I found the deformation below in a book. Could you explain how is it possible? \begin{equation} \int_\alpha^\beta \left|\frac{d\bf{e}}{ds}\right|ds = \int_\alpha^\beta \left|\frac{d\bf{e}}{dt}\right|\frac{dt}{ds}ds=\int_a^b \left|\frac{d\bf{e}}{dt}\right|dt \end{equation} where $a\leq t\leq b$ corresponds to $\alpha \leq s \leq \beta$.
I think it does not make a big difference, but in that book, $\left|\frac{d\bf{e}}{ds}\right|$ was written as $(\frac{d\bf{e}}{ds} \cdot \frac{d\bf{e}}{ds})^{\frac{1}{2}}$ instead.
The first transformation is a result of using the chain rule. Suppose that $t$ is a function of $s$, then, $$ \frac{d\textbf{e}}{ds} = \frac{d\textbf{e}}{dt}\frac{dt}{ds} \tag{1} $$
Taking the norm of (1) yields, $$ \begin{align} \left| \frac{d\textbf{e}}{ds} \right| =& \left( \frac{d\textbf{e}}{ds} \cdot \frac{d\textbf{e}}{ds} \right)^{\frac{1}{2}} \\ =& \left[\frac{d\textbf{e}}{dt} \cdot \frac{d\textbf{e}}{dt}\left(\frac{dt}{ds} \right)^2 \right]^{\frac{1}{2}} \\ =& \left| \frac{d\textbf{e}}{dt} \right|\frac{dt}{ds} \end{align} $$
The final part is a result of using Integration by Substitution, which gives that, $$ \int_{t(\alpha)}^{t(\beta)}\left| \frac{d\textbf{e}}{dt} \right| dt = \int_{\alpha}^{\beta} \left| \frac{d\textbf{e}}{dt} \right|\frac{dt}{ds} ds $$