Unable to understand usage of the Beta function in a previous question In this question, OP says they used the definition of the Beta function on, $$S := \sum_{n=0}^{\infty}(-1)^{n}\frac{\Gamma(n+\frac{3}{4})}{\Gamma(n+\frac{5}{4})}=\frac{\sqrt{\pi}}{2}$$ to get, $$\sqrt{\pi}S=\intop_0^1\frac{x^{-\frac{1}{4}}}{(1+x)\sqrt{1-x}}dx$$ I can't figure out how they ended up with this equation. I tried to think of it as choosing values for $x$ and $y$ to end up on this integral, but I can't figure out which ones they used. I do know the integral definition of the Beta function which is $$\beta(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt$$ Any hints would be helpful.

I'm the OP. This is how I did it:
$\sum_{n=0}^\infty (-1)^n\frac{\Gamma({n+\frac{3}{4}})}{\Gamma({n+\frac{5}{4}})}=\sum_{n=0}^\infty (-1)^n\frac{\Gamma({n+\frac{3}{4}})}{\Gamma({n+\frac{3}{4}+\frac{1}{2}})}$.
Now multipling by $\Gamma(\frac{1}{2})=\sqrt{\pi}$:
$$\sum_{n=0}^\infty (-1)^n\Gamma(\frac{1}{2})\frac{\Gamma({n+\frac{3}{4}})}{\Gamma({n+\frac{3}{4}+\frac{1}{2}})}$$
The Gamma terms are just $B(\frac{1}{2},n+\frac{3}{4})$
So we get:
$$\sum_{n=0}^\infty (-1)^nB(\frac{1}{2},n+\frac{3}{4})$$
Using the integral form of the Beta function:
$$\sum_{n=0}^\infty (-1)^n \intop_0^1 x^{n+\frac{3}{4}-1}(1-x)^{\frac{1}{2}-1}dx$$
Now, exchanging the sum and the integral (The sum defintely converges, and the integral does too)
$$ \intop_0^1 \sum_{n=0}^\infty (-1)^n x^n*x^{-\frac{1}{4}}(1-x)^{-\frac{1}{2}}dx$$
And finally using the formula for a geometric series:
$$ \intop_0^1 \frac{x^{\frac{1}{4}}}{(1+x)\sqrt{1-x}}dx$$
And since the only change we did to the original sum was Multiply by $\sqrt{\pi}$:
$$\sqrt{\pi}\sum_{n=0}^\infty (-1)^n\frac{\Gamma({n+\frac{3}{4}})}{\Gamma({n+\frac{5}{4}})}=\intop_0^1 \frac{x^{\frac{1}{4}}}{(1+x)\sqrt{1-x}}dx$$
