# Double integral of Dirac delta

Is this integral possible? Everything is positive and real. Maybe a substitution of variables?

$$\frac{1}{\Delta t\ \Delta \varepsilon} \int_t^{t+\Delta t} \int_{\varepsilon-\Delta \varepsilon/2}^{\varepsilon+\Delta \varepsilon/2} \delta\left(E - \frac{1}{2 m} \left( \frac{q V \tau}{d} \right)^2 \right) \mathrm{d}E\ \mathrm{d}\tau$$

• Why is $\epsilon$ both the dummy variable of integration and appearing at the upper lower limits? – velut luna Apr 23 '16 at 1:33
• Different variable. I'll change it for clarity. – user1543042 Apr 23 '16 at 1:37
• I think it's zero if $(t,t+\Delta t) \cap (\frac{d}{qV}\sqrt{2m(\epsilon-\Delta\epsilon/2)},\frac{d}{qV}\sqrt{2m(\epsilon+\Delta\epsilon/2)}) = \emptyset$. Otherwise it's one. Not sure. – velut luna Apr 23 '16 at 1:54
• it is of the form $C \int_a^b \int_c^d \delta(y-k x^2) dy dx$, can you compute that one ? – reuns Apr 23 '16 at 2:26

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.