Let $x(t)$ be a real-valued squared-integrable signal that have a beginning and an ending time (so, is of finite duration). Then I will have that its Fourier Transform: $$X(w) = U(w)+iV(w)$$ with $\{U(w),\ V(w)\}\in\mathbb{R}$ is such that:
- $X(w)$ is analytic
- $U(-w) = U(w)$
- $V(-w) = -V(w)$
- Kramers–Kronig relations: $U(w) = \displaystyle{\frac{\pi}{2}\,\text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}d\xi}$ and $V(w) = \displaystyle{-\frac{\pi}{2}\,\text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{U(\xi)}{\xi-w}d\xi}$
Now, since differentiating the Fourier Transform goes at follow: $\displaystyle{\frac{\partial X(w)}{\partial w}=iw X(w)} \equiv iwU(w)-wV(w)\tag{Prop. 1}$ $$\Rightarrow \frac{\partial X(w)}{\partial w}\; \overset{\text{linearity of}\,\frac{\partial}{\partial w}}{=}\; \frac{\partial U(w)}{\partial w}+i\frac{\partial V(w)}{\partial w} \overset{\text{Prop. 1}}{\equiv} iwU(w)-wV(w)$$ so by pairing the real and imaginary parts:
- $\frac{\partial U(w)}{\partial w} = -wV(w) \tag{Eq. 1}$
- $\frac{\partial V(w)}{\partial w} = wU(w) \tag{Eq. 2}$
Then, for example for the first one, I will have: $$V(w) = -\frac{1}{w}\frac{\partial U(w)}{\partial w} = -\frac{1}{w}\frac{\partial}{\partial w}\left(\frac{\pi}{2}\,\text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}d\xi\right) \tag{Eq. 3}$$
but I don't know how to differentiate $\frac{\partial}{\partial w}\left(\text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}d\xi\right)$: if I use Wolfram-Alpha it says that: $$\frac{\partial}{\partial w}\left(\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}d\xi\right)= \int\limits_{-\infty}^{\infty} \frac{V(\xi)}{(\xi-w)^2}d\xi$$ bit I am not sure if its right because the Principal Value implies is a complex integral.
- How to differentiate $\frac{\partial}{\partial w}\left(\frac{\pi}{2}\,\text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}d\xi\right)$?
- Does $\text{Eq. 3}$ set an equation to find a restricted form for $V(w)$ as a function of $w$? or it just will end in something like $V(w)=V(w)$? It is possible to solve this equation for $V(w)$?
- Does the equations $\text{Eq. 1}$ and $\text{Eq. 2}$ behave as similar restrictions as the Cauchy–Riemann equations?
As example, taking the second derivative of $X(w)$: $$\frac{\partial^2 X(w)}{\partial w^2} = (iw)^2X(w) = -w^2X(w) = -w^2U(w)-iw^2V(w) \tag{Eq. 4}$$
But if I use $\text{Eq. 1}$ and $\text{Eq. 2}$, by differentiating them I will have: $$ \frac{\partial^2 U(w)}{\partial w^2} = \frac{\partial}{\partial w}\left(\frac{\partial U(w)}{\partial w}\right) \overset{\text{Eq. 1}}{=} \frac{\partial}{\partial w}\left(-wV(w)\right) =-V(w)-w\frac{\partial V(w)}{\partial w} \overset{\text{Eq. 2}}{=} -w^2U(w)-V(w) \tag{Eq. 5} $$
$$\frac{\partial^2 V(w)}{\partial w^2} = \frac{\partial}{\partial w}\left(\frac{\partial V(w)}{\partial w}\right) \overset{\text{Eq. 2}}{=} \frac{\partial}{\partial w}\left(\; wU(w)\right) = \; U(w)+w \frac{\partial U(w)}{\partial w} \overset{\text{Eq. 1}}{=} -w^2V(w) +U(w) \tag{Eq. 6}$$
Now since, $$\begin{array}{r c l} \displaystyle{\frac{\partial^2 X(w)}{\partial w^2}} & = & \displaystyle{\frac{\partial^2 U(w)}{\partial w^2}+i\frac{\partial^2 V(w)}{\partial w^2}} \\ & \overset{\text{Eq. 5 & Eq. 6}}{=} & -w^2U(w)-V(w)+i\left(-w^2V(w) +U(w)\right) \\ & = & -w^2U(w)-iw^2V(w)+i\left(U(w) +iV(w)\right) \\ & = & -w^2X(w)+iX(w) = (i-w^2)X(w) \\ & \neq & \text{Eq. 4} \end{array}$$
Since I found a contradiction, please explain where and why I am making the mistake.
Added later___________________
I found in this answer that: $$\frac{\partial}{\partial w}\left(\text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}d\xi\right)= \text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{V(\xi)-V(w)}{(\xi-w)^2}d\xi$$ so it somehow solves point $(1)$, but I still don't know if and how to use it to make a differential equation for $V(w)$, so it still missing an answer for point $(2)$.
About the mentioned differential equation, I believe that the matching of $\text{Eq. 1}$ and $\text{Eq. 2}$ is an illegal operation, which leads to the contradiction, but I still don't know why and I would like to know were I am making a conceptual mistake... my intuition tells me is related with something about non-uniqueness of the complex derivative when Cauchy-Riemann Equations aren't hold: I believe this is the case of the Fourier Transform where only the imaginary axis is considered.
As example, following $\text{Eq. 1}$ and $\text{Eq. 2}$ I can make the following differential equations: $$\begin{array}{l} V''-\frac{V'}{w}+w^2V=0\\ U''-\frac{U'}{w}+w^2U=0\\ \Rightarrow y''-\frac{y'}{w}+w^2y=0 \Rightarrow y(w) = c_1\cos\left(\frac{w^2}{2}\right)+c_2\sin\left(\frac{w^2}{2}\right) \end{array}$$ where not just imply that $U \equiv V$ which is false, but also the solution is an even function which don't fulfill the properties of $V(w)$ which is odd - an also is a fixed solution when the transform can have multiple values.
The problem I think is that the transform $X(w)$ which is related to any function (so it could take many forms), cannot be considered as a function in their differentiation property:
$$\frac{\partial X(w)}{\partial w} = iw X(w) \overset{\text{as function}}{\Rightarrow} X'(w)-iwX(w)=0 \Rightarrow X(w)=X(0)e^{i\frac{w^2}{2}}=\int\limits_{-\infty}^{\infty}x(t)dt\ e^{i\frac{w^2}{2}}$$ which is a fixed functions differently from the transform, so it is a conceptual mistake, but I would like to know why it don't work: Does it imply is not possible to make a differential equation for $V(w)$?
As example using the Kramer-Kronig relation an $\text{Eq. 2}$: $$V(w) = \displaystyle{-\frac{\pi}{2}\,\text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{1}{\xi(\xi-w)}\cdot\frac{\partial V(\xi)}{\partial \xi}\,d\xi}$$
Is this conceptually wrong? Does having a differential equation fix the solution so is not applicable in the transforms framework?
My guess is $\frac{\partial X(w)}{\partial w} \neq \frac{\partial U(w)}{\partial w}+i\frac{\partial V(w)}{\partial w}$ but I don't know why is not plausible, since the derivative is a linear operator.
