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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.

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$\lim_{L \to \infty} \int_{-L}^L e^{ikx}dk$ 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 : $\lim_{L \to \infty} \int_{-L}^L \hat{\phi}(k)dk = \lim_{L \to \infty} \int_{-\infty}^\infty \phi(x) f_L(x)dx=2\pi \phi(0)$.

reuns
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  • To your first point, I should've noted that those integrals do exist within infinite integrals over $a$ and $b$. Is it then proper to perform outer integrals before taking the $L\rightarrow \infty$ limit, like $\lim_{L\rightarrow \infty} \int_{-\infty}^{\infty} e^{iLx}f(x)\ dx$, or do those infinite bound limits not commute? –  Nov 17 '17 at 22:31
  • @JoshMcK ?? What do you mean and what do you not understand in my answer ? $\int_a^b e^{ikx}dk = \frac{e^{ibx}-e^{iax}}{ix}$ and $\lim_{b \to \infty}\int_a^b e^{ikx}dk =\lim_{b \to \infty} \frac{e^{ibx}-e^{iax}}{ix}$ doesn't converge (it keeps oscillating) for any $x$. But $\lim_{b \to \infty}\int_a^b e^{ikx}dk$ does converge in the sense of distributions (remember $\delta(x)$ is not a function) – reuns Nov 17 '17 at 22:41
  • I mean that the integrals I provided are nested within integrals of the form $\int_{-\infty}^{\infty}da \int_{-\infty}^{\infty} db$. So, even though it indeed oscillates, it is later integrated, and my question is whether that integral could converge. –  Nov 17 '17 at 23:20
  • @JoshMcK The answer is : no. Again what do you not understand in my answer ? – reuns Nov 17 '17 at 23:50
  • I believe I understand your answer; I just don't see what it addresses beyond providing a definition of a delta function. I don't see a convincing reason based on your responses that an integral over the bounds I have described should not converge. –  Nov 19 '17 at 09:15
  • @JoshMcK It diverges in the usual sense, but it converges in the sense of distributions. – reuns Nov 19 '17 at 16:11