I was trying to take a very basic, differentiable function, and compute the Finite and Inverse Finite Fourier Transforms utilizing the properties of taking a Fourier Transform, but realized that there is an issue when dividing by the frequency term such that it produces a division by 0 where the frequency is 0.
So the rule is that:
$$F\left(\frac{d}{dt}u(t)\right) \longleftrightarrow i\omega\hat{u}, \ \ \ \ F^{-1}\hat{u} = u(t)$$
So, I thought it would follow that: $$i\omega\hat{u} \xrightarrow[]{divide} \frac{i\omega\hat{u}}{i\omega} = \hat{u} \xrightarrow[]{F^{-1}} F^{-1}\hat{u} = u(t)$$
Where $\hat{u} = u(\omega)$, I just like the hat notation more; $\omega = \frac{2\pi}{L_t}*t$, $L_t$ is the length of $t$ which is the number of points along its axis minus 1. But this did not work due to a division by zero issue as I mentioned already.
In my case I have a differentiable function:
$$p(z) = p_0 e^{-z/H}$$
where I have set $H=40$ and $p_0 = 1000$, the frequency variable is going to be $\zeta$ when we perform the Fourier Transform, and I have set $z = [0,20]$. And I use this to form a basic derivative that we will be working with below:
$$\frac{dp(z)}{dz}$$
My plan of action is the same as I showed in the first 2 equations.
$$F\left(\frac{dp(z)}{dz}\right) = i\zeta\hat{p} \xrightarrow[]{divide} \frac{i\zeta\hat{p}}{i\zeta}$$
And then ideally, I would take then Inverse Fourier Transform of $\hat{p}$ to get back to p, but the problem comes at the division step at $\zeta = 0$ which is also at $z=0$. How do I deal with this issue? I'm assuming take a limit of the Inverse Fourier Transform Integral or apply a boundary condition, but I don't know how to do that for Fourier Transforms. For the boundary condition, I have $p(z=0) = p_0 = 1000$ and I computed the derivative and got $p'(z = 0) = - 25$, but I just can't envision how to apply this to the Fourier integral.
Does anyone have any insight into this problem?
NOTE: I also had to do some other manipulations to treat this as a psuedo-periodic signal by extending the domain and some other steps necessary in order to obtain a proper Fourier Transform representations of the functions as well as their correct Inverse Fourier Transform counterparts, but we won't get into that unless someone asks, because I don't find it pertinent to the heart of the issue.
ADDED INFORMATION
The formula for the Discrete Time Fourier Transform is:
$$X(e^{j\omega}) = F(x)_{j\omega} = \sum_{n=N_0}^{N_1} xe^{-1j(\omega_n) t}$$
where $N_1$ is the number of points to the right of the line of symmetry of an aperiodic signal and $N_0$ is the number of points to the left. Here is an image below that should illustrate what this looks like (additionally this is what I am trying to use to recover $p(z)$):
Where I used a scatter plot over the line to illustrate each finite point which makes up the Discrete Time function.
So in our case, the computation at $\zeta = 0$ before dividing by $\zeta$ would look like:
$$F\left(\hat{\frac{dp}{dz}}_{N_0}\right) = i\zeta\hat{p}_{N_0} =i\zeta\Bigl( \\\frac{dp}{dz}_{N_0}e^{-1j(\zeta_{N_0}=0)*z_{N_0}} = \frac{dp}{dz}_{N_0} = -25 + \\ \frac{dp}{dz}_{N_0+1}e^{-1j(\zeta_{N_0} = 0)z_{N_0+1}} = \frac{dp}{dz}_{N_0+1} = -24.3827 + \\ \frac{dp}{dz}_{N_0+2}e^{-1j(\zeta_{N_0} = 0)z_{N_0+2}} = \frac{dp}{dz}_{N_0+2} = -23.7807 + \\ ... + \frac{dp}{dz}_{N_1}e^{-1j(\zeta_{N_0} = 0)z_{N_1}} = \frac{dp}{dz}_{N_1} = -15.1633 \\ \Bigl)$$
and when we divide, that's when the problem arises, and it occurs on the very first term in the summation as we'll see below:
$$F\left(\hat{\frac{dp}{dz}}_{N_0}\right) = \frac{i\zeta\hat{p}_{N_0}}{i\zeta} =\Bigl( \\\ \frac{\frac{dp}{dz}_{N_0}}{i(\zeta_{N_0} = 0)} = \lim_{\zeta = \pm \infty \rightarrow 0}{ \frac{\frac{dp}{dz}}{\zeta}} = ? + \\ \frac{\frac{dp}{dz}_{N_0+1}}{\zeta_{N_0+1} = \frac{1*2*pi}{N_0+1}} + \\ \frac{\frac{dp}{dz}_{N_0+2}}{\zeta_{N_0+2} = \frac{2*2*pi}{N_0+2}} + \\ ... + \frac{\frac{dp}{dz}_{N_1}}{\zeta_{N_1} = \frac{N_1*2*pi}{N_1+2}} \\ \Bigl)$$
Where you can see the issue arises on the first term in the sum when dividing. I do not know how to approach this problem.
ADDED PLOT
Where We can see at $\zeta = 0$, we have a maxima for $F(p(z))$ and minima for $F(\frac{dp}{dz})$ and this value repeats itself at $\zeta = 2\pi$. And again, we are starting from the assumption that we do not know what $p(z)$ is. We have $\frac{dp}{dz}$ and we have it's Discrete Time Fourier Transform which we can see plotted above. So, we only have information as it relates to the derivative and we are seeing if we either can't recover $F(p(z))$ and/or $p(z)$.
As an aside, I also made a note under the title that:
$$(-1)2(N-1)=40 \ \ * F(\frac{dp}{dz}) = F(p(z))$$
. I'd assume this comes from finite differentiation where the values are reduced by division by the scale length of the dimension of differentiation, so multiplying by the scale length can undo the difference equation or something like that. Ultimately I pulled this out of my ass, so I'd ignore it and treat it as an aside unless anyone can glean something from this. I tried it for different values of N, and recovered p(z), but I haven't tested it thoroughly for different spacing, functions, etc.