Parameter Integral Sin function, Gamma function Given $$ F: A \in\Bbb{R}\rightarrow\Bbb{R}$$ such that $$F(y)=\int\limits_{0}^{\pi/2} (sinx)^y(cosx)^{1-y} \; dx$$
Prove that $F(1/2)=\frac{1}{\sqrt\pi}(\Gamma(3/4))^2$ and then find the maximum domain $A$ such that $F$ exists.
$$\Gamma(s)= \int_0^\infty e^{-t}t^{s-1}dt$$
I arrived at $F(1/2)=\frac{\sqrt2}{2}\int\limits_{0}^{\pi/2}\sqrt\sin2x\;dx$ and that is about all that I can do.
 A: Define
$$
\Gamma(z)=\int_0^{+\infty} t^{z-1}e^{-t}\,dt
$$
and
$$
B(z,w)=\int_0^1 t^{w-1}(1-t)^{z-1}\,dt.
$$
Then you should be able to conclude what you want from the following two results.

It holds that
  $$
B(z,w)=\frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}.
$$

Proof 
We make a calculation, changing coordinates,
$$
\begin{align}
\Gamma(z+w)B(z,w)&=\Gamma(z+w)\int_0^1 t^{w-1}(1-t)^{z-1}\,dt\\
&=[t=u/(1+u)]\\
&=\Gamma(z+w)\int_0^{+\infty} u^{w-1}\Bigl(\frac{1}{1+u}\Bigr)^{z+w}\,du\\
&=\int_0^{+\infty}\int_0^{+\infty} u^{w-1}\Bigl(\frac{1}{1+u}\Bigr)^{z+w} e^{-v}\,dv\,du\\
&=[s=v/(1+u)]\\
&=\int_0^{+\infty}\int_0^{+\infty}u^{w-1}s^{z+w-1}e^{-s(u+1)}\,ds\,du\\
&=\int_0^{+\infty}s^ze^{-s}\int_0^{+\infty}(us)^{w-1}e^{-su}\,du\,ds\\
&=\int_0^{+\infty}s^{z-1}e^{-s}\Gamma(w)\,ds\\
&=\Gamma(z)\Gamma(w).
\end{align}
$$

The following is valid for $a>-1$ and $b>-1$
  $$
\int_0^{\pi/2}(\sin \phi)^a(\cos\phi)^b\,d\phi = \frac{1}{2}B\bigl(\tfrac{a+1}{2},\tfrac{b+1}{2}\bigr).
$$

Proof 
Just let $u=(\sin\phi)^2$.
