Using the Weierstrass definition for $\Gamma(x)$ and $\Gamma\Big(x + \frac12\Big)$, how can I prove the duplication formula? This is problem $10.7.3$ in the book Irresistible Integrals, by Boros and Moll.
Any help is highly appreciated.
[Math] Duplication formula for gamma function
gamma function
Related Solutions
As the comments (to Chandru1's answer) suggest, your two identities can be derived from the integral formula for the Beta function and its relation to the Gamma function. I'm not quite sure what Theo meant by "something non-trivial", but here's one way to prove that $ B(\alpha,\beta)=\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}.$
Starting with $ \Gamma(\alpha) = \int_0^{\infty} x^{\alpha} e^{-x} \frac{\mathrm dx}{x} $ and making the substitution $x=y^2$ gives $$ \Gamma(\alpha) = 2 \int_0^{\infty} y^{2\alpha-1} e^{-y^2}\mathrm dy. $$ Using this form of $\Gamma$, we get $$\Gamma(\alpha) \Gamma(\beta) = 4 \int_0^{\infty} \int_0^{\infty} x^{2\alpha-1} y^{2\beta-1} e^{-(x^2+y^2)}\mathrm dx\mathrm dy. $$ Switching to polar coordinates ($ x=r\cos(\theta), y=r\sin(\theta) $) gives $$ \Gamma(\alpha) \Gamma(\beta) = 4 \int_0^{\frac{\pi}{2}} \int_0^{\infty} r^{2(\alpha+\beta)-1} \cos(\theta)^{2\alpha-1} \sin(\theta)^{2\beta-1} e^{-r^2}\mathrm dr\mathrm d\theta $$ $$ = \left(2 \int_0^{\frac{\pi}{2}} \cos(\theta)^{2\alpha-1} \sin(\theta)^{2\beta-1} \mathrm d\theta \right) \left( 2\int_0^{\infty} r^{2(\alpha+\beta)-1} e^{-r^2}\mathrm dr \right) $$ $$ = \left(2 \int_0^{\frac{\pi}{2}} \cos(\theta)^{2\alpha-1} \sin(\theta)^{2\beta-1} \mathrm d\theta \right) \Gamma(\alpha+\beta). $$ Therefore, $$ \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)} = 2 \int_0^{\frac{\pi}{2}} \cos(\theta)^{2\alpha-1} \sin(\theta)^{2\beta-1}\mathrm d\theta $$ and the substitution $z=\sin(\theta)^2 $ yields $$ \frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)} = \int_0^1 (1-z)^{\alpha-1} z^{\beta-1}\mathrm dz.$$
Appropriate selections of $\alpha, \beta$ and observing some symmetry will get the desired identities.
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
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
$$\frac{\Gamma(x)\Gamma(x+\frac1{2})}{\Gamma(2x)}=\frac{\sqrt{\pi}}{2^{2x-1}}.$$