Derivative of Gamma function In Computing the integral of $\log(\sin x)$, user17762 provided a solution which requires differentiating $\displaystyle \frac{\Gamma(2z+1)}{4^z\Gamma^2(z+1)}\frac{\pi}{2}$ with respect to $z$. How is this done?
I had actually got $\displaystyle\int_0^{\pi/2}\sin^{2z}(x)dx = \frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}\frac{\sqrt\pi}{2}$ instead. But I am guessing they are equivalent and differentiating them would use the same technique. 

How is the derivative taken? If you have
$$
\int^{\pi/2}_0 \! \sin^{2z} (x) \ \mathrm{d}x = \frac{\pi}{2}\Gamma(2z+1)4^{-z}\Gamma^{-2}(z+1)
$$
then differentiating both sides with respect to $z$ gives 
\begin{align}
2\int^{\pi/2}_0 \! \sin^{2z}(x) \log(\sin(x)) \ \mathrm{d}x = 
 \frac{\pi}{2}&\left\{ 2\psi(2z+1)4^{-z}\Gamma^{-2}(z+1)  \right. \\
&\left. -2\Gamma(2z+1)4^{-z}\Gamma^{-3}(z+1)\psi(z+1)  \right. \\
&\left. -\log(4)\Gamma(2z+1)4^{-z}\Gamma^{-2}(z+1) \right\}
\end{align}
where $\psi$ is the digamma function. Set $z=0$ and note that $\Gamma(1)=1$, $\psi(1)=-\gamma$, where $\gamma$ is the Euler-Mascheroni constant, this gives
$$
2\int^{\pi/2}_0 \! \log(\sin(x)) \ \mathrm{d}x = \frac{\pi}{2}\left(-2\gamma+2\gamma-\log(4)\right) = -\frac{\pi}{2}\log(4) = -\pi\log(2) 
$$
So
$$
\int^{\pi/2}_0 \! \log(\sin(x)) \ \mathrm{d}x = -\frac{\pi}{2}\log(2)
$$
as confirmed by wolfram, http://www.wolframalpha.com/input/?i=integrate+log%28sin%28x%29%29+from+x%3D0+to+x%3Dpi%2F2.
