Fourier transform of hyperbolic functions

Question

I am looking for the Fourier transform of $x \tanh(x)$ in $\mathbb{R}$. The Fourier transform on $\mathbb{R}\setminus\{0\}$ is $$2\sum_{n=0}^{\infty}(2n+1)\exp\left(-(2n+1)\frac{|t|}{2}\right).$$ But there is singularity at $0$. How can I get the Fourier transform of $x \tanh(x)$ in $\mathbb{R}$?

Answer 1

The Fourier Transform of the distribution $t\tanh(t)$ is a tempered distribution. We can calculate its Fourier Transform in terms of the sum of the Fourier tranform of the $L^1$ funcion $f(t)=t\tanh(t)-|t|$ and the Fourier Transform of the tempered distribution $g(t)=|t|$. We now proceed.

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FOURIER TRANSFORM OF $\displaystyle |t|$:

In THIS ANSWER, I developed the Fourier Transform for $g_a(t)=|t|^a$ for all real values of $a$. In particular, for $a=1$, we have

$$\mathscr{F}\{ g_1\}(\omega)=-\frac2{\omega^2}$$

Note that the object $\mathscr{F}\{ g_1\}(\omega)=-2/\omega^2$ is not a function; it is a Tempered distribution. It does not assign a value for each $\omega$. Rather, it is defined as how it acts on any Schwartz function $\phi$ in the following way. We have

$$\langle \mathscr{F}\{ g_1\},\phi\rangle =\lim_{\delta\to0^+}\int_{|\omega|\ge\delta|} \left(\frac{-2}{\omega^2}\right)\left(\phi(\omega)-\phi(0)\right)\,d\omega $$

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FOURIER TRANSFORM OF $\displaystyle \tanh(t)-|t|$:

Note that the function $f(t)=\tanh(t)-|t|$ is in $L^1$. Therefore, its Fourier Transform is given by

$$\begin{align} \mathscr{F}\{f\}(\omega)&=\int_{-\infty}^\infty \left(t\tanh(t)-|t|\right)\,e^{i\omega t}\,dt\\\\ &=2\int_0^\infty t(\tanh(t) - 1)\cos(\omega t)\,dt\\\\ &=4\int_0^\infty t\left(\frac{e^{-2t}}{1+e^{-2t}}\right)\cos(\omega t)\,dt\\\\ &=4\frac{d}{d\omega}\int_0^\infty \left(\frac{e^{-2t}}{1+e^{-2t}}\right)\sin(\omega t)\,dt\\\\ &=4\frac{d}{d\omega}\Im\int_0^\infty \left(\frac{e^{-(2t-i\omega)t}}{1+e^{-2t}}\right)\,dt\\\\ &=4\frac{d}{d\omega}\Im\int_0^\infty e^{-(2t-i\omega)t} \sum_{n=0}^\infty (-1)^ne^{-2nt}\,dt\\\\ &=4\frac{d}{d\omega}\Im\sum_{n=0}^\infty (-1)^n\int_0^\infty e^{-(2t-i\omega)t} e^{-2nt}\,dt\\\\ &=4\frac{d}{d\omega}\Im\sum_{n=0}^\infty \frac{(-1)^n}{2(n+1)-i\omega}\\\\ &=4\frac{d}{d\omega}\sum_{n=0}^\infty \frac{(-1)^n\omega }{4(n+1)^2+\omega^2} \\\\ &=\frac{d}{d\omega}\sum_{n=1}^\infty \frac{(-1)^{n-1}\omega}{n^2+(\omega/2)^2}\\\\ &=\frac{d}{d\omega}\frac{\pi \omega \,\operatorname{csch}(\pi \omega/2)-2}{\omega}\\\\ &=-\frac{\pi^2}2 \frac{\cosh(\pi \omega /2)}{\sinh^2(\pi \omega/2)}+\frac2{\omega^2} \end{align}$$

where we exploited the partial fraction representation of the hyperbolic cosecant function given by

$$\operatorname{csch}(z)=\frac1z +2z\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(n\pi)^2+z^2}$$

The value of the Fourier Transform of $f$ at $\omega=0$ is equal to $-\pi^2/12$.

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Putting it all together, we find that the Fourier Transform of $t\tanh(t)$ is the tempered distribution

$$\mathscr{F}\{|f|\}(\omega)=-\frac{\pi^2}2 \frac{\cosh(\pi \omega /2)}{\sinh^2(\pi \omega/2)}$$

where (again), the object $\mathscr{F}\{|f|\}(\omega)$ is defined as a Tempered Distribution. Therefore, for any Schwartz Function $\phi$, we have

$$\langle \mathscr{F}\{|f|\}, \phi\rangle =\lim_{\delta\to 0^+}\int_{|\omega|\ge\delta} \left(-\frac{\pi^2}2 \frac{\cosh(\pi \omega /2)}{\sinh^2(\pi \omega/2)}\right)(\phi(\omega)-\phi(0))\,d\omega$$