# For What Values of $$p$$ does $$\int_{1}^{\infty} \frac{1}{x^{2} \sqrt{x^{p}-1}} dx$$ Converge?

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

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