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A derivation of the duplication formula for the Barnes-G function

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
In a commentary thread DreamWeaver asked if anyone knew of a derivation of the multiplication formula for the Barnes-G function other than the almost impossible to follow derivation in Barnes' original paper.

I suggested using the multiplication formula for the Hurwitz zeta function.

In this thread I'm going to derive the duplication formula.

But the approach could be generalized.



The Hurwitz zeta function satisfies the following multiplication formula:

$$ \displaystyle \zeta(s,nz) = n^{-s} \sum_{k=0}^{n-1} \zeta \Big(s,z+\frac{k}{n} \Big)$$

So for $n=2$,

$$ 2^{s} \zeta(s,2z) = \zeta(s,z) + \zeta \Big(s+ \frac{1}{2} \Big)$$

Differentiate both sides of the above equation with respect to $s$ and then let $s=-1$.

$$\frac{\ln 2}{2} \zeta(-1,2z) + \frac{\zeta'(s,2z)}{2} = \zeta(-1,z) + \zeta \Big(-1,z+ \frac{1}{2} \Big) \ \ \ \ \ (1)$$


For a positive integer $n$,

$$ \zeta(-n,z) = \frac{B_{n+1}(z)}{n+1} $$

where $B_{n+1}$ is the Bernoulli polynomial of order $n+1$.

The identity can be derived from the contour integral representation of the Hurwitz zeta function.

So

$$ \zeta(-1,2z) = - \frac{B_{2}(2z)}{2} = - \frac{1}{2} \Big(4z^{2}-2z+\frac{1}{6} \Big) = - 2z^{2} + z -\frac{1}{2} \ \ \ \ \ (2)$$


And for $\text{Re}(z) >0 $, the Barnes G function has the following closed form expression:

$$\log G(z+1) = z \log \Gamma(z) + \zeta'(-1) + \zeta'(-1,z)$$

$$ \implies \zeta'(-1,z) = (z-1) \log \Gamma(z) + \zeta'(-1) - \log G(z) \ \ \ \ \ (3)$$


Combining (1), (2), and (3) we have

$$\frac{\ln 2}{2} \Big( -2z^{2} +z - \frac{1}{12} \Big) + \frac{1}{2} (2z-1) \log \Gamma(2z) + \zeta'(-1) - \log G(2z) \Big) $$

$$ = (z-1) \log \Gamma(z) + \zeta'(-1) - \log G(z) + \Big(z- \frac{1}{2} \Big) \log \Gamma \Big( z + \frac{1}{2} \Big) + \zeta'(-1) - \log G \Big(z - \frac{1}{2} \Big)$$


$$ \implies \log G(2z) = -3 \zeta'(-1) + \Big(-2z^{2} +z - \frac{1}{12} \Big) \log 2 +(2z-1) \log \Gamma (2z) - +2 (1-z) \log \Gamma(z) $$

$$+2 \log G(z) +(1-2z) \log \Gamma \Big(z + \frac{1}{2} \Big) + 2 \log G \left(z + \frac{1}{2} \right)$$


Then using the duplication formula for the gamma function,

$$ \log G(2z) = -3 \zeta'(-1) + \Big(-2z^{2} +z - \frac{1}{12} \Big) \log 2 +(2z-1) \Big[ \log \Gamma (z) + \log \Gamma \Big(z + \frac{1}{2} \Big) + (2z-1) \log 2 $$

$$- \frac{\log \pi}{2} \Big] +2 (1-z) \log \Gamma(z) +2 \log G(z) +(1-2z) \log \Gamma \Big(z + \frac{1}{2} \Big) + 2 \log G \left(z + \frac{1}{2} \right)$$

$$ = -3 \zeta'(1) + \Big( 2z^{2} -3z + \frac{11}{12} \Big) \log 2 +(1-2z) \frac{\log \pi}{2} + \log \Gamma(z) + 2 \log G(z) + 2 \log \Big(z + \frac{1}{2} \Big) $$

$$ = -3 \zeta'(-1) + \Big( 2z^{2} -2z + \frac{5}{12} \Big) \log 2 + \frac{1-2z}{2} \log 2 \pi + \log \Gamma(z) + 2 \log G(z) + 2 \log G \left(z + \frac{1}{2} \right)$$


$$ \implies G(2z) = e^{-3 \zeta'(-1)} \ 2^{2z^{2}-2z +5/12} \ (2 \pi)^{1/2(1-2z)} \ \Gamma(z) \ G^{2}(z) \ G^{2} \left(z + \frac{1}{2} \right) $$

$$ = A^{3} \ e^{-1/4} \ 2^{2z^{2}-2z +5/12} \ (2 \pi)^{1/2(1-2z)} \ G(z) \ G^{2} \left(z + \frac{1}{2} \right) \ G(z+1)$$

where $A$ is the Glashier-Kinkelin constant
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Simply breath-taking...


Sincere thanks for sharing, RV... (Hug)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Thanks.

I can't quite figure out how to generalize other than to repeat the approach for other values of $n$ and try to notice a pattern.