I'm perfectly comfortable with the idea of different integral representations of the Dirac delta function when the bounds are infinite, such as $\int_{-\infty}^{\infty}dk\ e^{-ikx} = 2\pi \delta(x)$, which could also be written as $\lim\limits_{L \to \infty} \int_{-L}^Ldk\ e^{-ikx}$. In a physics context, I would like to enforce the ordering of fermionic particles in a 1-dimensional space. After Fourier transforming into momentum space, I attempted to do the ordering by using Heaviside step functions as $$\int_{-\infty}^{\infty} dx_2\ \int_{-\infty}^{\infty} dx_1\ e^{iax_1} e^{ibx_2}\ \delta(x_1-c)\ \Theta(x_2-x_1).$$ The presence of the delta function and theta function together would (perhaps naively) reduce this expression to $$ \lim\limits_{L \to \infty} \int_c^L dx_2\ e^{iac} e^{ibx_2}. $$ My hope was that, once combined with other terms not included here, factors involving $L$ would either cancel or group together to form legitimate integral representations of delta functions, i.e., both bounds infinite. I'm also aware that as a distribution rather than a proper function, the delta function behaves differently; I just don't know the subtleties involved here.
This post posed a similar question, only without the $\delta(x_1-c)$ term; in the responses, it was stated that products of distributions can be very tricky. Is there any hope in proceeding in this manner, or is this result simply divergent or ill-defined? It's not clear to me whether the presence of the step function entirely invalidates the interpretation of this expression as a Fourier transform, either.
$\lim_{L \to \infty} \int_{-L}^L e^{ikx}dk$ diverges for every $x$, forget about it.
Let $f_L(x) = \int_{-L}^L e^{ikx}dk = L\frac{2\sin(Lx)}{Lx}= L \,f_1(Lx)$ and $F_L(x) =\int_{-\infty}^x f_L(t)dt = F_1(Lx)$.
What is true is that $$\lim_{L \to \infty} F_L(x) =\lim_{L \to \infty} F_1(Lx)=2\pi 1_{x > 0}$$ where the convergence is locally uniform away from $x=0$, and the convergence is in $L^1$ around $x=0$.
Thus for any $\phi \in L^1,\phi' \in L^1$ :
$$\lim_{L \to \infty} \int_{-\infty}^\infty \phi(x) f_L(x)dx =-\lim_{L \to \infty} \int_{-\infty}^\infty \phi'(x) F_L(x)dx=-\int_{-\infty}^\infty \phi'(x) 2\pi 1_{ x >0}dx = 2\pi\phi(0)$$ which is precisely the definition of $f_L \to 2\pi \delta$ in the sense of distributions.
Note how this proves the Fourier inversion theorem : $\lim_{L \to \infty} \int_{-L}^L \hat{\phi}(k)dk = \lim_{L \to \infty} \int_{-\infty}^\infty \phi(x) f_L(x)dx=2\pi \phi(0)$.