Originally Posted by
Cairo Thanks for this Drexel.
I managed to find something similar on Wikipaedia, but it is the proof I am after.
Can you supply one?
Certainly. But, my original proof didn't work out. So I found an alternate...but it's fairly technical. You might find $\displaystyle \chi\sigma$'s suggestion more useful. Also, I did find an alternate proof on page 22 of this book (a very good book by the wayP may be found for a more general result. Regardless...
For $\displaystyle z>0$ define $\displaystyle \displaystyle f(z)=\frac{\Gamma(2z)}{\Gamma\left(z+\frac{1}{2}\r ight)}2^{1-2z}\sqrt{\pi}$. Note firstly then that $\displaystyle \displaystyle f(1)=\frac{\Gamma(2)}{\Gamma(\frac{1}{2})}2^{-1}\sqrt{\pi}=1$. Moreover,
$\displaystyle \displaystyle \begin{aligned}f(z+1) &=\frac{\Gamma(2z+2)}{\Gamma\left(\left(z+\frac{1} {2}\right)+1\right)}2^{1-2z-2}\sqrt{\pi}\\ &= \frac{1}{4}\frac{\left(2z+1\right)\left(2z\right)} {z+\frac{1}{2}}\frac{\Gamma(2z)}{\Gamma\left(z+\fr ac{1}{2}\right)}2^{1-2z}\sqrt{\pi}\\ &=z\frac{\Gamma(2z)}{\Gamma\left(z+\frac{1}{2}\rig ht)}2^{1-2z}\sqrt{\pi}\\ &=zf(z)\end{aligned}$
Finally, note that $\displaystyle \log(f(z))=\log\left(\Gamma(2z)\right)-\log\left(\Gamma\left(z+\frac{1}{2}\right)\right)+ \left(1-2z\right)\log(2)+\frac{1}{2}\log(\pi)$. So that $\displaystyle \frac{d^2}{dz^2}\log(f(z))=4\psi'(2z)-\psi\left(z+\frac{1}{2}\right)$ where $\displaystyle \psi$ is the digamma function. But, a quick check (using digamma's series expression) shows then that $\displaystyle 4\psi'(2z)-\psi\left(z+\frac{1}{2}\right)>0$ and thus $\displaystyle \log(f(z))$ is convex, and so $\displaystyle f(z)$ is log convex. The result then follows from the Bohr-Mollerup theorem.