Confusions when differentiating a Fourier Transform and using the Kramers-Kronig relations

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.

1. How to differentiate $$\frac{\partial}{\partial w}\left(\frac{\pi}{2}\,\text{P.V.}\!\!\!\int\limits_{-\infty}^{\infty} \frac{V(\xi)}{\xi-w}d\xi\right)$$?
2. 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 $$V(w)$$?
3. 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.

• There are some papers by Hardy on the differentiation of principal value integrals Commented Jul 12, 2022 at 6:46
• I don't know the right analysis to help answer your question. Hardy's paper and my upvote is the best I can do... Commented Jul 12, 2022 at 18:00
• I'm not so sure about $X'=U'+iV'$. $X$ is complex differentiable, but $U$ and $V$ are not Commented Jul 14, 2022 at 5:57
• No, $X’$ is definitely complex diff, this is because $x$ is compactly supported; this is the Paley Wiener theorem. So $U,V$ are harmonic. But they are definitely not complex differentiable (they are not open maps) Commented Jul 14, 2022 at 23:28
• Maybe if you pretend you have the formula for $X$, eg $X(w)=w$ then you can understand what went wrong. Then the real part $U(w)=\Re w$ which has a gradient but not a complex derivative. Also, you have used $X’=iwX$ but this is not true. What is true is that $\mathcal F (x’) = iw X$. ($x’$ is interpreted distributionally) Commented Jul 14, 2022 at 23:43

1 Answer

This answer is just for closing the question, since the answer was given by @CalvinKhor as a comment.

All the issue is due I have confused the derivative of the Fourier Transform $$\frac{\partial}{\partial w}\mathbb{F}\{x(t)\}(w)$$ with the Fourier Transform of the derivative $$\mathbb{F}\{\frac{\partial x(t)}{\partial t}\}(w)=iw \mathbb{F}\{x(t)\}(w)$$.