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I'd like to calculate $$int_0^1 frac{1}{sqrt{x(1-x)}} dx,$$ using residue calculus. I was given a hint to consider the function $$f(z) = frac{1}{zsqrt{1-frac{1}{z}}}. $$ I thought I was relatively comfortable solving residue calculus problems, until I saw this one. Because I am already stuck before I have even begun.

I'm thinking that $f(z)$ is analytic when $z neq 0$ and when $1-1/z > 0. $ I'm also assuming I should make a branch cut of some sort because of the aforementioned condition on $1-1/z$. But it occured to me that $1-1/z leq 0$ precisely when $0 < z < 1$, and therefore this makes no sense to me since the limits of the integral that I am trying to solve are $0$ and $1$...

So in conclusion, I need a lot of help.

$defth{theta}
defp{pi}
defe{varepsilon}
defg{gamma}
defG{Gamma}
DeclareMathOperator{sech}{sech}$Let
begin{align*}
z &= r_1 e^{ith_1} \
&= 1+r_2 e^{ith_2}
end{align*}
where $r_i>0$ and $0leth_i<2p$.
(Here we choose the branch cuts for the two roots and, with the restriction on $th_i$, are examining the principal branch.)
Then
begin{align*}
g(z) &= frac{1}{sqrt{z(1-z)}} \
&= frac{1}{sqrt{r_1 r_2}} e^{-i(th_1+th_2+p)/2}.
end{align*}
Let $0<ell 1$.
It is a straightforward exercise to show that the branch cut for $g(z)$ extends along the real line from $x=0$ to $x=1$ and that
begin{align*}
lim_{eto 0^+} g(x+ie) &= -lim_{eto 0^+} g(x-ie)
= -frac{1}{sqrt{x(1-x)}}
& textrm{for }0<x<1.
end{align*}
Let $g = sum_{i=1}^4 g_i$ be the counterclockwise contour with
begin{align*}
g_1 &: t-ie,
tin(0,1) \
g_2 &: 1+e e^{ith},
thin[-p/2,p/2] \
g_3 &: 1-t+ie,
tin(0,1) \
g_4 &: e e^{ith},
thin[p/2,3p/2].
end{align*}
One can show that $int_{g_2} = int_{g_4} = 0$ in the limit $eto 0^+$.
(Note that, $int_{g_2}, int_{g_4}sim sqrte$.)
Thus, in the limit $eto 0^+$,
begin{align*}
int_g &to int_{g_1}+int_{g_3} \
&= int_0^1 g(t-ie)dt + int_0^1 g(1-t+ie)dt \
&= int_0^1 g(t-ie)dt - int_0^1 g(t+ie)dt \
&to 2int_0^1 frac{dx}{sqrt{x(1-x)}}.
end{align*}
Thus,
$$int_0^1 frac{dx}{sqrt{x(1-x)}} = frac 1 2 int_g g(z)dz.$$
Now deform the contour to $G: R e^{ith}, thin[0,2p)$, where $R>1$, and consider the limit $Rtoinfty$.
One can show that, for large $|z|$,
$g(z) = -i/z + O(1/z^2)$
and so
$int_G g(z) dz = 2p i(-i) = 2pi$
in the limit $Rtoinfty$.
Thus,
$$int_0^1 frac{dx}{sqrt{x(1-x)}} = p.$$

Avoiding branch cuts

Let $x = frac{1}{2}(1+tanh t)$.
Then
$$int_0^1 frac{dx}{sqrt{x(1-x)}} = int_{-infty}^infty sech t,dt.$$
In the spirit of the question we evaluate this integral using residue calculus.
Note that
begin{align*}
int_{-infty}^infty sech t,dt
&= lim_{eto 0^+}int_{-infty}^infty e^{ie t}sech t,dt \
&= lim_{eto 0^+}int_g e^{ie z}sech z,dz,
end{align*}
where $g$ encircles the poles in the upper half plane in a counterclockwise manner.
The (simple) poles occur at $z = (2n+1)p i/2$ for $n=0,1,ldots$.
It is a straightforward exercise to find the residues,
$$mathrm{Res}_{z=(2n+1)p i/2} e^{ie z}sech z
= (-1)^{n+1}i e^{-(2n+1)p e/2}.$$
Then
begin{align*}
sum mathrm{Res}, e^{ie z}sech z
&= sum_{n=0}^infty (-1)^{n+1}i e^{-(2n+1)p e/2} \
&= -i e^{-pe/2} sum_{n=0}^infty (-e^{-pe})^n \
&= -frac{i e^{pe/2}}{1+e^{pe}} \
&to -i/2.
end{align*}
Thus,
$$int_0^1 frac{dx}{sqrt{x(1-x)}} = 2pi i(-i/2) = p.$$