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It is possible to make an approximation of $\int_{-\infty}^\infty iwF(iw)e^{iwt}dw$ through the Stationary-Phase Method [1] or the Laplace's Method [2].

Here $F(iw)$ is the Fourier transform of a function $f(t)$ that is a time-limited real one-variable function, so is of unlimited bandwidth. Also assume that the Fourier Transform $F(iw)$ fulfill the requirement of the Riemann-Lebesgue Lemma [3], and if required, also the Paley–Wiener_theorem [4].

I am trying to make an approximation to find a useful bound $\max_t |df(t)/dt|$ on this another question here.

I have already tried it with unsuccessful results, it shows to be "over" my mathematical skill, so, if you find an answer, please explain also how you get it.

Beforehand, thanks you very much.

Added later

Atack Plan: What I am trying to do is the following: since on Wikipedia page for the Stationary-Phase Approximation [1] is said that as $k \to \infty$: $$ \int\limits_\mathbb{R} g(w)\,e^{j k h(w)}\,dw \approx \sum\limits_{w_i \in \Lambda} g(w_i)\,\sqrt{\frac{2\pi}{k|h''(w_i)|}}e^{jkh(w_i)+j\frac{\pi}{4}\text{sign}(h''(w_i))} + o\left(\frac{1}{\sqrt{k}}\right)$$ with $\Lambda$ the set of critical points of the function $h(w)$ (i.e. points where $\frac{d}{dw}h(w)=0$), with all critical points been non-degenerate (i.e. $\det\left(\mathrm{Hess}\left(h(w_i)\right)\right)\neq 0, \,\,\forall w_i \in \Lambda$, and $g(w)$ is either compactly supported or has exponential decay.

I want to use it for: $$ \max_t \left|\frac{d}{dt}f(t) \right| = \max_t \left|\frac{1}{2\pi} \int\limits_{-\infty}^\infty jw F(jw)\,e^{j w t}\,dw\right| $$ where $F(jw) = \int\limits_{t_0}^{t_F} f(t)\,e^{-j w t}\,dt$ is the Fourier transform of the compacted-supported function $f(t)$, so I am trying to find an upper bound "before moving" the absolute value inside the integrand.

If I try to directly match both integrals, I will choose $k \cong t$ since time could be sent to infinity (even if $f(t)$ is non-zero only for $t_0 \leq t \leq t_F$), but this will lead to choosing $h(w) = w$ for which $h''(w) = 0\,\forall w$, so the Stationary-Phase approximation formula will be indeterminate.

For trying to avoid it, I am thinking in expanding the integrand in their amplitude and phase components in the following way: $$ \max_t \left|\frac{d}{dt}f(t) \right| = \max_t \left|\frac{1}{2\pi} \int\limits_{-\infty}^\infty \left| jw F(jw)\right|\,e^{jt\left( \frac{\sphericalangle\{ jwF(jw) \}}{t} + w \right)}\,dw \right| $$ so by matching: $$\begin{array}{c} k \cong t \\ g(w) \cong \left| jw F(jw)\right| \\ h(w) \cong \frac{\sphericalangle\{ jwF(jw) \}}{t} + w \\ \rightarrow h'(w) = \frac{\frac{d}{dw}\left(\sphericalangle\{ jwF(jw) \}\right)}{t} +1 \\ \rightarrow h''(w) = \frac{\frac{d^2}{dw^2}\left(\sphericalangle\{ jwF(jw) \}\right)}{t} \\ \end{array}$$ I will have that: $$ \max_t \left|\frac{d}{dt}f(t) \right| \approx \max_t \left|\frac{1}{2\pi} \sum\limits_{w_i \in \Lambda} \left| jw_i F(jw_i)\right|\,\sqrt{\frac{2\pi}{t\,\left| \frac{\frac{d^2}{dw^2} \left(\sphericalangle\{ jwF(jw) \}\right)|_{w=w_i}}{t}\right|}}\,e^{jt\left( \frac{\sphericalangle\{ jw_iF(jw_i) \}}{t} + w_i \right)+j\frac{\pi}{4}\text{sign}\left(\frac{\frac{d^2}{dw^2}\left(\sphericalangle\{ jwF(jw) \}\right)|_{w=w_i}}{t}\right)} \right| $$ Now because I am assuming that $t$ goes to "positive" infinite, using that $w^* = \text{argmax}\left\{ \left| jw F(jw)\right| \right\}$, remembering that $|e^{j\phi}| = 1\,\,\forall \phi \in \mathbb{R}$, and also, I will arbitrarily assume there is only one critical point, then I will have: $$\begin{array}{r c l} \max_t \left|\frac{d}{dt}f(t) \right| & \overset{?}{\approx} & \max_t \left|\frac{1}{2\pi} \left| jw_i F(jw_i)\right|\,\sqrt{\frac{2\pi}{\left|\frac{d^2}{dw^2} \left(\sphericalangle\{ jwF(jw) \}\right)|_{w=w_i}\right|}}\,e^{jt\left( \frac{\sphericalangle\{ jw_iF(jw_i) \}}{t} + w_i \right)+j\frac{\pi}{4}\text{sign}\left(\frac{\frac{d^2}{dw^2}\left(\sphericalangle\{ jwF(jw) \}\right)|_{w=w_i}}{t}\right)} \right| \\ & \approx & \frac{1}{2 \pi} \max_t \left| \left| jw_i F(jw_i)\right|\,\sqrt{\frac{2\pi}{\left|\frac{d^2}{dw^2} \left(\sphericalangle\{ jwF(jw) \}\right)|_{w=w_i}\right|}} \right| \,\,\,\textit{indep. of}\,\,t \\ & \leq & \frac{1}{2 \pi} \max_w \left\{ \left| jw F(jw)\right|\right\}\,\sqrt{\frac{2\pi}{\left|\frac{d^2}{dw^2} \left(\sphericalangle\{ jwF(jw) \}\right)|_{w=w_i}\right|}} \\ & \overset{?}{\approx} & \frac{1}{2 \pi} \max_w \left\{ \left| jw F(jw)\right|\right\}\,\sqrt{\frac{2\pi}{\left|\frac{d^2}{dw^2} \left(\sphericalangle\{ jwF(jw) \}\right)|_{w=w^*}\right|}} \end{array} $$ Were in the last step I have arbitrarily assume that I am "so lucky" that the critical point $w_i$ is just the same point as $w^*$ ($w_i \cong w^*$) where $|jwF(jw)|$ attain its maximum value.

