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.
 A: $\newcommand{\B}{\mathfrak{B}}\newcommand{\d}{\,\mathrm{d}}\newcommand{\res}{\operatorname{Res}}$It is known that $\B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ whenever $\Re x,y>0$ (and possibly for a more general case, I don't know). I use $\B$ rather than $\beta$ as $\beta$ can be confused with the Dirichlet Beta function.
Specifically, you have: $$\frac{\Gamma\left(n+\frac{3}{4}\right)}{\Gamma\left(n+\frac{5}{4}\right)}=\frac{1}{\Gamma\left(\frac{1}{2}\right)}\cdot\frac{\Gamma\left(n+\frac{3}{4}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(n+\frac{3}{4}+\frac{1}{2}\right)}=\pi^{-1/2}\cdot\B\left(n+\frac{3}{4},\frac{1}{2}\right)$$Which evaluates to: $$\pi^{-1/2}\int_0^1t^{n-1/4}(1-t)^{-1/2}\d t$$(for $n\in\Bbb N_0$).
So: $$\sqrt{\pi}\cdot S=\sum_{n\ge0}(-1)^n\int_0^1t^{n-1/4}(1-t)^{-1/2}\d t=\int_0^1(1-t)^{-1/2}\left(\sum_{n\ge0}(-1)^nt^{n-1/4}\right)\d t$$If you accept the sum-integral interchange, this is e.g. justified by uniform convergence of geometric series. Evaluating the geometric series finds: $$\sqrt{\pi}\cdot S=\int_0^1(1-t)^{-1/2}\cdot t^{-1/4}\frac{1}{1-(-t)}\d t=\int_0^1\frac{t^{-1/4}}{(1+t)\sqrt{1-t}}\d t$$
As claimed.

This integral can be evaluated using complex analysis. Unfortunately I can't add this answer to the original post since that one was closed. Firstly, we need to have good symmetry - the fourth root does not match with the singular square root. So I let $t\mapsto t^2$ to instead focus on: $$2\int_0^1\frac{t^{1/2}}{(1+t^2)\sqrt{1-t^2}}\d t$$To use complex analysis, we need to carefully define the two square roots to exploit certain symmetries. Let $\log_1$ denote the logarithm based on $0\le\arg z<2\pi$, and $\log_2$ the principal logarithm based on $-\pi<\arg z\le\pi$. Let $t^{1/2}$ denote (for complex $t$, now) $\exp\left(\frac{1}{2}\log_1(t)\right)$ and $(1-t^2)^{-1/2}$ denote $\exp\left(-\frac{1}{2}\log_2(1-t^2)\right)$. Assemble these two different definitions of square root to $f:z\mapsto\frac{2}{1+z^2}\cdot z^{1/2}(1-z^2)^{-1/2}$, as a function $\Bbb C\setminus\{\pm i\}\to\Bbb C$. I won't check this here, but it is a straightforward-to-verify claim that $f$ is meromorphic on $\Bbb C\setminus[0,1]$. The point is, the two square roots individually have branch cuts $[0,\infty)$ and $[1,\infty)$, but the "jumps" (as the argument goes $-\pi\to\pi$ or $0\to2\pi$) that cause these discontinuities cancel each other out on $(1,\infty)$. This is the most important point.
Given all this, let $D_\epsilon$ (for $\epsilon>0$) denote the "dogbone" contour that runs clockwise around $[0,1]$ as follows: $i\epsilon\to1+i\epsilon$ as a straight line, $1+i\epsilon\to1-i\epsilon$ as an outward clockwise arc of radius $\epsilon$, $1-i\epsilon\to-i\epsilon$ as a straight line, and $-i\epsilon\to i\epsilon$ as an outward clockwise arc of radius $\epsilon$. Since $f$ has zero "residue at infinity" (it's straightforward to check that $\lim_{|z|\to\infty}z\cdot f(z)=0$) I know: $$2\pi i\cdot\res_{z=i}f+2\pi i\cdot\res_{z=-i}f=\oint_{D_\epsilon}f(z)\d z$$But it's also straightforward to check that the integrals on the semicircular arcs vanish, e.g.: $$\left|\int_{-\pi/2}^{\pi/2}f(\epsilon e^{-it})\cdot(-i\epsilon e^{-it})\d t\right|\le\int_{-\pi/2}^{\pi/2}\epsilon^{1/2}\cdot\epsilon\cdot\mathcal{O}(1)\d t\to0$$
Now fix $x\in(0,1)$. As $z=x+i\epsilon\to x$, $f(z)\to\frac{2}{1+x^2}\cdot\frac{x^{1/2}}{\sqrt{1-x^2}}$. But as $z=x-i\epsilon\to x$, $f(z)\to\frac{2}{1+x^2}\cdot\frac{x^{1/2}}{\sqrt{1-x^2}}\color{red}{(-1)}$, where the $-1$ term arises from $\exp\left(\frac{1}{2}(2\pi i)\right)=-1$, recalling that $z^{1/2}$ was defined via $0\le\arg z<2\pi$.
As the poles are simple, the residue calculation is also simple (being careful with branch definitions) and you can find the left hand side to equal $\pi$. So: $$\pi=\int_{0+i\epsilon}^{1+i\epsilon}f(z)\d z-\int_{0-i\epsilon}^{1-i\epsilon}f(z)\d z+o(1)\to\int_0^1f(x)\d x-\int_0^1f(x)(-1)\d x$$That is: $$\pi=2\int_0^1\frac{2x^{1/2}}{(1+x^2)\sqrt{1-x^2}}\d x\\\int_0^1\frac{x^{-1/4}}{(1+x)\sqrt{1-x}}\d x=\frac{\pi}{2}$$
A: 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 $\sqrt{\pi}$:
$$\sqrt{\pi}\sum_{n=0}^\infty (-1)^n\frac{\Gamma({n+\frac{3}{4}})}{\Gamma({n+\frac{5}{4}})}=\intop_0^1 \frac{x^{\frac{1}{4}}}{(1+x)\sqrt{1-x}}dx$$
