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.
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 $u=(\sin\phi)^2$.
