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I am trying to calculate the FT of $\frac{d}{dx}\log({\sinh{x}})$. I did not calculate it directly but I used the result of its derivative. This is a tentative: I found the following result in the literature but they did not mention how they get the result except adding a small positive $\epsilon$ which I believe is considered as the imaginary part of $x$ : $\hat{F}_k\Biggl\{\frac{d^2}{dx^2}\log({\sinh{x}})\Biggr\}=\frac{k}{1-e^{-\pi k}}$ then $\hat{F}_k\Biggl\{\frac{d}{dx}\log({\sinh{x}})\Biggr\}=\frac{1}{ik}\hat{F}_k\Biggl\{\frac{d^2}{dx^2}\log({\sinh{x}})\Biggr\}=\frac{i}{e^{-\pi k}-1}$ First, is my reasoning correct ? I want to understand how they get the result by adding $ i \epsilon$ and if it is possible how do I calculate the FT of $\frac{d}{dx}\log({\sinh{x}})$ without using FT properties.

HJames
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  • In terms of distributions, $$(\ln \sinh x)' = \mathcal P(\coth x) - \pi i \delta(x), \ \mathcal F[(\ln \sinh x)'] = -\frac i 2 \mathcal P {\left( \coth \frac {\pi k} 2 \right)} - \frac i 2 = i \mathcal P {\left( \frac 1 {e^{-\pi k} - 1} \right)},$$ where $\ln$ is the principal branch. The FT of $\coth$ can be found by showing that $(1 + 2 \sum_{k \geq 1} e^{-2 k |x|}) \operatorname {sgn} x$ converges to $\mathcal P(\coth x)$ and the FT of this series converges to $-i \mathcal P(\coth(\pi k/2))/2$. – Maxim Jan 17 '21 at 18:35
  • I don't think the extra $-i/2$ term is correct. In particular from the integral formula you have ${\cal F}\coth(x)$ is an odd function of $k$, while yours isn't. – Diger Jan 18 '21 at 13:37
  • @Diger The line above clearly says that it's not the FT of $\coth x$, it's the FT of the principal value distribution $\mathcal P(\coth x)$ minus $\pi i \delta(x)$. – Maxim Jan 18 '21 at 14:13
  • Then tell me why it is $-i\pi \delta(x)$ and not $+i\pi\delta(x)$. – Diger Jan 18 '21 at 14:43
  • @Diger The line below clearly says that $\ln$ is the principal branch, corresponding to $-\pi < \arg z \leq \pi$. – Maxim Jan 18 '21 at 15:26
  • Nobody stops you from using another branch which the result shouldn't depend upon. – Diger Jan 18 '21 at 15:50
  • @Diger That is not correct. $\pm \pi i H(-x)$ have different distributional derivatives, which in turn have different Fourier transforms. – Maxim Jan 18 '21 at 16:51
  • What is $H(x)$? – Diger Jan 18 '21 at 17:12
  • @Diger It's the unit step function. – Maxim Jan 18 '21 at 18:23
  • I'm confused. I see that $(i\pi H(-x))'=-i\pi \delta(x)$, but where is that $i\pi H(-x)$ part of $\ln \sinh x$? – Diger Jan 19 '21 at 01:00
  • @Diger The distributional derivative of $\ln \sinh x - \ln x$ is (the regular distribution induced by) the ordinary function $\coth x - 1/x$. It is identical to $\mathcal P(\coth x) - \mathcal P(1/x)$. It remains to find the distributional derivative of $\ln x$. – Maxim Jan 19 '21 at 15:11
  • You are a funny guy... Should I have asked directly where is the $i\pi H(-x)$ part of $\ln x$? I want to know how you think you arrive at it and not for instance on the other hand $-i\pi H(-x)$. Yeah, I get it, it is the principal branch so you would say $$\ln(x)=\ln(|x|e^{i\phi})=\ln(|x|) + i\phi$$ and $\phi=\pi H(-x)$ if you only consider $x$ to be real. So choosing the principal branch to be $-\pi \leq \phi < \pi$ you would get a different solution, namely $\phi = -\pi H(-x)$. Which one is the correct solution now? I expect there to be only one... – Diger Jan 19 '21 at 22:44

