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What is a good starting point for approaching the following integral?

$$\int_t^z \frac{\mathrm{d}x}{\left(z - x\right)^{1 - \alpha} (x - t)^\alpha} $$

I have tried several change of variables (with the intent of changing the integral bounds to $0$ and $1$), but none of them seemed to have helped. The intuition behind this approach was that the integrand appeared to look somewhat like the beta function.

Another tactic that I tried was to rewrite the integral as follows:

$$\int_0^z \frac{\mathrm{d}x}{\left(z - x\right)^{1 - \alpha} (x - t)^\alpha} - \int_0^t \frac{\mathrm{d}x}{\left(z - x\right)^{1 - \alpha} (x - t)^\alpha} $$

I then proceeded to apply a separate change of variable to each integral. However, I have been unable to make much progress.

The value is apparently $\frac{\pi}{\sin{\pi \alpha}}$, but I have been unable to show it. A hint, rather than a complete solution, would be appreciated.

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  • $\begingroup$ Try approaching from the reflection formula for the $\Gamma$ function. $\endgroup$ – Henricus V. Mar 19 '16 at 16:12
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    $\begingroup$ Hint: What substitution transforms $(t,z)$ into $(0,1)$ ? $\endgroup$ – Lucian Mar 19 '16 at 16:14
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    $\begingroup$ @Lucian $u = \frac{x - t}{z - t}$ seems to work $\endgroup$ – ra1nmaster Mar 19 '16 at 16:16
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$$u=\frac{x-t}{z-t}\implies du=\frac1{z-t}dx\implies$$

$$\int_t^z(z-x)^{\alpha-1}(x-t)^{-\alpha}dx=\int_0^1(1-u)^{\alpha-1}\color{red}{(z-t)^{\alpha-1}}u^{-\alpha}\color{red}{(z-t)^{-\alpha}}\,du\color{red}{(z-t)}=$$

$$=\int_0^1 u^{-\alpha}(1-u)^{1-\alpha}du=B(1-\alpha,\,2-\alpha))$$

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    $\begingroup$ You changed $(1-u)^{\alpha-1}$ to $(1-u)^{1-\alpha}$ in your solution. $\endgroup$ – user5713492 Mar 20 '16 at 3:30

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