# Determining for which values of $p$ the improper integral $\int_1^{\infty}\frac{dx}{ x^2\sqrt{x^p -1}}$ converges

Determining for which values of $p$ the improper integral $\int_1^{\infty}\frac{dx}{ x^2\sqrt{x^p -1}}$ converges. i did comparison test and got for every p > -2 why its wrong?

• Compare "for which positive values of p" and "got for every p > -2". – Did Dec 31 '16 at 8:38
• Ordinarily, the "minus 1" in the expression won't affect convergence. So we can compare it to $\frac{1}{x^2\sqrt{x^p}}=\frac{1}{x^{2+p/2}}$ which converges precisely when $2+p/2>1$, i.e. $p>-2$. However, as Sangchul observed, we also require $p>0$, otherwise $x^p-1\leq 0$ which would make $\frac{1}{x^2\sqrt{x^p-1}}$ undefined. So the answer is $p>0$. – Ben W Dec 31 '16 at 8:42

We assume $p>0$.

Since the integrand is continuous over $(1,\infty)$, potential issues are near $1^+$ and near $\infty$.

As $x \to 1^+$, one has $$\frac{1}{ x^2\sqrt{x^p -1}}=\frac{1}{ x^2\sqrt{(1+(x-1))^p -1}}\sim\frac{1}{\sqrt{p} \sqrt{x-1}}$$ giving the convergence of the given integral.

As $x \to \infty$, one has $$\frac{1}{ x^2\sqrt{x^p -1}}=\frac{1}{ x^{2+p/2}}\cdot \frac{1}{ \sqrt{1-1/x^p}}\sim \frac{1}{ x^{2+p/2}}$$ giving the convergence of the given integral for $2+p/2>1$.

• yes but the answer is for p > 0 – ameen ali Dec 31 '16 at 8:41
• why assuming p > 0 , and you want to find for which p ? – ameen ali Dec 31 '16 at 8:43
• If you consider $p<0$, then how do you define $\sqrt{x^p -1}$ when $x\ge1$? – Olivier Oloa Dec 31 '16 at 8:46

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