Answer
the integral converges for $p\lt1$
the integral diverges for $p\geq1$
Work Step by Step
$\int_{R}^{1}\frac{dx}{x^{p}}$ = $\frac{x^{-p+1}}{-p+1}|_{R}^{1}$ = $\frac{1}{-p+1}(1-R^{-p+1})$
$p\lt1$ then $-p+1 \gt 0$
$\int_{0}^{1}\frac{dx}{x^{p}}$ = $\lim\limits_{R \to {0^{+}}}$$\frac{1}{-p+1}(1-R^{-p+1})$ = $\frac{1}{-p+1}(1-0)$ = $\frac{1}{-p+1}$
$p\gt1$ then $-p+1 \lt 0$
$\int_{0}^{1}\frac{dx}{x^{p}}$ = $\lim\limits_{R \to {0^{+}}}$$\frac{1}{-p+1}(1-R^{-p+1})$ = $\infty$
$p=1$
$\int_{R}^{1}\frac{dx}{x^{p}}$ = $\int_{R}^{1}\frac{dx}{x}$ = $\ln{x}|_{R}^{1}$ = $-\ln{R}$
$\int_{0}^{1}\frac{dx}{x^{p}}$ = $\lim\limits_{R \to {0^{+}}}$$-\ln{R}$ = $\infty$
