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I came across this integral:

\begin{align} \int_{-\infty}^{\infty} \mathrm{d}k_1 |f(k_1)| \int_{-\infty}^{\infty} \mathrm{d} k_2 |g(k_2)|& \nonumber \times\bigg\{\lim _{V \to \infty}\frac{1}{2V}\int_{-V}^V \mathrm{d} x e^{-i(k_2-k_1) x}\bigg\} & \end{align} Assuming $f,g\in\mathcal{L}^1(\mathbb{R})$, I was really unsure about how to evaluate the last integral since it seems like if $k_1 \ne k_2$ the integral is $0$ (since it becomes something like $\text{sinc}(V(k1-k2))\to 0$ as $V\to\infty$), whereas if $k_1=k_2$ the integral is 1, which makes it seem like the entire integral acts like $\int\delta(k_2-k_1)$. On the other hand, $k_1$ and $k_2$ both have integrals of their own, so setting $k_1=k_2$ might not be completely legitimate. I also know that $\sin(V(k_1-k_2))/(k_1-k_2))\to\delta(k_2-k_1)$ as $V\to\infty$, so it would seem weird perhaps that this integral also evaluates to a delta function.

I should say I am not looking for much formality, I just want confirmation that this behaves as the integral of the delta function I mentioned before, if that is the case.

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I'm finding it a bit hard to understand what you mean, in particular what "the entire integral acts like $\int\delta(k_2-k_1)$" means or what the two instances of "this" refer to later on. Are you basically asking whether $\lim_{V \to \infty} \frac{1}{2V} \int_{-V}^V e^{-i(k_2 - k_1)x}\,dx$ equals $\delta(k_2 - k_1)$, so that the entire expression at the top becomes $\int_{\mathbb R^2} \lvert f(k_1) g(k_2) \rvert \delta(k_2-k_1) \, dk_2 \, dk_1$? –  epimorphic Feb 15 at 22:17

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