The duplication formula can be written as
$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}= \frac{\Gamma(\frac1{2})}{2^{2x-1}}= \frac{\sqrt{\pi}}{2^{2x-1}}.$$
We want to derive this formula using the Weierstrass definition for the gamma function,
$$\frac1{\Gamma(x)}=xe^{\gamma x}\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)e^{-x/k}.$$
We have
$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}=\frac{2xe^{2\gamma x}}{xe^{\gamma x}(x+\frac1{2})e^{\gamma x}e^{\gamma/2}}\frac{\prod_{k=1}^{\infty}\left(1+\frac{2x}{k}\right)e^{-2x/k}}{\prod_{k=1}^{\infty}\left(1+\frac{x}{k}\right)e^{-x/k}\prod_{k=1}^{\infty}\left(1+\frac{x}{k}+\frac{1}{2k}\right)e^{-x/k}e^{-1/2k}}\\ =\frac{1}{e^{\gamma/2}}\lim_{n \rightarrow \infty}\frac{2x\prod_{k=1}^{2n}\left(1+\frac{2x}{k}\right)}{x(x+\frac1{2})\prod_{k=1}^{n}\left(1+\frac{x}{k}\right)\prod_{k=1}^{n}\left(1+\frac{x}{k}+\frac{1}{2k}\right)}\frac{\prod_{k=1}^{2n}e^{-2x/k}}{(\prod_{k=1}^{n}e^{-x/k})^2\prod_{k=1}^{n}e^{-1/2k}}\\ =\frac{1}{e^{\gamma/2}}\lim_{n \rightarrow \infty}P_n(x)Q_n(x).$$
First simplify $P_n(x)$ as follows:
$$P_n(x)=\frac{2x\prod_{k=1}^{2n}\left(1+\frac{2x}{k}\right)}{x(x+\frac1{2})\prod_{k=1}^{n}\left(1+\frac{x}{k}\right)\prod_{k=1}^{n}\left(1+\frac{x}{k}+\frac{1}{2k}\right)}\\=\frac{(n!)^2}{(2n)!\left(x+n+\frac1{2}\right)}\frac{\prod_{k=0}^{n}\left(2x+2k\right)\prod_{k=0}^{n-1}\left(2x+2k+1\right)}{\prod_{k=0}^{n}\left(x+k\right)\prod_{k=0}^{n-1}\left(x+k+\frac1{2}\right)}\\=\frac{(n!)^22^{2n+1}}{(2n)!\left(x+n+\frac1{2}\right)}$$
Next consider $Q_n(x)$:
$$Q_n(x)=\frac{\prod_{k=1}^{2n}e^{-2x/k}}{(\prod_{k=1}^{n}e^{-x/k})^2\prod_{k=1}^{n}e^{-1/2k}}\\=\frac{n^{1/2}}{2^{2x}}\frac{(2n)^{2x}\prod_{k=1}^{2n}e^{-2x/k}}{(n^x\prod_{k=1}^{n}e^{-x/k})^2n^{1/2}\prod_{k=1}^{n}e^{-1/2k}}$$
Reassembling we get
$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}=\frac{1}{e^{\gamma/2}}\lim_{n \rightarrow \infty}\frac{(n!)^22^{2n+1}}{(2n)!\left(x+n+\frac1{2}\right)}\frac{n^{1/2}}{2^{2x}}\frac{(2n)^{2x}\prod_{k=1}^{2n}e^{-2x/k}}{(n^x\prod_{k=1}^{n}e^{-x/k})^2n^{1/2}\prod_{k=1}^{n}e^{-1/2k}}\\=\frac{1}{2^{2x-1}}\lim_{n \rightarrow \infty}\frac{n}{\left(x+n+\frac1{2}\right)}\frac{(n!)^22^{2n}}{(2n)!n^{1/2}}\frac{(2n)^{2x}\prod_{k=1}^{2n}e^{-2x/k}}{e^{\gamma/2}(n^x\prod_{k=1}^{n}e^{-x/k})^2n^{1/2}\prod_{k=1}^{n}e^{-1/2k}}.$$
We can evaluate the limit in three parts.
First,
$$\lim_{n \rightarrow \infty}\frac{n}{\left(x+n+\frac1{2}\right)}=1.$$
Second, using a well-known identity for the Euler-Mascheroni constant,
$$\lim_{n \rightarrow \infty}\frac{(2n)^{2x}\prod_{k=1}^{2n}e^{-2x/k}}{e^{\gamma/2}(n^x\prod_{k=1}^{n}e^{-x/k})^2n^{1/2}\prod_{k=1}^{n}e^{-1/2k}}=\frac{e^{-2\gamma x}}{(e^{-\gamma x})^2e^{-\gamma /2}e^{\gamma /2}}=1.$$
Third using Stirlings's asymptotic formula $n! \sim \sqrt{2\pi}n^{n+1/2}e^{-n},$
$$\lim_{n \rightarrow \infty}\frac{(n!)^22^{2n}}{(2n)!n^{1/2}}=\sqrt{\pi},$$
and finally we get
$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}=\frac{\sqrt{\pi}}{2^{2x-1}}.$$
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.
Best Answer
Interesting question. We may start from the definition of the Beta function:
$$B(m, n) = \int_0^1 t^{m-1}(1-t)^{n-1}\ \text{d}t$$
and rewrite it when $m\to 2m$:
$$B(2m, n) = \int_0^1 t^{2m-1}(1-t)^{n-1}\ \text{d}t$$
$$B(2m, n) = \int_0^1 t^{2m-1} \frac{(1-t)^n}{1-t}\ \text{d}t$$
Now, since the range of integration is $[0, 1]$, we are allowed to make use of the geometric series
$$\frac{1}{1-t} = \sum_{k = 0}^{+\infty} t^k$$
Hence
$$B(2m, n) = \int_0^1 t^{2m-1} (1-t)^n \sum_{k = 0}^{+\infty} t^k\ \text{d}t = \sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1} (1-t)^n t^k\ \text{d}t$$
And easily write:
$$\sum_{k = 0}^{+\infty} \int_0^1 t^{2m-1+k} (1-t)^n\ \text{d}t$$
Calling now
$$2m-1+k = a ~~~~~~~~~~~ n = b$$
We notice that the integral is well known:
$$ \int_0^1 t^a (1-t)^b\ \text{d}t \equiv B(a+1, b+1)$$
Then we end up with the partial result (re-expanding $a$ and $b$):
$$B(2m, n) = \sum_{k = 0}^{+\infty} B(2m+k, n+1)$$
That series does exist and it does converge to a known result:
$$\sum_{k = 0}^{+\infty} B(2m+k, n+1) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
What you end up with is a sort of recursive relation for the Beta function:
$$B(2m, n) = \frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)}$$
BUT
The above expression can be simplified!
$$\frac{\Gamma (n) \Gamma (2 m+n+1) B(2 m,n+1)}{\Gamma (n+1) \Gamma (2 m+n)} \equiv \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
What we obtained is actually nothing than what we would have obtained by simply substituting at the beginning $m\to 2m$ in the Gamma function / Beta function definition.
$$B(2m, n) = \frac{\Gamma (2 m) \Gamma (n)}{\Gamma (2 m+n)}$$
This really suggest that such a particular duplication formula for the Beta function may not exist at all, since all you need is the Gamma function and ITS duplication formula, through which you can evaluate $\Gamma(2m)$.
Seems like that this is the only "duplication formula" for the beta function.
(Also, I found nothing on reviews or literature).