Improper Integral $\int_0^1\frac{dx}{x^p}$ Is this integral convergent only for $p<1$?

$$\int_0^1\frac{dx}{x^p}$$


Yes. 
\begin{align}
\int \dfrac{dx}{x^p} = \begin{cases} \dfrac{x^{1-p}}{1-p} + \text{constant} & \text{ for } p \neq 1\\ \log(x) & \text{ for }p=1\end{cases}
\end{align}
From the above, note that for $p\geq1$ as $x \to 0^+$, we have $x^{1-p} \to \infty$ and for $p<1$ as $x \to 0^+$, we have $x^{1-p} \to 0$.
Hence,
\begin{align}
\int_0^1 \dfrac{dx}{x^p} = \begin{cases} \dfrac{1}{1-p} & \text{ for } p < 1\\ \text{does not exist} & \text{ for }p\geq1\end{cases}
\end{align}
