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I am trying to find an specific Fourier Transform to be used for trying to solve this another question, and in this answer another user recommend me to consider an approximation for the Signum function given by an Hyperbolic Tangent.

However, in order to be useful, I need to find its Fourier Transform on an a finite interval, let say between times $t_0<t_f$, and I have not been able of find it through Wolfram-Alpha, neither by separating the interval as a squared function in time with a Sinc function in frequencies, and using the convolution theorem with the full-time Fourier Transform of the Hyperbolic Tangent which is known.

I hope you could help me to find: $$I(iw)=\mathbb{F}\left\{\tanh\left(\dfrac{x}{\varepsilon}\right)\right\}\Biggr|_{t_0}^{t_f}:=\int\limits_{t_0}^{t_f} \tanh\left(\dfrac{x}{\varepsilon}\right)e^{-iwx}dx$$

Joako
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    Assuming $t_0<t_f\land \epsilon>0$, Mathematica gives the result $$\int\limits_{t_0}^{t_f} \tanh\left(\frac{x}{\epsilon}\right), e^{-i \omega x} , dx=\frac{\omega \epsilon e^{-\frac{1}{2} \pi \omega \epsilon} \left(B_{-e^{\frac{2 t_0}{\epsilon}}}\left(-\frac{1}{2} i \epsilon \omega ,0\right)-B_{-e^{\frac{2 t_f}{\epsilon}}}\left(-\frac{1}{2} i \epsilon \omega ,0\right)\right)-i e^{-i t_0 \omega}+i e^{-i t_f \omega}}{\omega}.$$ – Steven Clark Oct 08 '24 at 19:12
  • @StevenClark Thanks for the comment, is interesting since the last terms matches with those of other questions I did. The mentioned $B()$ function is the Beta function, right? – Joako Oct 08 '24 at 19:22
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    $B_z(a,b)$ is the incomplete beta function. – Steven Clark Oct 08 '24 at 19:38
  • @StevenClark I think that maybe is not $100%$ correct: if I make $\varepsilon\to 0$ I woulg get only $\mathbb{F}{1}|_{t_0}^{t_f}$ and a missing Principal Value part from the signum function transform will be missing, or I got lost in the formula? – Joako Oct 08 '24 at 20:12
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    I'm not sure the formula is correct as I seem to be getting some inconsistent results from Mathematica which I posted in an answer below. – Steven Clark Oct 08 '24 at 22:55
  • @StevenClark thanks, I will look at them... I also found inconsistent result when doing similar things by hand, my suspicious is those are due some nuances from distributions come within the jumps ficticiously introduced by the integration limits (the transform don't know is a whole function and only looks a section), and maybe some Principal Values or Direac's Delta functions rises in between, I am not sure, but in general I saw that $\int_{[a,b]} fe^{-iwt} dt = \int_{(a,b)} fe^{-iwt} dt + \text{jump related terms}$ as shown here for classic transforms – Joako Oct 08 '24 at 23:13
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    I tried a few values of $t_0$, $t_f$, and $\epsilon$ and the two formulas seem to evaluate similar to each other and to evaluation of the Fourier transform using numerical integration. Also, as $\epsilon\to 0$ the two formulas seem to approximate the Fourier transform of $f(x)=\text{sgn}(x), (\theta(x-x_0)-\theta(x-x_f)$ which is consistent with $\underset{\epsilon\to 0^+}{\text{lim}}\tanh\left(\frac{x}{\epsilon}\right)=\text{sgn}(x)$. So far I haven't seen any inconsistencies in the evaluations, but please let me know if you have any specific cases where you're seeing inconsistencies. – Steven Clark Oct 09 '24 at 01:49
  • @StevenClark Maybe I was doing a mistake, but I got trouble when trying to verify if the solutions were fulfilling Plancherel theorem, somehow without the effect of the jumps the integral diverge, when my intuition tells that by removing energy it should had being bounded by $\int_{t_0}^{t_f} |f(t)|^2 dt$ – Joako Oct 09 '24 at 02:01
  • I think the problem may be that the Planchel theorem assumes the Fourier transform is defined as $$\hat{f}(\omega)=\mathcal{F}xf(x)=\int\limits{-\infty}^{\infty} f(x), e^{-i 2 \pi \omega x} , dx$$ whereas you're using the definition $$\hat{f}(\omega)=\mathcal{F}xf(x)=\int\limits{-\infty}^{\infty}, f(x) e^{-i \omega x} , dx$$ which gives a different result when evaluating $$\int\limits_{-\infty}^{\infty} |\hat{f}(\omega)|^2 , d\omega.$$ – Steven Clark Oct 09 '24 at 03:59
  • @StevenClark that change in definition at much should change a constant value, when I tried it for $\int_{-\infty}^{\infty}|\int_{(t_0,,t_f)} fe^{iwt}dt|^2 dw\to\infty$... hope it is not just my mistake – Joako Oct 09 '24 at 04:33
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    You can compensate by evaluating $$\int\limits_{-\infty}^{\infty} |\hat{f}(2 \pi \xi)|^2 , d\xi$$ when using the definition $$\hat{f}(\omega)=\mathcal{F}xf(x)=\int\limits{-\infty}^{\infty}, f(x) e^{-i \omega x} , dx.$$ – Steven Clark Oct 09 '24 at 14:23

