Kronecker delta in integral

Question

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 $$I=\iint_0^\infty dE\;dE' \delta^\star(E-E')f(E,E').$$

Answer 1

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 $0$.