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This question is based on a doubt in the response given here. Since the answer posted is quite old, I am asking my doubt as a separate question.

We have,

$$\begin{align}\tag 1 F'(\omega)&=-\sqrt{\frac 2\pi}\int_0^\infty \frac{x\sin(\omega x)}{1+x^2}\,dx\\\\ &=-\sqrt{\frac 2\pi}\int_0^\infty \frac{(1+x^2-1)\sin(\omega x)}{x(1+x^2)}\,dx\\\\ &=-\sqrt{\frac 2\pi}\left(\frac\pi 2 -\int_0^\infty \frac{\sin(\omega x)}{x(1+x^2)}\right)\,dx\tag 2 \end{align}$$

Now letting $\omega\to 0$ in (1) gives $0$. But letting $\omega\to 0$ in (2) gives $-\sqrt{\frac{\pi}{2}}$. Why is the former wrong and the latter correct?

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    Because 2 is not correct since the integral of $\frac{\sin(\omega x)}{x}$ is $0$ if $\omega =0$ – Conrad Apr 25 '19 at 11:59
  • Can you explain that further please? It is because $\frac{\sin (\omega x)}{x}$ is $0$ that I conclude that (2) gives $-\sqrt{\pi/2}$. –  Apr 25 '19 at 12:03
  • made it an answer as it gets too long to detail all steps – Conrad Apr 25 '19 at 12:28
  • I apologize but I was hasty and the answer in the original question was correct; the issue is that both 1 and 2 are correct only for $\omega >0$, but in 1 the convergence is non-uniform on the full interval $(0, \infty)$ and in 2 it is – Conrad Apr 25 '19 at 12:49
  • edited my answer to reflect the above with proof for both assertions – Conrad Apr 25 '19 at 13:31

1 Answers1

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I apologize as my original answer was incorrect; the issue is that actually the integral in 1 is not uniformly convergent near $0$ but the one in 2 is, so we cannot switch limit and integral in 1, but we can in 2.

Proof for 1: note that $\sin(\frac{\pi x}{4N}) \ge \frac{\sqrt 2}{2} \ge .6, N \le x \le 2N$, so for $\omega_N=\frac{\pi }{4N} \to 0$ and $A_N=N, B_N=2N \to \infty$, $\int_N^{2N} \frac{x\sin(\frac{\pi x}{4N})}{1+x^2}\,dx \ge .6\int_N^{2N} \frac{x}{1+x^2}\,dx \ge .3 \log 2$ since $\frac{x}{1+x^2} \ge \frac{1}{2x}$ there, hence the convergence is not uniform for $\omega \to 0$

Proof for 2: For $0 \le x \le 1, 0 \le \omega \le 1, 0 \le \frac{\sin(\omega x)}{\omega x} \le 1$,

so $\int_0^\infty|{\frac{\sin(\omega x)}{x(1+x^2)}}|dx \le \int_0^1{\frac{\omega}{(1+x^2)}}dx+\int_1^\infty{\frac{1}{x(1+x^2)}}dx < 1+1<2$, so the integral in 2 is absolutely (independent of $\omega$ near zero) hence uniformly convergent for $\omega \to 0$

Conrad
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  • Thank you for your detailed response. Can you confirm then that the procedure outlined in the following link wrong: https://math.stackexchange.com/questions/2236261/find-the-fourier-cosine-transform-of-the-function-defined-by-displaystyle-fx/2236287#2236287 –  Apr 25 '19 at 12:38
  • I think I am actually wrong and it is the integral in 1 that doesn't converge uniformly and the one in two that does, so indeed the answer given to the original question is correct – Conrad Apr 25 '19 at 12:45
  • Thank you. I have another question. Why can't we take $s=0$ instead of taking limit $s\to 0$? –  Apr 25 '19 at 16:30
  • The original problem has $F(\omega)$ defined as an integral in $x$ for which $F'(\omega)$ computed by formal differentiation within the $x$ integral is absolutely and uniformly convergent only for $\omega > \omega_0 >0$, so by the usual theorems of calculus regarding differentiation within an integral, $F'(\omega)$ is equal to the formal differentiation within the integral only for $\omega > \omega_0 >0$ and we need the manipulation above to be able to take the limit at zero and compute $F'(0)$ – Conrad Apr 25 '19 at 18:26
  • It is definitely true that substituting $0$ within equation 1 above makes sense and gives zero, but it is not true that $F'(0)=0$ since the limit of the integral in equation 1 as $\omega \to 0$ is not the integral of the limit (namely, $0$) but $-\sqrt{\frac{\pi}{2}}$ as is proved rigorously in equation 2 – Conrad Apr 25 '19 at 18:31
  • So to summarize: The integral given in (1) is uniformly convergent for $(\omega_0,\infty) $ while that in (2) is uniformly convergent for $(0,\infty) $. Right? If this is correct I have another question. Since (1) defines $F'(\omega)$, it does so only for $(\omega_0,\infty)$,. How is it possible that the domain gets extended to $(0,\infty) $ in (2) and to $[0,\infty) $ when we take the limit $\omega\to 0$? –  Apr 27 '19 at 12:36
  • $\omega_0 >0$ arbitrary so 1 defined on $(0,\infty)$ since for any $a>0$ there is $0<\omega_0<a$; for 2 the integral is uniformly convergent at 0 so we can take the limit; this is standard procedure when you have functions defined by integrals that are not convergent at limit points, manipulating them in a way or another to make the potentially singular behaviour explicit – Conrad Apr 27 '19 at 13:00