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. 
 A: 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.
A: Regarding the two expressions and your doubt about their equality: 
The equality of $\displaystyle \frac{\Gamma(2z+1)}{4^z\Gamma^2(z+1)}\frac{\pi}{2}$ and $\displaystyle \frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}\frac{\sqrt\pi}{2}$ can be shown by using the fact that $\Gamma(z)\Gamma(z+\frac12)=2^{1-2z}\sqrt{\pi}\Gamma(2z)$ (see wiki):
$$\Gamma(2z+1)=\Gamma\left(2\left(z+\frac12\right)\right) = \frac{\Gamma\left(z+\frac12\right)\Gamma\left(z+1\right)}{2^{-2z}\sqrt{\pi}}$$
Thus,
$$\frac{\Gamma(2z+1)}{4^z\Gamma^2(z+1)}\frac{\pi}{2}=\frac{\Gamma\left(z+\frac12\right)\Gamma\left(z+1\right)}{2^{-2z}\sqrt{\pi}4^z\Gamma^2(z+1)}\frac{\pi}2=\frac{\Gamma(z+\frac{1}{2})}{\Gamma(z+1)}\frac{\sqrt\pi}{2}$$
A: 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
$$
\int_0^{\pi/2}\log(\sin(x))\,dx=-\frac{\pi}{2}\log(2).
$$
