{\,{\large\alpha}}\right)}{1 + t}\,{\rm d}t $$ has a closed-form representation? Is there a general formula giving results for such cases?

Are there such algebraic $\alpha$ of degree $> 2$ ?

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Here is a list of some of these integrals: $$ \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} $$

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    $\begingroup$ could you give a reference for th article where all this is proved? $\endgroup$ – Norbert Nov 18 '13 at 20:17
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    $\begingroup$ @alexjo Thank you for this list. $\endgroup$ – Piotr Shatalin Nov 18 '13 at 22:33
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    $\begingroup$ Thanks for pulling this together! There should be equally nice answers for $(3+\sqrt{5})/2$, $(5+\sqrt{21})/2$, $(7+\sqrt{45})/2$, $(9+\sqrt{77})/2$, etc. (The general term is $(2k+1 + \sqrt{4k^2+4k-3})/2$.) Can you find them just as easily? $\endgroup$ – David E Speyer Nov 19 '13 at 18:32
  • $\begingroup$ @PiotrShatalin Here is it $\endgroup$ – Shivam Patel Apr 27 '14 at 16:31
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One that follows from the one above is $$ \int_0^1\frac{\log\left(1+t^{4-\sqrt{15}}\right)}{1+t}\,\mathrm{d}t =\log(2)^2-\int_0^1\frac{\log\left(1+t^{4+\sqrt{15}}\right)}{1+t}\,\mathrm{d}t $$ I will look for others.

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