The question is from the paper http://arxiv.org/abs/1312.7115 (A curious formula related to the Euler Gamma function, by Bakir Farhi): is it possible to express the integral $$\eta=2\int\limits_0^1 \ln{(\Gamma(x))}\cdot \sin{(2\pi x)}\,dx= 0.7687478924\ldots$$ in terms of the known mathematical constants as $\pi,\,e,\,\gamma,\;\ln{\pi},\, \ln{2},\,\Gamma{(1/4)}\ldots$? It is shown in the paper that $$\frac{\ln{1}}{1}\frac{\ln{3}}{3}+\frac{\ln{5}}{5}\ldots=\pi\ln{(\Gamma{(1/4)})}\frac{\pi^2}{4}\eta\frac{\pi}{2}\ln{\pi}\frac{\pi}{4}\ln{2}.$$ So the question actually asks whether one can give a closedform expression for the series in the l.h.s. in terms of the known mathematical constants.

Let's introduce a notation $$\alpha_k := \intop_0^1 \sin(2 \pi k z) \log \Gamma(z) dz$$ Let me also remind of the duplication formula: $$\log \Gamma(2z) = \log \Gamma(z) + \log \Gamma(z + 1/2) + 2\log 2 \cdot z  \log(2 \sqrt \pi)$$ Now apply that to the calculation of $\alpha_k$: $$\alpha_k = 2 \intop_0^{1/2} \sin(4 \pi k z) \log \Gamma(2z) dz =$$ $$= 2 \intop_0^{1/2} \sin(4 \pi k z) (\log \Gamma(z) + \log \Gamma(z + 1/2)) dz + 4 \log 2 \cdot \intop_0^{1/2} z \sin(4 \pi k z) dz =$$ $$= 2 \alpha_{2k}  \frac{\log 2} {2 \pi k}$$ In particular, this implies that $\alpha_1 = 2^n \alpha_{2^n}  \frac{\log 2}{2 \pi} \cdot n$. The limit of that as $n \to \infty$ can be calculated, at least in principle, since this kind of asymptotics of Fourier coefficients depends only on the point where the function has a singularity, which is the endpoint here. The singularity here comes from the $\log$ term in expansion $\Gamma(z) = \log z  \gamma z + \dots$ at zero, so the asymptotics of Fourier coefficients must be the same as that of $\log$ (up to lower order terms that are irrelevant). So we can relate to a constant that, I presume, must be better known by $$ \intop_0^1 \sin(2 \pi k z) \log z^{1} dz = \frac{1}{2\pi} k^{1} \log k + \frac{1}{2} \eta k^{1} + \dots $$ Upd. That, in turn, may be expressed in terms of the cosine integral: $$ \intop_0^1 \sin(2 \pi k z) \log z^{1} dz = \frac{1}{2 \pi k} \intop_0^{2 \pi k} (1  \cos z) z^{1} dz =$$ $$ = \frac{1}{2 \pi k} (\log k + \gamma + \log (2 \pi)  \mathrm {Ci}(2 \pi k))$$ Since $\mathrm{Ci}(x) = O(x^{1})$ at $+\infty$, we have $$\eta = \frac{\gamma + \log (2 \pi)}{\pi}$$ 


Experimentally with mpmath this holds to precision $1000$ decimal digits: $$ \frac{\ln{1}}{1}\frac{\ln{3}}{3}+\frac{\ln{5}}{5}\ldots = 1/4\,\pi \, \left( \gamma+2\,\ln \left( 2 \right) +3\,\ln \left( \pi \right) 4\,\ln \left( {\frac {\pi \,\sqrt {2}}{\Gamma \left( 3 /4 \right) }} \right) \right) $$ and $$ \eta = {\frac {\gamma+\ln \left( 2 \right) +\ln \left( \pi \right) }{\pi }} $$ Program:


