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. 

An excellent discussion of this topic can be found in the book The Gamma Function by James Bonnar. Consider just two of the provably equivalent definitions of the Beta function:
$$
\begin{eqnarray}
B(x,y)&=& 2\int_0^{\pi/2}\sin(t)^{2x-1}\cos(t)^{2y-1}\,dt\\
      &=& \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}.
\end{eqnarray}
$$
Directly from this definition we have 
$$
B(n+\frac{1}{2},\frac{1}{2}): \int_0^{\pi/2}\sin^{2n}(x)\,dx=\frac{\sqrt{\pi}  \cdot\Gamma(n+1/2)}{2(n!)}
$$
$$
B(n+1,\frac{1}{2}): \int_0^{\pi/2}\sin^{2n+1}(x)\,dx=\frac{\sqrt{\pi}  \cdot n!}{2 \Gamma(n+3/2)}
$$
Hence the quotient of these two integrals is
$$
\begin{eqnarray}
\frac{ \int_0^{\pi/2}\sin^{2n}(x)\,dx}{\int_0^{\pi/2}\sin^{2n+1}(x)\,dx}&=& \frac{\Gamma(n+1/2)}{n!}\frac{\Gamma(n+3/2)}{n!}\\
&=& \frac{2n+1}{2n}\frac{2n-1}{2n}\frac{2n-1}{2n-2}\cdots\frac{3}{4}\frac{3}{2}\frac{1}{2}\frac{\pi}{2}
\end{eqnarray}
$$
where the quantitiy $\pi/2$ results from the fact that
$$
\frac{\int_0^{\pi/2}\sin^{2\cdot 0}(x)\,dx}{\int_0^{\pi/2}\sin^{2\cdot 0+1}x\,dx}=\frac{\pi/2}{1}=\frac{\pi}{2}.
$$
So we have that
$$
\int_0^{\pi/2}\sin^{2n}(x)\,dx=\frac{2n-1}{2n}\frac{2n-3}{2n-2}\cdots\frac{1}{2}\frac{\pi}{2}=\frac{(2n)!}{4^n (n!)^2}\frac{\pi}{2}.
$$
Hence an analytic continuation of $\int_0^{\pi/2}\sin^{2n}(x)\,dx $ is
$$
\int_0^{\pi/2}\sin^{2z}(x)\,dx=\frac{\pi}{2}\frac{\Gamma(2z+1)}{4^z \Gamma^2(z+1)}=\frac{\pi}{2}\Gamma(2z+1)4^{-z}\Gamma^{-2}(z+1).
$$
Now differentiate both sides with respect to $z$ which yields
$$
\begin{eqnarray}
2\int_0^{\pi/2}\sin^{2z}(x)\log(\sin(x))\,dx =\frac{\pi}{2} \{2\Gamma'(2z+1)4^{-z}\Gamma^{-2}(z+1)\\
+2\Gamma(2z+1)4^{-z}\Gamma^{-3}(z+1)\Gamma'(z+1)\\
-\log(4)\Gamma(2z+1)4^{-z}\Gamma^{-2}(z+1)\}.
\end{eqnarray}
$$
Finally set $z=0$ and note that $\Gamma'(1)=-\gamma$ to complete the integration:
$$
\begin{eqnarray}
2\int_0^{\pi/2}\log(\sin(x))\,dx&=&\frac{\pi}{2}(-2\gamma+2\gamma-\log(4))\\
&=& -\frac{\pi}{2}\log(4)=-\pi\log(2).
\end{eqnarray}
$$
We conclude that
$$
\int_0^{\pi/2}\log(\sin(x))\,dx=-\frac{\pi}{2}\log(2).
$$