2 Answers2

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This might not be an answer, but is too long for a comment.

Assuming $t_0<t_f\land \epsilon>0$, the Mathematica Integrate function gives the result

$$\int\limits_{t_0}^{t_f} \tanh\left(\frac{x}{\epsilon}\right)\, e^{-i \omega x} \, dx=\frac{\omega \epsilon e^{-\frac{1}{2} \pi \omega \epsilon} \left(B_{-e^{\frac{2 t_0}{\epsilon}}}\left(-\frac{1}{2} i \epsilon \omega ,0\right)-B_{-e^{\frac{2 t_f}{\epsilon}}}\left(-\frac{1}{2} i \epsilon \omega ,0\right)\right)-i e^{-i t_0 \omega}+i e^{-i t_f \omega}}{\omega}\tag{1}$$

whereas the Mathematica FourierTransform function gives the result

$$\mathcal{F}_x\left[\tanh\left(\frac{x}{\epsilon}\right)\, \left(\theta\left(x-t_0\right)-\theta\left(x-t_f\right)\right)\right](\omega)=\frac{1}{2} \epsilon e^{-\frac{1}{2} \pi \omega \epsilon} \left(B_{-e^{\frac{2 t_0}{\epsilon}}}\left(1-\frac{i \epsilon \omega}{2},0\right)+B_{-e^{\frac{2 t_0}{\epsilon}}}\left(-\frac{1}{2} i \epsilon \omega ,0\right)-B_{-e^{\frac{2 t_f}{\epsilon}}}\left(1-\frac{i \epsilon \omega}{2},0\right)-B_{-e^{\frac{2 t_f}{\epsilon}}}\left(-\frac{1}{2} i \epsilon \omega ,0\right)\right)\tag{2}$$

where $\theta(x)$ is the Heaviside step function and $B_z(a,b)$ is the incomplete beta function.

The Fourier transform result in formula (2) above was evaluated using FourierParameters$\to\{1, -1\}$ which is equivalent to $a=1$ and $b=-1$ in formula (15) at Wolfram MathWorld: Fourier Transform, and consequently the Fourier transform result above should be equivalent to

$$\int\limits_{=\infty}^{\infty} \tanh\left(\frac{x}{\epsilon}\right)\, \left(\theta\left(x-t_0\right)-\theta\left(x-t_f\right)\right)\, e^{-i \omega x} \, dx=\int\limits_{t_0}^{t_f} \tanh\left(\frac{x}{\epsilon}\right)\, e^{-i \omega x} \, dx$$

and so the results in formulas (1) and (2) above should be equivalent, but I haven't confirmed their equivalence.

