Fourier Transform of product of trigonometric and polynomial function

Question

I am trying to calculate the correction to thermal conductance in a physics problem. It includes finding the Fourier transform of the following function: $$f(x)=\frac{x^3}{\sinh(x)(1+\sin^2x)}$$ Anyone has any idea how to approach this?

Answer 1

The transform can be written as an infinite sum. With $\operatorname{csch}$ understood in the p.v. sense and writing $1/(1 + \sin^2 x)$ as a Poisson kernel, we have $$\mathcal F[\operatorname{csch}] = \frac i 2 \tanh \frac {\pi w} 2, \\ \mathcal F\!\left[ \frac 1 {1 + \sin^2 x} \right] = \mathcal F \!\left[ \frac 1 {\sqrt 2} \sum_{k \in \mathbb Z} (3 + 2 \sqrt 2)^{-|k|} e^{2 i x k} \right] = \frac 1 {\sqrt 2} \sum_{k \in \mathbb Z} (3 + 2 \sqrt 2)^{-|k|} \delta(w + 2 k), \\ \mathcal F\!\left[ \frac {\operatorname{csch} x} {1 + \sin^2 x} \right] = \frac i {2 \sqrt 2} \sum_{k \in \mathbb Z} (3 + 2 \sqrt 2)^{-|k|} \tanh \left( \frac {\pi (w + 2 k)} 2 \right).$$

Answer 2

To perform this Fourier Transform, I'm going to use the following relations

$$\begin{align*}F(s) =\mathscr{F}\left\{f(x)\right\} &= \int_{-\infty}^{\infty} f(x) e^{-2\pi i sx} dx \quad \mbox{Fourier Transform convention}\\ \\ \mathscr{F}\left\{f(x)g(x)\right\} &= \mathscr{F}\left\{f(x)\right\} *\mathscr{F}\left\{g(x)\right\} = F(s) * G(s) \quad \mbox{Convolution theorem}\\ \\ \mathscr{F}\left\{F(x)\right\} &= f(-s) \quad \mbox{Repeated forward transform}\\ \\ \mathscr{F}\left\{\left(-2\pi i x\right)^nf(x)\right\} &=\dfrac{d^n}{ds^n}F(s) \quad \mbox{Derivative theorem}\\ \\ \mathscr{F}\left\{f(ax)\right\} &= \dfrac{1}{|a|}F\left(\dfrac{s}{a}\right)\quad \mbox{Similarity theorem}\\ \\ \mathscr{F}\left\{af(x)+bg(x)\right\} &= a F(s) + b G(s) \quad \mbox{Linearity}\\ \\ aF(s)*bG(s) &= ab\left[F(s) * G(s)\right] \quad \mbox{Scalar multiplication in convolution}\\ \\ \delta(as+b) &= \dfrac{1}{|a|}\delta\left(s + \dfrac{b}{a}\right) \quad \mbox{Scaling of Dirac delta independent axis}\\ \\ F(s) * \delta(s-a) &= F(s-a) \quad \mbox{Shift property of Dirac delta}\\ \\ F(s)^{*n} &= \underbrace{F(s) * F(s) * \dots * F(s)}_{n \space \mathrm{times}} \quad \mbox {Notation for self-convolution}\\ \\ \mathscr{F}\left\{\sin(\pi x)\right\} &= \dfrac{1}{2i}\left[\delta\left(s-\dfrac{1}{2}\right) - \delta\left(s+\dfrac{1}{2}\right)\right] \quad \mbox{From Bracewell, 1986}\\ \\ \mathscr{F}\left\{\tanh(\pi x)\right\}&=\dfrac{1}{i\sinh(\pi s)} \quad \mbox{From Bracewell, 1986}\\ \\ \dfrac{1}{1+\sin^2(x)} &= 1 + \sum_{n=1}^{\infty} (-1)^n \sin^{2n}(x) \quad \mbox{From long division}\\ \end{align*}$$

