This was a question posed at Mathematics Stack Exchange. The obvious solutions were posted before I saw the question so this is an alternative solution that is interesting yet inefficient.

Let \(y=1/x\), then \(z=y^{-p}-1\)

\begin{align}

\int\limits_{1}^{\infty} \frac{1}{x^{2} \sqrt{x^{p}-1}} dx &= \int\limits_{0}^{1} \frac{1}{\sqrt{y^{-p}-1}} dy \\

&= \frac{1}{p} \int\limits_{0}^{\infty} \frac{z^{-1/2}}{(z+1)^{1+1/p}} dz \\

&= \frac{1}{p} \mathrm{B}\left(\frac{1}{2},\frac{1}{p} + \frac{1}{2} \right) \\

&= \frac{\sqrt{\pi}\Gamma \left( \frac{1}{p} + \frac{1}{2} \right)}{\Gamma \left( \frac{1}{p} \right)}

\end{align}

The integrand and the limits of integration require that \(p \gt 0\). From the

beta function we have \(\mathrm{Re} (\frac{1}{p} + \frac{1}{2}) \gt 0\), \(p \lt -2\) and \(p \gt 0\). Thus we

must have \(p \gt 0\).

We have used the following integral definition of the beta function

\begin{equation}

\mathrm{B}(x+1,y+1) = \int\limits_{0}^{\infty} \frac{z^{x}}{(1+z)^{x+y+2}} dz

\end{equation}