My plan is to check if this bound, or a "constant" number of times this bound, could "do the magic" when reviewing it with the functions of the second table of this question here.

The setback:

When trying to test this bound: $$ \frac{\max_{w = w^*} \left| jw F(jw)\right|}{\sqrt{2\pi\, \left|\frac{d^2}{dw^2} \left(\sphericalangle\{jwF(jw) \}\right)|_{w=w^*}\right|}}$$ I am finding through using Wolfram-Alpha that the part: $$\left|\frac{d^2}{dw^2} \left(\sphericalangle\{jwF(jw) \}\right)|_{w=w^*}\right|$$ Is giving $\text{indeterminate}$ as results, or functions with Dirac's delta distributions like $\frac{\pi \delta'(0)}{4}$, or just $0$, all values with which I cannot evaluate the bound, so certainly I am committing a conceptual mistake here.

I don't really understand the meaning of $\left|\frac{d^2}{dw^2} \left(\sphericalangle\{jwF(jw) \}\right)|_{w=w^*}\right|$: since real even functions will have real even Fourier transform, ¿Does it means it will have complex argument $\sphericalangle\{jwF(jw) \} = 0 \,\,\forall w$ (making useless the Stationary-Phase approximation)???.... Also, I don't really understand the Stationary-Phase method, so any help finding my mistake will be appreciated.

Hope this could make you have some ideas of what I am trying to do, and maybe gives a kick-start for your own attempts.

Joako
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    Assuming $t$ is large, consider http://dlmf.nist.gov/2.3.E20 – Gary Oct 28 '21 at 11:40
  • @Gary thanks, I will review it.. I didn't know this website. – Joako Oct 28 '21 at 12:11
  • @Gary I see the approximation, but since the integration limits aren´t $(-\infty,,\infty)$, and also it use some constants $s,,\lambda$ and $\mu$ which doesn´t show how to get them, I get stuck. I added some details to the question, so maybe yo can elaborate an answer. Thanks again for your help. – Joako Oct 28 '21 at 14:10
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    This integral is just $∇ f$ – LL 3.14 Oct 28 '21 at 16:47
  • @LL3.14 I already know that, Is which I am trying to approximate with an upper bound. Thanks any way. – Joako Oct 28 '21 at 16:56
  • Then it is not clear what you are really looking for. $\mathrm df/\mathrm dt$ is usually a simple enough object that is not difficult to compute / approximate. You have to explain to the reader why it is easier in your case to consider its Fourier transform, why the stationary-phase method or the Laplace's method seem to you methods that are easier to implement in your project, ... you will never receive a good answer if you do not ask a good question – LL 3.14 Oct 28 '21 at 17:18
  • @Joako Split the integral at $0$ into to parts and perform a change of variables $w=-v$ in the one from $-\infty$ to $0$. In this way you have two integrals from $0$ to $+\infty$. You just have to be patient and read the text on the website and it tells you what $s$, $\lambda$ and $\mu$ are. $s$ is just the running index of the sum. The other two appear in http://dlmf.nist.gov/2.3.E7 above the formula I sent you. – Gary Oct 28 '21 at 22:05
  • @Gary I have tried a second time and I think I understood what is going on: I have chosen $\lambda = \mu = 1$ so the coefficients $a_k$ will becomes the coefficients of the Maclaurin/Taylor series of $jwF(jw)$ (actually, the only scenario for which I have enough mathematical inside to manage). Then, noting that the first element of the series is $jwF(jw)|{w=0} = 0$ (since, assuming now $F(jw)$ is Lebesgue integrable), and noting that $d^k/dw^k (jwF(jw))|{w=0}$ is equivalent to $j k, d^{k-1}/dw^{k-1} (F(jw))|_{w=0}$ }$, I can use the formula as: – Joako Oct 31 '21 at 16:46
  • @Gary ...: $\int_0^\infty jwF(jw),e^{jwt},dw \approx \sum\limits_{k=1}^\infty \frac{\Gamma((k+1)/2)}{(k-1)!}\cdot\frac{e^{j\pi((k+1)/2)}}{t^{k+1}}\cdot d^{k-1}/dw^{k-1} (F(jw))|_{w=0}$, but with this formula I still have the problem of moving the absolute value inside a summation, and it diverges when looking for $\max_t$. Also, I wasn´t able to match this formula with the Stationary-Phase formula shown in Wikipedia (now incorporated to the question). So, I don´t know how to use it for my purpose right now. – Joako Oct 31 '21 at 16:47
  • @LL3.14 I have extend what I am trying to do, hope it will help to explain myself better. – Joako Oct 31 '21 at 16:48
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    @Joako Your original Fourier transform does not have a stationary phase so the method is not applicable. – Gary Oct 31 '21 at 22:24

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