Contour integration around a closed loop By considering the contour integral $$\int z^{5/4}(z-1)^{-1/4} dz$$
around a closed loop C that encircles the real interval $[0,1]$, show that $$\int^1_0 x^{5/4}(x-1)^{-1/4} dx = \frac{5\pi}{16\sqrt{2}}$$
Previously I have that the function $$f(z)  = z^p(z-1)^q$$ where the branch of the function is chosen such that $-\pi < arg(z) \le \pi$ and $-\pi < arg(z-1) \le \pi$ and that $p + q =$ integer and a branch is not needed on the negative real axis.
I have calculated that $f(x+i0) = x^p((1-x)e^{-i\pi})^q$ and $f(x-i0) = x^p((1-x)e^{i\pi})^q$.
Please can someone give me a hint on how to approach the contour integration. I'm not sure how to start or where to begin! How will it involve using $f(x+i0)$ and $f(x-i0)$ as I had to calculate them at the start so I'm sure it's relevant to the integral.
 A: METHOD 1: Using Real Analysis
Let $I$ be the integral given by 
$$\begin{align}
I&=\int_0^1 x^{5/4}(1-x)^{-1/4}\,dx\\\\
&=B(9/4,3/4)\\\\
&=\frac{\Gamma(9/4)\Gamma(3/4)}{\Gamma(3)}\\\\
&=\frac{\frac54 \frac14 \Gamma(1/4)\Gamma(3/4)}{2}\\\\
&=\frac5{32}\Gamma(1/4)\Gamma(3/4)\\\\
&=\frac5{32}\frac{\pi}{\sin(\pi/4)}\\\\
&=\frac{5\pi}{16\sqrt 2}
\end{align}$$

METHOD 2: Using Complex Analysis
We will analyze the contour integral $J$ given by 
$$J=\oint_C z^{5/4}(1-z)^{-1/4}\,dz$$
where $C$ is the classical dog-bone contour around the line segment from $(0,0)$ to $(1,0)$.  We can then write
$$\begin{align}
J&=(1-i)\int_0^1 x^{5/4}(1-x)^{-1/4}\,dx \tag 1\\\\
&=2\pi i \text{Res}\left(z^{5/4}(1-z)^{-1/4},z=\infty\right)\\\\
&=-2\pi i \text{Res}\left(\frac{z^{-5/4}(1-z^{-1})^{-1/4}}{z^2},z=0\right)\\\\
&=-2\pi i \text{Res}\left(\frac{(z-1)^{-1/4}}{z^3},z=0\right)
\end{align}$$
Note that in a neighborhood of $z=0$, we can write
$$(z-1)^{-1/4}=e^{i\pi/4}-\frac14 e^{i5\pi/4}z+\frac5{32}e^{i9\pi/4}z^2+O(z^3)$$
Therefore, we have
$$J=-2\pi i \left(\frac{5}{32\sqrt 2}(1+i)\right) \tag 2$$
Setting $(1)$ and $(2)$ equal reveals
$$I=\frac{5\pi}{16\sqrt 2}$$
as expected!
A: By considering a keyhole-type contour of outer radius $R$ about the branch cuts and applying Cauchy's theorem, one gets that
$$e^{-i \pi/4} \int_0^1 dx \, x^{5/4} (1-x)^{-1/4} + e^{i \pi/4} \int_1^0 dx \, x^{5/4} (1-x)^{-1/4} \\+ i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \, R^{5/4} e^{i (5/4) \theta} (R e^{i \theta} - 1)^{-1/4}  = 0$$
Consider the limit as $R \to \infty$:
$$i \sqrt{2}  \int_0^1 dx \, x^{5/4} (1-x)^{-1/4} = i R^2 \int_{-\pi}^{\pi} d\theta \, e^{i 2 \theta} \, \left (1 - \frac1{R e^{i \theta}} \right )^{-1/4}$$
We Taylor expand the factor in parentheses in the integrand on the RHS.  Only the second-order term makes any contribution to the integral, which is
$$\left (1 + \frac1{R e^{i \theta}} \right )^{-1/4} = 1 -\frac1{4 R e^{i \theta}} + \frac1{2!} \left (-\frac14 \right ) \left (-\frac54 \right ) \frac1{R^2 e^{i 2 \theta}} + \cdots$$
Thus,
$$ \int_0^1 dx \, x^{5/4} (1-x)^{-1/4} = \frac{5}{32 \sqrt{2}} 2 \pi = \frac{5 \pi}{16 \sqrt{2}}$$
