How can I prove that: $$\sum_{n=0}^{\infty}(-1)^{n}\frac{\Gamma(n+\frac{3}{4})}{\Gamma(n+\frac{5}{4})}=\frac{\sqrt{\pi}}{2}$$
The closest I've gotten is that if I call the sum S, then using the definition of the Beta function: $$\sqrt{\pi}S=\intop_0^1\frac{x^{-\frac{1}{4}}}{(1+x)\sqrt{1-x}}dx$$ But I don't really know where to go from here.
First we substitute $x=u^4$ to obtain $$\int_0^1{x^{-1/4}\over(1+x)\sqrt{1-x}}\,dx =4\int_0^1{u^2\over(1+u^4)\sqrt{1-u^4}}\,du.$$ Then we substitute $u=1/v$ to get
$$4\int_1^\infty{v^2\over(1+v^4)\sqrt{v^4-1}}\,dv.$$
Then substitute $v=\sqrt{\tan y}$:
$$\begin{align*} 4\int_{\pi/4}^{\pi/2}{\tan y\over(1+\tan^2y)\sqrt{\tan^2y-1}} {\sec^2y\over2\sqrt{\tan y}}\,dy &=2\int_{\pi/4}^{\pi/2}{\sqrt{\tan y}\over\sqrt{\tan^2y-1}}\,dy \\ &=2\int_{\pi/4}^{\pi/2}{\sqrt{\sin y\cos y}\over\sqrt{\sin^2y-\cos^2y}}\,dy\\ &=\sqrt2\int_{\pi/4}^{\pi/2}{\sqrt{\sin(2y)}\over\sqrt{-\cos(2y)}}\,dy\\ &=\sqrt2\int_{\pi/4}^{\pi/2}\sqrt{-\tan(2y)}\,dy. \end{align*}$$ Now we substitute $w=-2y$ to get $${\sqrt2\over2}\int_{-\pi}^{-\pi/2}\sqrt{\tan w}\,dw ={1\over\sqrt2}{\pi\over\sqrt2} ={\pi\over2}$$ as needed. (The integral $\int\sqrt{\tan x}\,dx$ is standard; see for example here.)
