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In the top answer of this post here why can we interchange the two limits $\lim_{R\rightarrow \infty}$ and $\lim_{\epsilon \rightarrow 0^+}$?

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Note that due to the asymmetry of the integrand, the expression is equal to \begin{align}\lim_{R\to \infty} &\lim_{\epsilon \to 0^+}\int_\mathbb{R} 1_{\{|k| >\epsilon\}}\frac{\phi(k)}{ik}(e^{ikR}+e^{-ikR})\,dk\\ &=\lim_{R\to \infty} \lim_{\epsilon \to 0^+}\int_\mathbb{R} 1_{\{|k| >\epsilon\}}\frac{\phi(k)-1_{\{|k|\leq1\}}\phi(0)}{ik}(e^{ikR}+e^{-ikR})\,dk \\ &=\lim_{R\to \infty} \int_\mathbb{R}\frac{\phi(k)-1_{\{|k|\leq1\}}\phi(0)}{ik}(e^{ikR}+e^{-ikR})\,dk=0,\end{align} where in the end we use the Riemman-Lebesgue Lemma for the function $k\mapsto \frac{\phi(k)-1_{\{|k|\leq1\}}\phi(0)}{ik}$ , which is integrable.

Hidde
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    How is $\frac{\phi(k)-\phi(0)}{k}$ integrable? Replace $\phi(0)$ with $\phi(0) 1_{|k|\le 1}$ and you would be correct. – Mark Viola Apr 09 '23 at 17:22
  • @MarkViola you're totally right, I fixed it now – Hidde Apr 09 '23 at 17:34
  • And now (+1). Well done. – Mark Viola Apr 09 '23 at 18:17
  • @helloworld00, exactly, since $\frac{\phi-1_{|k|\leq1}\phi(0)}{k}$ is in $L^1$, we simply take the $\epsilon$ limit inside the integral in the second equality (dominated convergence theorem), and in the end we use Riemann Lebesgue for this whole function. – Hidde Apr 10 '23 at 10:49