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The question is from the paper (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.

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Here there is a good historical review on Euler Gamma function,…. In this thesis you can find a lot of relation about your question –  Hassan Jolany Feb 19 '14 at 13:17

2 Answers 2

up vote 6 down vote accepted

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

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

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Experimentally with mpmath this holds to precision $1000$ decimal digits:

$$ \frac{\ln{1}}{1}-\frac{\ln{3}}{3}+\frac{\ln{5}}{5}-\ldots = -1/4\,\pi \, \left( \gamma+2\,\ln \left( 2 \right) +3\,\ln \left( \pi \right) -4\,\ln \left( {\frac {\pi \,\sqrt {2}}{\Gamma \left( 3 /4 \right) }} \right) \right) $$


$$ \eta = {\frac {\gamma+\ln \left( 2 \right) +\ln \left( \pi \right) }{\pi }} $$


import mpmath
from mpmath import pi,gamma,euler,log
pre=100 #precision, adjust
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)
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The displayed equation has a more complicated expression in place of the $\Gamma(1/4)$ that appears in the code. Is there a reason for that? –  S. Carnahan Feb 19 '14 at 10:14
@S.Carnahan Indeed. This was produced by latex() in Maple, don't know why... –  joro Feb 19 '14 at 10:29
According to Euler's reflection formula $\Gamma(1-1/4)\Gamma(1/4)=\frac{\pi}{\sin{\pi/4}}=\sqrt{2}\pi$. So $\Gamma(3/4)$ can be expressed through $\Gamma(1/4)$. If I didn't make a mistake in a hurry, joro's result implies $\eta=(\gamma+\ln{2}+\ln{\pi}+8\ln{(\Gamma(1/4))}$. –  Zurab Silagadze Feb 19 '14 at 10:37
@ZurabSilagadze I edited the answer with different result for eta which holds at least to 1000 decimal digits. It doesn't depend on $\Gamma$. –  joro Feb 19 '14 at 10:40
$\eta = \frac{\gamma + \log(2 \pi)}{\pi}$. Please see my answer below. –  Alexander Shamov Feb 19 '14 at 11:36

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