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 closed-form expression for the series in the l.h.s. in terms of the known mathematical constants. 



An integral related to the Euler gamma function
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


$$\eta = \frac{\gamma + \log (2 \pi)}{\pi}$$