1 Answers1

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I guess you should make use of the identity $$\coth(x)=\frac{1}{x} + \sum_{n=1}^\infty \frac{2x}{x^2+n^2\pi^2} \, .$$ The Fourier-Transform $${\cal F}\left[\coth(x)\right](k)=PV\int_{-\infty}^\infty {\rm d}x \, e^{-ikx} \coth(x)$$ (where $PV$ stands for the principal-value at $x=0$) is not convergent due to the behaviour of the cotangens hyperbolicus at $\pm \infty$. The integral can be regularized by introducing a dampening factor $e^{-\epsilon |x|}$ and eventually taking the limit $\epsilon \rightarrow 0^+$. This gives rise to the regularized Fourier-Transform $$ {\cal F}\left[\coth(x)\right](k) = \lim_{\epsilon \rightarrow 0} PV \int_{-\infty}^\infty {\rm d}x \, e^{-ikx-\epsilon|x|} \coth(x) = I_1(k) + I_2(k)$$ where $$I_1(k) = \lim_{\epsilon \rightarrow 0} PV\int_{-\infty}^\infty \frac{e^{-ikx-\epsilon|x|}}{x} \, {\rm d}x = PV\int_{-\infty}^\infty \frac{e^{-ikx}}{x} \, {\rm d}x = -i\pi \, {\rm sign}(k) \\ I_2(k) = \lim_{\epsilon \rightarrow 0}\int_{-\infty}^\infty {\rm d}x \sum_{n=1}^\infty \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx-\epsilon|x|} = \lim_{\epsilon \rightarrow 0} \sum_{n=1}^\infty \int_{-\infty}^\infty {\rm d}x \, \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx-\epsilon|x|} \\ = \sum_{n=1}^\infty \int_{-\infty}^\infty {\rm d}x \, \frac{2x}{x^2+n^2\pi^2} \, e^{-ikx} = \sum_{n=1}^\infty -2\pi i \,{\rm sign}(k) \, e^{-\pi|k|n} = \frac{-2\pi i \, {\rm sign}(k)}{e^{\pi|k|}-1} \, .$$ The integral $I_1(k)$ converges even for $\epsilon=0$ so that the limit can be taken immediately. For $I_2(k)$ the principal-value is redundant as the integrand is well-behaved at $x=0$. Summation and integration can be interchanged by dom. convergence, since the RHS of $$\int_{-\infty}^\infty {\rm d}x \, \underbrace{\left|\sum_{n=1}^N \frac{2x}{x^2+n^2\pi^2}\right|}_{\leq 1 \,\forall N\in{\mathbb N}} \, e^{-\epsilon|x|} \leq \int_{-\infty}^\infty {\rm d}x \, e^{-\epsilon|x|}$$ converges for $\epsilon>0$. Since the integral and sum converge even for $\epsilon=0$, the limit can be taken immediately. The resulting integral can be calculated by means of the residue theorem for the 2 cases $k>0$ and $k<0$. ${\rm sign}(k)$ is the signum function.

Finally you can easily check that $$I_1(k)+I_2(k)=-i\pi \, {\rm sign}(k) \left(1+\frac{2}{e^{\pi|k|}-1}\right) = -i\pi \coth\left(\frac{\pi k}{2}\right) \, .$$

