Prove that:

\begin{equation} \int_0^1 \frac{\ln\left(1+t^{4+\sqrt{15}}\right)}{1+t}\mathrm dt= -\frac{\pi^2}{12}(\sqrt{15}-2)+\ln (2) \ln(\sqrt{3}+\sqrt{5})+\ln(\phi) \ln(2+\sqrt{3}) \end{equation}

where $\phi$ is the golden ratio.

My friend gave me this challenging problem but I cannot prove it into the given expression. She doesn't know either how to approach this problem. I have tried to use a substitution to obtain the improper integral so that I can use the derivative of beta function but it failed. I also tried to use Taylor series for $\frac{1}{1+t}$ and $\ln\left(1+t^{4+\sqrt{15}}\right)$ but I am unable to derive the result from double sums of series. Could anyone here please help me how to prove it? Any method is welcome and also any help would be greatly appreciated. Thank you.

  • $\begingroup$ Are you familiar with contour integration? If so, give it a try. $\endgroup$ – Asier Calbet Aug 20 '14 at 15:22
  • $\begingroup$ @Assaultous2 No, I am not. But if you familiar with contour integration, you may give a try $\endgroup$ – Anastasiya-Romanova 秀 Aug 20 '14 at 15:23
  • $\begingroup$ This is old problem, you can find it on there $\endgroup$ – china math Aug 20 '14 at 15:36
  • $\begingroup$ @chinamath 'there' where? $\endgroup$ – Anastasiya-Romanova 秀 Aug 20 '14 at 15:42
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    $\begingroup$ A similar integral has been evaluated here: math.stackexchange.com/questions/426325/… $\endgroup$ – SuperAbound Aug 20 '14 at 15:52

See second integral in theorem 7 here.

  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. $\endgroup$ – rlartiga Aug 22 '14 at 13:45
  • $\begingroup$ I don't have time for retyping a 30-pages paper here. $\endgroup$ – Norbert Aug 22 '14 at 13:53
  • $\begingroup$ Maybe just put a commentary? $\endgroup$ – rlartiga Aug 22 '14 at 13:54
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    $\begingroup$ No, because this answers OP's question $\endgroup$ – Norbert Aug 22 '14 at 16:14
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    $\begingroup$ This is a valid answer. In case the original link invalidates, here is the archived version. $\endgroup$ – Vladimir Reshetnikov Aug 22 '14 at 17:17

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