According to wikipedia on the page Glaisher–Kinkelin constant $$\int_0^{1/2} \ln\Gamma(x) dx=\frac32\ln \text{A}+\frac5{24}\ln 2+\frac14\ln\pi$$ I got interested in how you possibly could prove something like that, but couldn't find any citations about it on the wiki page.

  • $\begingroup$ Use integration by parts then you need to use some psi function identities. $\endgroup$ Jul 30, 2014 at 14:57

3 Answers 3


In general, $$ \begin{align} \int_{0}^{z} \log \Gamma(x) \ dx &= \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z) \\ &= \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} + \ln A - \frac{1}{12} + \zeta^{'}(-1,z) \end{align}$$

where $\zeta(a,z)$ is the Hurwitz zeta function and where I used the fact that $\zeta'(-1) = \frac{1}{12} - \log A$.

You can find a derivation here.

Then $$ \int_{0}^{1/2} \log \Gamma(x) \ dx = \frac{\log 2 \pi}{4} + \frac{1}{24} + \log A + \zeta' \left( -1, \frac{1}{2} \right) .$$

But using the identity $$\zeta \left( z, \frac{1}{2} \right) = (2^z-1) \zeta(z) \ , \tag{1}$$

we have

$$\zeta' \left( z, \frac{1}{2} \right) = \log(2) 2^{z} \zeta(z) +(2^z-1) \zeta'(z) . $$

And replacing $z$ with $-1$, $$\begin{align} \zeta' \left( -1, \frac{1}{2} \right) &= \frac{\log 2}{2} \zeta(-1) - \frac{1}{2} \zeta'(-1) \\ &= - \frac{\log 2}{24} + \frac{\log A}{2} - \frac{1}{24} \end{align}$$

where I used the fact that $\zeta(-1) = -\frac{1}{12}$.


$$ \begin{align} \int_{0}^{1/2}\log\Gamma (x) \ dx &= \frac{\log 2 \pi}{4} - \frac{\log 2}{24} + \frac{3}{2} \log A \\ &= \frac{3}{2} \log A + \frac{5}{24} \log 2 + \frac{\log \pi}{4} . \end{align}$$

$ $

$(1)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (11)


A proof showing that $\zeta'(-1) = \frac{1}{12} - \ln A$ can be found here.


Recall Kummer´s fourier expansion for LogGamma $0<x<1$

$$\ln\left(\Gamma(x)\right)=\frac{\ln\left(2 \pi\right)}{2}+ \sum_{k=1}^{\infty}\frac{1}{2k}\cos\left(2 \pi k x\right) + \sum_{k=1}^{\infty}\frac{\gamma+\ln\left(2 \pi k\right)}{ \pi k}\sin\left(2 \pi k x \right) \tag{1}$$

Integrating $(1)$ we obtain

$$\int_0^x\ln\left(\Gamma(u)\right)du=\frac{x \ln\left(2 \pi\right)}{2}+\frac{1}{4 \pi}\sum_{k=1}^{\infty}\frac{\sin\left(2 \pi k x \right)}{k^2} +\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}-\frac{\gamma}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\cos \left(2 \pi k x \right)}{k^2}+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right)}{k^2}-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( 2 \pi k\right) \cos\left(2 \pi k x \right)}{k^2} \tag{2}$$

Letting $x=\frac12$ in $(2)$ we obtain

$$ \begin{aligned} \int_0^{1/2}\ln\left(\Gamma(x)\right)dx&=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{2 \pi^2}\zeta(2)+\frac{\gamma}{2 \pi^2}\eta(2)+\frac{\ln(2\pi)}{2 \pi^2}\zeta(2)+\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k\right)}{k^2}+\frac{\ln \left( 2 \pi \right)}{2 \pi^2}\eta(2)-\frac{1}{2 \pi^2}\sum_{k=1}^{\infty}\frac{\ln \left( k \right) \left(-1 \right)^{k}}{k^2}\\ &=\frac{ \ln\left(2 \pi\right)}{4} +\frac{\gamma}{12 }+\frac{\gamma}{24}+\frac{\ln(2\pi)}{12 }-\frac{1}{2 \pi^2}\zeta^{\prime}(2)+\frac{\ln \left( 2 \pi \right)}{24}-\frac{1}{2 \pi^2}\eta^{\prime}(2)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{1}{2 \pi^2} \cdot \frac{\pi^2}{6}\left(\ln(2 \pi)+\gamma-12 \ln A \right)-\frac{1}{2 \pi^2} \left(\frac{\pi^2 \ln(2 )}{12} +\frac{\pi^2}{12}\left(\ln(2 \pi)+\gamma-12 \ln A\right)\right)\\ &=\frac{\gamma}{8}+\frac{3 \ln(2 \pi)}{8}-\frac{\ln(2 \pi)}{12}-\frac{\gamma}{12}+\ln A-\frac{\ln (2)}{24}-\frac{\ln(2 \pi)}{24}-\frac{\gamma}{24}+\frac{\ln A}{2}\\ &=\left(\frac{1}{8}-\frac{1}{12}-\frac{1}{24} \right)\gamma+\left(\frac38-\frac{1}{12}-\frac{1}{24} \right)\ln(2 \pi)-\frac{\ln(2)}{24}+\ln A+\frac{\ln A}{2}\\ &=\frac{3}{2}\ln A+ \frac{6}{24}\ln (2 \pi)-\frac{\ln (2)}{24}\\ &=\frac{3}{2}\ln A+\frac{5}{24}\ln 2+\frac{1}{4}\ln \pi \qquad \blacksquare \end{aligned} $$

We used that $\zeta^{\prime}(2)=\frac{\pi^2}{6}\left(\ln 2 \pi + \gamma -12 \ln A \right)$ and that $\eta^{\prime}(s)=2^{1-s}\ln (2) \zeta(s)+(1-2^{1-s})\zeta^{\prime}(s)$


It can be derived from Fourier series expansion of $\ln\Gamma(x)$ (see here) and formula for $-\zeta'(2)$ on the page that you cited (this value appears after integrating of Fourier series term-by-term).


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