 Solution

On an attracting branch of the critical manifold, $w=f(v)$ to leading order. The chain rule therefore givesThe initial fast segment ends at $v=1$, and the first slow phase ends at the right fold point of a critical manifold, $v_R$. HenceOn this interval $f^{\prime }(v)<0$ and $\gamma v-f(v)>0$. The integrand is negative, but the upper limit is smaller than the lower limit, so $T_1>0$. The numerator tends to zero at the fold while the denominator remains positive; the leading integral is finite. Fast segments and the fold delay give vanishing corrections as $\epsilon \to 0^{+}$.