2

All people know expressions G-BARNES FUNCTION for example G(1/2), G(3/2) etc ... or G(1/4), G(3/4). But someone know G(1/8), G(3/8), G(5/8) or G(7/8) in terms of Psi(1,1/8) ? Thanks.

Raptor
  • 87
  • 1
    Just funny that you wrote "All people know ..." - I honestly had to google the Barnes G-function :) – Matti P. Jul 25 '19 at 09:12
  • @MattiP. I was tempted to edit it, but figured it was more quaint that way. Here is another of the rather few occurrences of Barnes G-function in MSE. The function seems a bit exotic that even the notation and terminology (aka "double gamma function") needs caution. – Tito Piezas III Jul 25 '19 at 09:25
  • You might find this useful: http://www.cs.cmu.edu/~adamchik/articles/issac/issac01.pdf – uhhhhidk Jul 26 '19 at 08:10

1 Answers1

2

Closed-forms (in terms of other constants) of the Barnes G-function $G(z)$ for fractional $z = \frac{p}{q}$ with $0<p<q$ are known only (for now) for SEVEN special values. Given the Clausen function $\operatorname{Cl}_2(z)$ and,

$$\begin{aligned} A \;&= \text{Glaisher–Kinkelin constant}\\ \operatorname{Cl}_2\left(\frac\pi2\right) &=\text{Catalan's constant}\\ \operatorname{Cl}_2\left(\frac\pi3\right) &=\text{Gieseking's constant} \end{aligned}$$

then,

$$\begin{aligned} \ln G\left(\frac12\right) &= -\frac32\ln A -\frac12\ln\Gamma\left(\frac12\right)+\frac1{24}\ln 2+\frac1{8}\\ \ln G\left(\frac13\right) &= -\frac43\ln A -\frac23\ln\Gamma\left(\frac13\right)-\frac{1}{6\pi}\operatorname{Cl}_2\left(\frac\pi3\right)+\frac1{72}\ln 3+\frac1{9}\\ \ln G\left(\frac23\right) &= -\frac43\ln A -\frac13\ln\Gamma\left(\frac23\right)+\frac{1}{6\pi}\operatorname{Cl}_2\left(\frac\pi3\right)+\frac1{72}\ln 3+\frac1{9}\\ \ln G\left(\frac14\right) &= -\frac98\ln A -\frac34\ln\Gamma\left(\frac14\right)-\frac{1}{4\pi}\operatorname{Cl}_2\left(\frac\pi2\right)+\frac3{32}\\ \ln G\left(\frac34\right) &= -\frac98\ln A -\frac14\ln\Gamma\left(\frac34\right)+\frac{1}{4\pi}\operatorname{Cl}_2\left(\frac\pi2\right)+\frac3{32}\\ \ln G\left(\frac16\right) &= -\frac56\ln A -\frac56\ln\Gamma\left(\frac16\right)-\frac{1}{4\pi}\operatorname{Cl}_2\left(\frac\pi3\right)-\frac1{72}\ln 2-\frac1{144}\ln3+\frac5{72}\\ \ln G\left(\frac56\right) &= -\frac56\ln A -\frac16\ln\Gamma\left(\frac56\right)+\frac{1}{4\pi}\operatorname{Cl}_2\left(\frac\pi3\right)-\frac1{72}\ln 2-\frac1{144}\ln3+\frac5{72}\\ \end{aligned}$$

That's all. However, for ratios of Barnes G-function, we have,

$$\ln\left( \frac{G(1-z)}{G(z)} \right)= z\ln\left(\frac{\sin\pi z}{\pi} \right)+\ln\Gamma(z)+\frac{1}{2\pi}\operatorname{Cl}_2(2\pi z)$$

Since there's a relationship between $\operatorname{Cl}_2(m)$ and polygamma $\psi^{(1)}(n)$, one can indeed use $\psi^{(1)}(\frac18)$ etc to express the ratio $G(\frac78)/G(\frac18)$, but it is harder when numerator and denominator are taken separately.


$\color{red}{\text{Update, July 26}}$. From this post, we find,

$$\psi^{(-2)}(z)=\int_0^z\ln\Gamma(t)~dt=\frac{z(1-z)}2+\frac z2\ln(2\pi)+z\ln\Gamma(z)-\ln G(z+1)$$

Since $G(1+z)=\Gamma(z)\, G(z)$ then,

$$\qquad\color{red}{\psi^{(-2)}(z)} = \frac{z(1-z)}{2}+\frac{z}{2}\ln 2\pi -(1-z)\ln\Gamma(z) -\ln \color{red}{G(z)}$$

Thus, it is possible to express every Barnes G-function $G(z)$ in terms of the polygamma function, but one has to use negative order and analytic continuation.