So one may prove the second equality of the question (the so-called Weyl integral formula) in the following way:

For every 
$H\in\mathfrak{h}=Lie(T)$ we denote
$$
Q(H):=\prod_{\alpha\in\Delta^+}(e^{\pi i(\alpha,H)}-e^{-\pi i(\alpha,H))}).
$$
We think of the roots of $G$ as elements of a fixed Weyl chamber (which we denote by $C_0$) of $\mathfrak{h}$.

The Weyl group $W$ is generated by reflexions $r_j$. For every $r_j$ there is 
an associated simple root $\alpha_j$ such that $r_j(\Delta^+)=(\Delta^+-\{\alpha_j\})\cup\{-\alpha_j\}$. Using this observation one sees that $Q(r_jH)=-Q(H)$ and hence
$Q(\sigma H)=sign(\sigma)Q(H)$. Now since $Q(H)$ is an alternating function with respect to the $W$-action one has that
$$
Q(H)=\sum_{\sigma\in W}sign(\sigma)e^{2\pi i(\sigma(\delta),H)}+\mbox{possible other terms}.
$$
Now the key observation is to note that $\delta$ is the only vector of the form $\frac{1}{2}\sum \pm\alpha$ which is in 
$C_0$. It follows from this that there is in fact no other terms!