Show $\int_{0}^{\epsilon}\rho(x)^{-2}dx= +\infty, \hspace{4mm} \forall \epsilon >0$

Question

let $\rho(x)=\sqrt{x}, \hspace{4mm} \forall x \in \mathbb{R}$

Show : $$ \int_{0}^{\epsilon}\rho(x)^{-2}dx= +\infty, \hspace{4mm} \forall \epsilon >0. $$

My attempt:

\begin{align*} \int_{0}^{\epsilon}\rho(x)^{-2}dx &= \int_{0}^{\epsilon}{1\over x}\,dx\\ &=[ \ln(x) ]_{0}^{\epsilon} \hspace{2mm} \forall \epsilon >0 \\ &=\ln(\epsilon)-\ln(0), \hspace{2mm} \forall \epsilon >0\\ &=\ln(\epsilon), \hspace{2mm} \forall \epsilon >0 \end{align*} ln(x) :Natural logarithm so i'm stuck help please

Answer 1

$$\int_{0}^{\epsilon}{1\over x}dx = \lim_{y \rightarrow 0^{+}}\int_{y}^{\epsilon}{1\over x}dx = \lim_{y \rightarrow 0^{+}} \ (\log \epsilon -\log y) = + \infty$$