Double integral of Dirac delta Is this integral possible? Everything is positive and real. Maybe a substitution of variables?
$$\frac{1}{\Delta t\ \Delta \varepsilon} \int_t^{t+\Delta t} \int_{\varepsilon-\Delta \varepsilon/2}^{\varepsilon+\Delta \varepsilon/2} \delta\left(E - \frac{1}{2 m} \left( \frac{q V \tau}{d} \right)^2 \right) \mathrm{d}E\ \mathrm{d}\tau$$
 A: Perform first the $E$-integral, which is by definition equal to $1$ if 
$$
\epsilon-\Delta\epsilon/2<\frac{1}{2m}\left(\frac{qV\tau}{d}\right)^2<\epsilon+\Delta\epsilon/2
$$
and zero otherwise. Solving the inequality for $\tau>0$ (assuming $\epsilon-\Delta\epsilon/2$>0)
$$
\sqrt{(\epsilon-\Delta\epsilon/2)2md^2/q^2V^2}<\tau<\sqrt{(\epsilon+\Delta\epsilon/2)2md^2/q^2V^2}\ .
$$
Then you just need to compare $
\sqrt{(\epsilon\pm\Delta\epsilon/2)2md^2/q^2V^2}$ with $t,t+\Delta t$. According to different values of the parameters, you will have to solve different integrals, for example
$$
\int_{
\sqrt{(\epsilon-\Delta\epsilon/2)2md^2/q^2V^2}}^{
\sqrt{(\epsilon+\Delta\epsilon/2)2md^2/q^2V^2}}d\tau\ ,
$$
or 
$$
\int_{t}^{t+\Delta t}d\tau\ ,
$$
or others. The $\tau$-integral to be performed is $\int_\sigma d\tau$, where $$\sigma=[
\sqrt{(\epsilon-\Delta\epsilon/2)2md^2/q^2V^2},
\sqrt{(\epsilon+\Delta\epsilon/2)2md^2/q^2V^2}]\cap [t,t+\Delta t]\ ,$$ so the support $\sigma$ may change in different regions of parameter space.
