The Duplication Formula for the Gamma Function by logarithmic derivatives.

Question

I was reading Ahlfors' "Complex Analysis" (second edition) and in Chapter 5, section 2.4, where he studies the Gamma Function, he proves Legendre's Duplication Formula: $$\sqrt{\pi}\Gamma(2z)=2^{2z-1}\Gamma(z)\Gamma\left(z+\frac{1}{2}\right).$$His deduction begins with the fact that$$\frac{d}{dz}\left(\frac{\Gamma'(z)}{\Gamma(z)}\right)=\sum_{n=0}^{\infty}\frac{1}{(z+n)^2}$$(where I don't have any problem deducing) and considers the sum $$ \frac{d}{dz}\left(\frac{\Gamma'\left(z\right)}{\Gamma\left(z\right)}\right)+\frac{d}{dz}\left(\frac{\Gamma'\left(z+\frac{1}{2}\right)}{\Gamma\left(z+\frac{1}{2}\right)}\right) = \sum_{n=0}^{\infty}\frac{1}{\left(z+n\right)^{2}}+\sum_{n=0}^{\infty}\frac{1}{\left(z+n+\frac{1}{2}\right)^{2}} = 4\left[\sum_{n=0}^{\infty}\frac{1}{\left(2z+2n\right)^{2}}+\sum_{n=0}^{\infty}\frac{1}{\left(2z+2n+1\right)^{2}}\right] = 4\sum_{m=0}^{\infty}\frac{1}{\left(2z+m\right)^{2}}. $$I fully understand these simple calculations, however, Ahlfors then states that the last series is equal to $$ 2\frac{d}{dz}\left(\frac{\Gamma'(2z)}{\Gamma{2z}}\right). $$ According to the first equation, shouldn't the coefficient be 4? My other question is after the differential equation is integrated two times and then exponentiated. According to Ahlfors, after integrating twice and exponentiating, the expression turns to $$ \Gamma(z)\Gamma\left(z+\frac{1}{2}\right)=e^{az+b}\Gamma(2z) $$ where $a$ and $b$ are constants, my question is: shouldn't to equation be $$\Gamma(z)\Gamma\left(z+\frac{1}{2}\right)=e^{az+b}\left(\Gamma(2z)\right)^2$$becuase of the coefficient 2 preceding the left hand side of the differential equation? (Or elevated to the fourth power as my first question would suggest?)

I know that the duplication formula is correctly stated in Ahlfors' book and I know I'm either missing something or (perhaps less probably) Ahlfors is missing something.

Answer 1

First question:
We note that $$\frac{d}{dz}\left(\frac{\Gamma'(\zeta )}{\Gamma(\zeta )}\right)=\frac{d}{d\zeta}\left (\frac{\Gamma'(\zeta )}{ \Gamma(\zeta )}\right)\cdot \frac{d\zeta }{dz}=\sum_{n=0}^{\infty} \frac{1}{(\zeta +n)^2} \cdot \frac{d\zeta }{dz}.$$ Let $\zeta =2z$, then we have$$\frac{d}{dz}\left(\frac{\Gamma'(2z)}{\Gamma(2z)}\right) =\sum_{n=0}^{\infty} \frac{1}{(2z +n)^2}\cdot \frac{d\zeta }{dz}=2\sum_{n=0}^{\infty}\frac{1}{(2z +n)^2}.$$ Thus $$4\sum_{m=0}^{\infty}\frac{1}{\left(2z+m\right)^{2}}=2\frac{d}{dz}\left(\frac{\Gamma'(2z)}{\Gamma(2z)}\right).$$

Second question:
You need only to note that $$\frac{d}{dz}\log \Gamma(2z)=2\ \frac{\Gamma^\prime(2z)}{\Gamma(2z)}.$$ Ahlfors is correct.

Answer 2

As shown at the end of this answer, we have Euler's Reflection Formula: $$ \Gamma(x)\Gamma(1-x)=\pi\csc(\pi x)\tag{1} $$ As shown at the beginning of this answer, $$ \prod_{k=1}^{n-1}\sin(k\pi/n)=\frac{n}{2^{n-1}}\tag{2} $$ Define $$ f(x)=\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)\tag{3} $$ then $f$ is log-convex and $$ \begin{align} x\,f(x) &=x\,\prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right)\\ &=x\,\Gamma(x)\,\prod_{k=1}^{n-1}\Gamma\left(x+\frac kn\right)\\ &=\Gamma(x+1)\,\prod_{k=0}^{n-2}\Gamma\left(x+\frac{k+1}n\right)\\ &=\prod_{k=1}^{n-1}\Gamma\left(x+\frac1n+\frac kn\right)\\[6pt] &=f\left(x+\frac1n\right)\tag{4} \end{align} $$ Plugging $\frac xn$ into $(4)$ gives $$ \frac xnf\left(\frac xn\right)=f\left(\frac{x+1}n\right)\tag{5} $$ $(5)$ and log-convexity implies that $$ f\left(\frac xn\right)=C_n\frac{\Gamma(x)}{n^x}\tag{6} $$ Using $(1)$ and $(2)$ yield $$ \begin{align} f\left(\frac1n\right)^2 &=\prod_{k=1}^{n-1}\Gamma\left(\frac kn\right)\Gamma\left(1-\frac kn\right)\\ &=\prod_{k=1}^{n-1}\pi\csc(k\pi/n)\\ &=\frac1n2^{n-1}\pi^{n-1}\tag{7} \end{align} $$ $(6)$ and $(7)$ yield $$ C_n=\sqrt{n2^{n-1}\pi^{n-1}}\tag{8} $$ Thus, $(6)$ and $(8)$ yield Gauss' Multiplication Theorem $$ \prod_{k=0}^{n-1}\Gamma\left(x+\frac kn\right) =\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{9} $$ The Duplication Formula is the Multiplication Theorem with $n=2$.