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$\textbf{Problem statement}$ I was able to make some progress in calculating the integral at this Arathoon

$$I=\int_{0}^{\frac{\pi }{2}}x\left( \log \tan x\right) ^{2n+1}dx$$

with the great answer of [Jack D'Aurizio]:

$$\mathcal{I}\left( n\right) =\int_{0}^{\infty }\frac{\arctan u}{1+u^{2}}% \left( \log u\right) ^{2n+1}du=\underset{\alpha \rightarrow 0}{\lim }\frac{% d^{2n+1}}{d\alpha ^{2n+1}}\int_{0}^{\infty }u^{\alpha }\frac{\arctan u}{% 1+u^{2}}du$$

and the help of Mathematica: $$\mathcal{B}\left( \alpha \right) =\int_{0}^{\infty }u^{\alpha }\frac{\arctan u}{1+u^{2}}du=-\frac{1}{4}\pi \csc \left[ \frac{\pi \,\alpha }{2}\right] \left( \gamma +\psi \left( \frac{1-\alpha }{2}\right) +2\ln 2\right)$$

$\gamma $ is the Euler--Mascheroni constant, $\psi $ is the diagamma function. Computing $\mathcal{I}\left( n\right) $ for $n=3$ reads%

$$ \mathcal{I}\left( 3\right) =\underset{\alpha \rightarrow 0}{\lim }\frac{% d^{2x3+1}}{d\alpha ^{2x3+1}}\mathcal{B}\left( \alpha \right) =$$

$$ =\frac{217}{1536}\pi ^{6}\zeta \left( 3\right) +\frac{1519}{256}\pi ^{4}\zeta \left( 5\right) +\frac{13335}{64}\pi ^{2}\zeta \left( 7\right) +% \frac{160965}{32}\zeta \left( 9\right)$$

The solution $\mathcal{I}\left( 3\right) $ may be presented as a function of Dirichlet $\lambda $ function, as allready suspected by Arathoon

$$\mathcal{I}\left( 3\right) =155\;\lambda \left( 3\right) \lambda \left( 6\right) +588~\lambda \left( 5\right) \lambda \left( 4\right) +1680~\lambda \left( 7\right) \lambda \left( 2\right) +7!\lambda \left( 9\right) $$

Note for $\mathcal{I}\left( n\right) ~$the last term reads:

$$\mathcal{I}\left( n\right) =\;...+~\left( 2n+1\right) !\lambda \left( 2n+3\right)$$

The mixed terms could be expressed by an definite sum of the form: $$ \sum_{k=0}^{n-1}...\lambda \left( 2\left( n-k\right) +1\right) \lambda \left( 2\left( k+1\right) \right) $$

but I can't figure out the rule of the misterious prefactors.

$\textbf{1st Question}$ $\textit{How can I write the solution for $\mathcal{I}\left( n\right)$ in terms of a definite sum of the Dirichlet $\lambda $ function.} $

$\textbf{2st Question}$

$\textit{Is there another direct way to prove that this is the solution of $\mathcal{I}\left( n\right) $} $

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  • $\begingroup$ can it be that the result is given by $$-1/16\,{\pi }^{2}{4}^{-n}\cos \left( 2\,n\pi \right) -i/16{\pi }^{2}{ 4}^{-n}\sin \left( 2\,n\pi \right) $$ $\endgroup$ – Dr. Sonnhard Graubner Dec 27 '17 at 18:53
  • $\begingroup$ @Dr.SonnhardGraubner I have to check this $\endgroup$ – stocha Dec 27 '17 at 19:06

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