Diger
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  • $\operatorname {v. ! p.} \int_{-\infty}^\infty e^{-i k x} \coth x , dx$ doesn't exist though (because $\operatorname {v. ! p.} \int_{-\infty}^\infty e^{-i k x} (\coth x - \operatorname {sgn} x) , dx$ exists and $\operatorname {v. ! p.} \int_{-\infty}^\infty e^{-i k x} \operatorname {sgn} x , dx$ doesn't). – Maxim Jan 18 '21 at 00:29
  • Are you worried about $\coth(x)$ being $\pm 1$ at $\pm\infty$? You can continue it analytically: $$\int_{-\infty}^{\infty} e^{-ikx} , {\rm sgn}(x) , {\rm d}x = -\int_{-\infty}^0 e^{-ikx} , {\rm d}x +\int_0^\infty e^{-ikx} , {\rm d}x \=-2i , \lim_{\epsilon \rightarrow 0} \int_0^{\infty} \sin(kx) , e^{-\epsilon x} , {\rm d}x = \lim_{\epsilon \rightarrow 0} \frac{-2ik}{\epsilon^2+k^2} = \begin{cases} \frac{-2i}{k} \quad k\neq 0 \ 0 \quad k=0 \end{cases}, .$$ Basically that's the same as how you would do it with the "undefined" integral $$\int_{-\infty}^\infty e^{-ikx} , {\rm d}x , .$$ – Diger Jan 18 '21 at 12:56
  • Besides: You can check that the result is correct by using the inverse Fourier Transform. – Diger Jan 18 '21 at 13:01
  • $\lim_{\epsilon \to 0^+} \int_{-\infty}^\infty e^{-\epsilon |x|} e^{-i k x} \operatorname {sgn} x , dx$ is not called the principal value integral of $e^{-i k x} \operatorname {sgn} x$. The principal value integral $\lim_{A \to \infty} \int_{-A}^A e^{-i k x} \operatorname {sgn} x , dx$ doesn't exist (as an ordinary function; the distributional limit exists). Re the use of regularizing factors, consider your own example $\int_{-\infty}^\infty e^{-\epsilon |x|} e^{-i k x} , dx$. You cannot find the FT of $1$ by taking the pointwise limit of this expression. – Maxim Jan 18 '21 at 14:07
  • When I was using PV I was refering to $\int_{-\infty}^{-\epsilon} + \int_\epsilon^\infty$, because that is the only way to make sense of the FT in the first place. A complex continuous contour doesn't give a unique value here. Regarding your other statement: What do you mean you can not find the FT of 1 by taking the pointwise limit? – Diger Jan 18 '21 at 14:37
  • That doesn't change the fact that $\operatorname {v. ! p.} \int_{-\infty}^\infty e^{-i k x} \coth x , dx$ doesn't exist. Since you're not using regularizing factors or distributions in the answer, it's not clear what is being computed. Regarding $\int_{-\infty}^\infty e^{-\epsilon |x|} e^{-i k x} , dx$, you can do the calculation or just consider that $\delta(k)$ is not an $\mathbb R \to \mathbb R$ function, you cannot obtain it as a pointwise limit. – Maxim Jan 18 '21 at 15:21
  • $$ \lim_{\epsilon \rightarrow 0} \int_{-\infty}^\infty e^{-ikx-\epsilon|x|} , {\rm d}x = \lim_{\epsilon \rightarrow 0} \frac{2\epsilon}{\epsilon^2+k^2}=\begin{cases} 0 \quad k\neq 0 \ \infty \quad k=0 \end{cases}$$ so at least we already know that the integral is $0$ for $k\neq 0$. To find out what happens at $0$ there is more work necessary. Take any sufficiently benign function $f(k)$ and consider for any $\epsilon>0$ the integral $$\int_{-\epsilon}^\epsilon{\rm d}k,f(k)\int_{-\infty}^\infty e^{-ikx},{\rm d}x=\int_{-\infty}^\infty{\rm d}x\int_{-\epsilon}^\epsilon f(k)e^{-ikx},{\rm d}k$$ – Diger Jan 18 '21 at 15:27
  • $$=f(k_0) \int_{-\infty}^\infty {\rm d}x \int_{-\epsilon}^\epsilon e^{-ikx} , {\rm d}k = f(k_0) \int_{-\infty}^\infty {\rm d}x , \frac{2\sin(\epsilon x)}{x} = 2\pi f(k_0)$$ where $k_0 \in (-\epsilon,\epsilon)$. At the very end take $\epsilon \rightarrow 0$ s.t. $k_0=0$ and this is nothing but what people call the delta-function. – Diger Jan 18 '21 at 15:29
  • The inner integral $\int_{-\infty}^\infty e^{-i k x} dx$ is already a problem, that's precisely the object which we haven't defined yet. If you want to consider the integral of $1 \cdot \mathcal F[f]$ instead of the integral of $\mathcal F[1] f$ (which, when formalized, is essentially how the FT of distributions is defined), note that you wouldn't get just $\int_{-\infty}^\infty f(k)/k , dk$ in the previous example with $\operatorname {sgn}$, because the integral of $\operatorname {sgn}(x) \mathcal Ff$ exists whether or not $f(0) = 0$. – Maxim Jan 18 '21 at 16:47
  • The inner integral is not defined? I'm not defining anything. I'm just deducing it from the $\epsilon \rightarrow 0$ "prescription" as you claim is not possible. But apparently it is... But maybe I'm also missunderstanding you. Are you asking me now to do the same procedure again for $1\cdot {\cal F}[f]$ instead? Why should it be different? I edited my answer for better understanding. Maybe it is clearer now; let me know. – Diger Jan 18 '21 at 17:20
  • In the previous example you have to consider the integral as a principal-value. That way $\int_{-\infty}^\infty \frac{f(k)}{k} , {\rm d}k$ exists regardless of $f(0)=0$ or not. That's why I think it makes sense to also define the Fourier-Transform as a principal-value integral at $0$. It is irrelevant to functions that are well-behaved at $0$ and assigns a unique value to those functions that are not well-behaved s.a. $\coth(x)$ or $1/x$ as mentioned before. – Diger Jan 18 '21 at 17:56
  • If you're not defining it, then it stays undefined. Everything from that point on (starting with interchanging the integration order) is also undefined. The main issue is that to make the regularization argument rigorous, you have to take the distributional limit for $\epsilon \to 0$. There are sequences for which the pointwise limit is $0$ everywhere but the distributional limit is $\delta(x)$, you cannot expect the pointwise limit to give the correct result. – Maxim Jan 18 '21 at 18:18