Kronecker delta in integral

I am interested in calculating the follwing integral $$I=\lim_{T\to\infty}\frac{1}{T}\int_0^Tdt\iint_0^\infty dE\;dE'e^{i(E-E')t}f(E,E'),$$ for a complicated function $$f$$. One might initially calculate the $$t$$ integral $$I_t(E-E')=\lim_{T\to\infty}\frac{1}{T}\int_0^Tdt\;e^{i(E-E')t}$$ and by first assuming $$E\neq E'$$: $$I_t(E-E')=\lim_{T\to\infty}\frac{1}{T}\frac{e^{i(E-E')T}-1}{i(E-E')}=0.$$ And if $$E=E'$$: $$I_t(0)=\lim_{T\to\infty}\frac{1}{T}\int_0^Tdt=1$$ so $$\lim_{T\to\infty}\frac{1}{T}\int_0^Td t\;e^{i(E-E')t}=\Bigg\{\begin{matrix}1,\;&\text{if }E-E'=0\\0,&\text{else}\end{matrix}\equiv\delta^\star(E-E')$$ which acts as a continous version of the Kronecker delta. My question is if this version of a delta inside an integral has the same properties as a usual dirac delta i.e. if $$I=\int_0^\infty dE\;f(E,E)$$ If not, is the calculation of $$I_t$$ wrong, or does $$I$$ not evaluate to anything 'nice' after all, meaning I am left with $$I=\iint_0^\infty dE\;dE' \delta^\star(E-E')f(E,E').$$

No, if $$\delta^\star$$ was the Dirac delta, then your limit would not be $$1$$ when $$E=E'$$ but $$\infty$$.
• In this case, if your function $$f$$ is sufficiently nice, then since $$\delta^\star$$ is a function that is $$0$$ almost everywhere, you would get $$I = \iint \delta^\star(x-y)\,f(x,y)\,\mathrm d x\,\mathrm d y = 0$$ because the $$\delta^\star(x-y)\,f(x,y)$$ function is $$0$$ exacpt on the set $$\{(x,y)\in\Bbb R^2, x-y=0\}$$ that is a set of measure $$0$$ in $$\Bbb R^2$$. When I say sufficiently nice, it means that you can indeed take the limit in the integral. This would work for instance as soon as $$f$$ is integrable on $$\Bbb R^2$$ by the theorem of dominated convergence.
• Actually, one can notice also that your integral is $$I = \lim_{T\to\infty} \frac{1}{T} \int_0^T \widehat{f}(-t,t)\,\mathrm d t$$ and so if $$f$$ is nice enough, that is for example such that $$\int |\widehat{f}(-t,t)|^p\,\mathrm d t \leq C^p$$ for some constant $$C>0$$, then by Hölder's inequality $$\left|\frac{1}{T} \int_0^T \widehat{f}(-t,t)\,\mathrm d t\right| \leq \frac{C}{T} \left(\int_0^T \mathrm d t\right)^{1-1/p} = \frac{C}{T^{1/p}}$$ and so your integral converges again to $$0$$.