As a physicist sometimes I do operations without checking if the conditions to do such operations are satisfied. Most of the times all functions that we interact with are very well-behaved so it goes well. This time though I wanted to check because this operations seems a little bit fishy.
What I wanted to evaluate was the following integral
$$ \int_{\mathbb R}\int_0^t e^{ikx}e^{-Dk^2(t-t')}f(x,t')\delta(k)\,\mathrm dt'\,\mathrm dk $$
where
$$ f(x,t) =\begin{cases}1&0<t<\tau \\ 0 &t>\tau\end{cases} $$
I've got this integral as part of the solution of the heat equation PDE with an external force $f(x,t)$, doing a Fourier transform and then anti-transforming.
To evaluate it I've changed the oder of integration to get rid of the Dirac delta which simplifies a lot the calculation
$$ \int_0^t\int_{\mathbb R} e^{ikx}e^{-Dk^2(t-t')}f(x,t')\delta(k)\,\mathrm dk\,\mathrm dt' = \int_0^t f(x,t')\,\mathrm dt' $$
and proceded to evaluate my solution for the PDE.
Question: Is this operation allowed? Does the function satisfie the requests of Fubini's theorem?
Try it directly, $$ \int_0^t \int_{\mathbb{R}}e^{ikx}e^{-Dk^2(t-t')}\delta(k)dk f(t')dt' = \int_0^t 1\cdot f'(t')dt' \\ \int_{\mathbb{R}}\left(\int_{0}^{t}e^{-Dk^2(t-t')}f(t')dt'\right) e^{ikx}\delta(k)dk= \left.\int_{0}^{t}e^{-Dk^2(t-t')}f(t')dt' \right|_{k=0} $$ It looks like it works out just fine.