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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.

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  • $\begingroup$ There is other way to prove the identity. $\endgroup$ – Mhenni Benghorbal Jan 28 '15 at 1:31
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    $\begingroup$ I know. You can use the beta function or Stirling's formula, but this method (if correct) yields a beautifully simple proof. I might also be confusing some fundamental fact which, if is the case, I must correct. $\endgroup$ – Alejandro De Las Peñas Jan 28 '15 at 2:37
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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.

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    $\begingroup$ the chain rule... $\endgroup$ – Alejandro De Las Peñas Jan 29 '15 at 7:15
  • $\begingroup$ Yes, just the chain rule! $\endgroup$ – ts375_zk26 Jan 29 '15 at 20:30
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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 $n=2$.

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  • $\begingroup$ How does $(5)$ and log convexity imply $(6)$? $\endgroup$ – D.R. Sep 11 '17 at 2:04
  • $\begingroup$ @D.R.: Applying the Bohr-Mollerup Theorem to $f\left(\frac xn\right)$. $\endgroup$ – robjohn Sep 11 '17 at 3:37
  • $\begingroup$ I don't understand how the fact that the gamma function is the only function that satisfies the three properties the Bohr-Mollerup Theorem has leads to $(6)$. Where does the $C_n$ and $n^x$ come from? $\endgroup$ – D.R. Sep 11 '17 at 18:15
  • $\begingroup$ @D.R.: $(5)$ guarantees that $n^x\,f\!\left(\frac xn\right)\frac1{n\,f\left(\frac1n\right)}$ satisfies the recurrence for $\Gamma(x)$; i.e. $$ \begin{align} \color{#C00}{n^{x+1}\,f\!\left(\frac{x+1}n\right) \frac1{n\,f\left(\frac1n\right)}} &=n^{x+1}\frac xn\,f\!\left(\frac xn\right) \frac1{n\,f\left(\frac1n\right)}\\ &=x\,\color{#C00}{n^x\,f\!\left(\frac xn\right)\frac1{n\,f\left(\frac1n\right)}} \end{align} $$ $\endgroup$ – robjohn Sep 11 '17 at 18:40
  • $\begingroup$ Since $\log(f(x))$ is convex, $\log\left(n^x\,f\!\left(\frac xn\right)\frac1{n\,f\left(\frac1n\right)}\right)=x\log(n)+\log\left(f\!\left(\frac xn\right)\right)-\log\left(n\,f\left(\frac1n\right)\right)$ is also. Trivially, we have that $n^1\,f\!\left(\frac 1n\right)\frac1{n\,f\left(\frac1n\right)}=1$. So we have $(6)$ with $C_n=n\,f\left(\frac1n\right)$. $\endgroup$ – robjohn Sep 11 '17 at 18:57

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