An integral related to the Euler gamma function 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. 
 A: 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:
import mpmath
from mpmath import pi,gamma,euler,log
pre=100 #precision, adjust
mpmath.mp.dps=pre
mpmath.mp.pretty=True
A=mpmath.nsum(lambda n: (-1)**n*log(2*n+1)/(2*n+1) ,[0,mpmath.inf])
B= -(pi/4)*(euler+2*log(2)+3*log(pi)-4*log(gamma(1/4)))
print mpmath.chop(A-B)

A: In fact, both the experimental result by Joro and the (very nice) asymptotic analysis by A. Shamov, above, are correct. I'd like to advice those more interested readers that I've included this topic and some generalizations in my most recent work on integrals and series involving the log-Gamma function (to appear in arXiv website). F. M. S. Lima
A: 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
$$\eta = \frac{\gamma + \log (2 \pi)}{\pi}$$