Steven Clark
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  • I think there is a difference: think of $$(f\Delta\theta)'=f'\Delta\theta+f\Delta\delta$$ there you could see if I apply the Fourier Transmorm there would be differences – Joako Oct 12 '24 at 18:15
  • @Joako I don't understand your comment. What is $\Delta$? Are you questioning the validity of $$\int\limits_{-\infty}^{\infty} f(x), (\theta (x-a)-\theta(x-b)), e^{-i w x}, dx=\int\limits_a^b f(x), e^{-i w x}, dx,,\quad a<b?$$ – Steven Clark Oct 13 '24 at 00:44
  • @Joako Assuming by $f\Delta\theta$ you mean $f(x), \theta(x)$, then $$\int\limits_{-\infty}^{\infty} \frac{\partial (f(x) \theta(x))}{\partial x}, e^{-i w x} , dx=\int\limits_{-\infty}^{\infty} (\delta(x), f(x)+\theta(x), f'(x)), e^{-i w x} , dx\=\int\limits_{-\infty}^{\infty} \delta(x), f(x), e^{-i w x} , dx+\int_{-\infty}^{\infty} \theta(x), f'(x), e^{-i w x} , dx=f(0)+\int\limits_0^{\infty} f'(x), e^{-i w x} , dx.$$ I don't see any contradiction. – Steven Clark Oct 13 '24 at 00:44
  • I mean with $\Delta\theta:=\theta(t-t_0)-\theta(t-t_f)$ and $\Delta\delta:=\delta(t-t_0)-\delta(t-t_f)\equiv (\Delta\theta)'$ – Joako Oct 13 '24 at 00:58
  • @Joako Ok, the result would depend on how you define $\int\limits_{-\epsilon}^0 \delta(x), dx$ and $\int\limits_0^{\epsilon} \delta(x), dx$ for any $\epsilon>0$ and also how you define $\theta(0)$. But I don't see how any of this is relevant to your question and my answer here since $\tanh(x)$ is an analytic function for $x\in\mathbb{R}$ and $$\underset{\epsilon\to 0}{\text{lim}} \left(\ \int\limits_{c-\epsilon}^{c+\epsilon} \tanh(x) , dx\right)=0,,\quad \forall c\in\mathbb{R}.$$ – Steven Clark Oct 13 '24 at 01:47
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You find the the indefinite Fourier integral of the derivative

$$ \partial_x \ \tanh (a x) \ = \frac{a}{ \cosh^2(a x)}$$

as

$$F[\text{sech (a x)}^2](k,x)=\frac{4 a}{2 a+i k}\cdot e^{2 a x+i k x}\cdot \, _2F_1\left(2,\frac{i k}{2 a}+1;\frac{i k}{2 a}+2;-e^{2 a x}\right)$$

that yields with the integral wrt to $x$ the Fourier integral as multiplication by ${(i k)}^{-1} $ as an algebraically simplification in the best case a Beta function

$$F[\tanh^{-1} (a x)](k,x) \ = \ 2 i\cdot \frac{e^{\frac{\pi k}{2 a}}}{k}\cdot B_{-e^{2 a x}}\left(\frac{i k}{2 a}+1,-1\right)$$

This is to be expected because the $\tanh^{-1}$ is a $\log$ function and the indefinite integrals of the log series yield Beta series, a special case of hypergeometric series.

Of course such a result has to be checked numerically for the correctness of coefficients. But that's in most cases not a problem for a linear integral transform, if the algebraic kernel is functionally correct.

Roland F
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  • Thanks for taking your time for answering. I am trying to understand the answer: in your notation, If the integral I am looking for is $I(iw)$ what I think your plan is the following:$$I(iw)=F\tanh(ax)-F\tanh(ax)$$ right? Does not that get me into troubles due $\infty - \infty = \text{finite}$?... Also, do you have a source for the $\text{sech}^2()$ formula?... I think there is a typo since you are using $\tanh^{-1}$ the inverse function notation. – Joako Oct 08 '24 at 11:32