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$$f(\omega,u):=\frac1{\omega+iu}$$ where $i$ is the imaginary unit number. We see that the integral of a Fourier transform $$\int_1^\infty du\int_{-\infty}^\infty d\omega\,f(\omega,u)\,e^{-i\omega x}=2\pi i\int_1^\infty du\,e^{-ux}=2\pi i\frac{e^{-x}}x$$ for $u>0$, $x>0$, while interchanging the order of integration, $$\int_{-\infty}^\infty d\omega \,e^{-i\omega x}\int_1^\infty duf(\omega,u)$$ is not even integrable in the first integration. Is there a theory, say distribution or generalized function, to treat this kind of double integral involving the Fourier transform?

On the other hand, if we integrate by part the Fourier transform $$\int_{-\infty}^\infty d\omega\,f(\omega,u)\,e^{-i\omega x}=\frac1 {ix}\int_{-\infty}^\infty d\omega\,\partial_\omega f(\omega,u)\,e^{-i\omega x},$$ $$\partial_\omega f(\omega,u)=-\frac1{(\omega+iu)^2}.$$ We interchange the order of integration and obtain a finite value $$\int_1^\infty du\,\partial_{iu} f(\omega,u)=if(\omega,u=1)=\frac i{\omega+i}$$ which is already shown to be Fourier transformable. It is easy to check the two orders of integration agree.

How do we turn this trick into a coherent theory? Does, say tempered distribution, help?

Hans
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  • Yes Fubini's Theorem arising in measure theory gives you criteria which if met, allow you to safely interchange the order of integration. Essentially if the integrand is absolutely convergent then you can switch the orders. – FofX Feb 18 '19 at 00:01
  • @FofX: Have you even checked whether the condition of the Fubini's theorem applies or the integrand is absolutely convergent? – Hans Feb 18 '19 at 01:48
  • In your case it does not converge absolutely. You can see this by noting that $\vert{e^{i\omega x}}\vert = 1$ and $\vert{f(\omega,u)}\vert=1/\sqrt{\omega^2+u^2}$ so the absolute value of the integrand as whole is $\vert{e^{i\omega u}f(\omega,u)}\vert=1/ \sqrt{\omega^2+u^2}$, and the integral over this does not converge. – FofX Feb 18 '19 at 02:59
  • Check out: https://en.wikipedia.org/wiki/Fubini%27s_theorem, and take a look at the section titled Fubini-Tonelli Theorem. Hopefully that will be helpful to you. – FofX Feb 18 '19 at 03:09
  • @FofX: Exactly. My question originates from the fact, which you have now just discovered, that the Fubini-Tonelli theorem does not apply, at least directly. Bear in mind the thoerem is only a sufficient not necessary condition for interchanging the order of integration. My question is thus how one treats these cases, particularly for Fourier transform, beyond the reign of the Fubini-Tonelli theorem. Integration by parts, like I have done in my question, twice would have pushed the problem into the theorem's realm. But I wonder if there is a better and more systematic theory for these cases. – Hans Feb 18 '19 at 04:32
  • @FofX: Besides, oscillation of $e^{i\omega}x$ for real $\omega$ and $x$ allows the companion function, say, $f$ to not be absolutely integrable but still makes the whole integrand integrable. You can see that the integration by parts once, that I have performed in the question, establishes the order interchange yet still does not make the integrand absolutely integrable. That means the oscillation gives more leeway than what the Fubini-Tonelli theorem demands directly. – Hans Feb 18 '19 at 04:51
  • I think it has something to do with $\frac{\partial }{\partial\omega} f(\omega,u)$ being square integrable. – Random Variable Feb 01 '22 at 13:54
  • @RandomVariable: Could you please elaborate? – Hans Feb 01 '22 at 19:51
  • It might be possible top prove that if $ \int_{\mathbb{R}^{2}} g(\omega,u) , \mathrm d \omega , \mathrm d u $ and $ \int_{\mathbb{R}^{2}} g(\omega,u) , \mathrm du , \mathrm d \omega $ are square integrable (but not necessarily absolutely integrable), then $$\int_{\mathbb{R}^{2}} g(\omega,u) e^{-i \omega x} , \mathrm d\omega , \mathrm du = \int_{\mathbb{R}^{2}} g(\omega,u) e^{-i \omega x} , \mathrm d u , \mathrm d \omega. $$ I vaguely remember reading something about this a long time ago, but I don't know if I'm remembering correctly what I read. – Random Variable Feb 01 '22 at 20:37
  • @RandomVariable: Well, this can not be right. As my question shows that it does not apply to $f$ which is square-integrable --- so long as $\omega+iu\ne0$. – Hans Feb 01 '22 at 21:32
  • $$\int_{1}^{\infty} \int_{-\infty}^{\infty} \left|\frac{1}{\omega + i u} \right|^{2} , \mathrm dw , \mathrm du =\int_{1}^{\infty} \int_{-\infty}^{\infty} \frac{1}{\omega^{2}+u^{2}} , \mathrm d\omega , \mathrm d u = \int_{1}^{\infty} \frac{\pi}{u} , \mathrm du > \infty $$ – Random Variable Feb 01 '22 at 21:57
  • @RandomVariable: Oh, I was thinking of one integration variable which was not sufficient. I agree that we should talk about integrability in both variables as you have done. It would be greatly appreciated if you could find the specifics, source, and references of that proposition. – Hans Feb 02 '22 at 10:19
  • The closet thing I can find is Plancherel's theorem for the Fourier transform written in the form $$\int_{\mathbb{R}^{2}} f(x) g(t) e^{-ixt} , \mathrm dt , \mathrm dx = \int_{\mathbb{R}^{2}} f(x) g(t) e^{-ixt} , \mathrm dx , \mathrm dt.$$ (This is sometime referred to as the multiplication formula.) Plancherel's theorem can be proven for functions in $L^{1}(\mathbb{R}) \cap L^{2}(\mathbb{R})$ and then apparently extended to functions in $L^{2}(\mathbb{R})$. – Random Variable Feb 21 '22 at 10:39
  • Maybe the reason you say it is "the closest thing" is the integrand assumes a product instead of the general form. I thought the Plancherel's theorem regards a single integral over the time and frequency domains as shown in https://en.wikipedia.org/wiki/Plancherel_theorem instead of a double integral in the time domai as in your equation. – Hans Feb 22 '22 at 19:06
  • I didn't know you had responded to my last comment. The above equation can be written as $$\int_{-\infty}^{\infty} f(x) \hat{g}(x) , \mathrm dx = \int_{-\infty}^{\infty} \hat{f}(t) g(t) , \mathrm dt, $$ which as explained here is another way of expressing Plancherel's/Parseval's theorem. – Random Variable May 01 '22 at 17:57
  • @RandomVariable: I see you expanded out the Fourier transform. I do not know why I was confused when I wrote my last comment. That said, it does not resolve the original question, as the Plancherel theorem requires the integrand separable into a product rather than be a general form as in the question. I suppose that was the reason you said "the closest thing". – Hans May 01 '22 at 22:08

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