With those preliminaries out of the way, for your problem, you want to use the convolution theorem to break down the problem:

$$\mathscr{F}\left\{\dfrac{x^3}{\sinh(x)\left(1+\sin^2(x)\right)}\right\} = \mathscr{F}\left\{\dfrac{x^3}{\sinh(x)}\right\} * \mathscr{F}\left\{\dfrac{1}{1+\sin^2(x)}\right\}$$

We need an expression for $\mathscr{F}\left\{\dfrac{1}{\sinh(x)}\right\}$, so starting with $\mathscr{F}\left\{\tanh(\pi x)\right\}$

$$\begin{align*}\mathscr{F}\left\{\tanh(\pi x)\right\}&=\dfrac{1}{i\sinh(\pi s)}\\ \\ \mathscr{F}\left\{i \tanh(\pi x)\right\}&=\dfrac{1}{\sinh(\pi s)}\\ \\ \mathscr{F}\left\{i \tanh(\pi^2 x)\right\}&=\dfrac{1}{\pi}\dfrac{1}{\sinh(s)}\\ \\ \mathscr{F}\left\{i \pi \tanh(\pi^2 x)\right\}&=\dfrac{1}{\sinh(s)}\\ \\ \mathscr{F}\left\{\dfrac{1}{\sinh(x)}\right\}&= i \pi \tanh(-\pi^2 s)= -i \pi \tanh(\pi^2 s)\\ \\ \end{align*}$$

Now to factor in the $x^3$ using the derivative theorem

$$\begin{align*}\mathscr{F}\left\{\dfrac{x^3}{\sinh(x)}\right\} &= \dfrac{1}{(-2\pi i)^3}\mathscr{F}\left\{ \dfrac{(-2\pi i x)^3}{\sinh(x)} \right\} \\ \\ &= \dfrac{-i\pi}{(-2\pi i)^3} \dfrac{d^3}{ds^3} \tanh(\pi^2 s)\\ \\ &= -\dfrac{1}{8\pi^2} \dfrac{d^3}{ds^3} \tanh(\pi^2 s)\\ \\ \mathscr{F}\left\{\dfrac{x^3}{\sinh(x)}\right\}&= -\dfrac{\pi^4}{4} \mathrm{sech}^4\left(\pi^2 s\right)\left[2\sinh^2(\pi^2 s)-1\right]\\ \end{align*}$$

That gives us the result for the first term of the convolution.

Now we need to find $\mathscr{F}\left\{\sin(x)\right\}$, so starting with $\mathscr{F}\left\{\sin(\pi x)\right\}$

$$\begin{align*}\mathscr{F}\left\{\sin(\pi x)\right\} &= \dfrac{1}{2i}\left[\delta\left(s-\dfrac{1}{2}\right) - \delta\left(s+\dfrac{1}{2}\right)\right]\\ \\ \mathscr{F}\left\{\sin(x)\right\} &= \dfrac{\pi}{2i}\left[\delta\left(\pi s-\dfrac{1}{2}\right) - \delta\left(\pi s+\dfrac{1}{2}\right)\right]\\ \\ &= \dfrac{1}{2i}\left[\delta\left( s-\dfrac{1}{2\pi}\right) - \delta\left( s+\dfrac{1}{2\pi}\right)\right]\\ \end{align*}$$

Now for the second term

$$\begin{align*}\mathscr{F}\left\{\dfrac{1}{1+\sin^2(x)}\right\} &= \mathscr{F}\left\{1 + \sum_{n=1}^{\infty} (-1)^n \sin^{2n}(x)\right\} \\ \\ &= \delta(s) + \sum_{n=1}^{\infty} (-1)^n \left(\dfrac{1}{2i}\left[\delta\left( s-\dfrac{1}{2\pi}\right) - \delta\left( s+\dfrac{1}{2\pi}\right)\right]\right)^{*2n} \\ &= \delta(s) + \sum_{n=1}^{\infty} \left(\dfrac{1}{2}\right)^{2n} \left[\delta\left( s-\dfrac{1}{2\pi}\right) - \delta\left( s+\dfrac{1}{2\pi}\right)\right]^{*2n} \\ \\ &= \delta(s) + \sum_{n=1}^{\infty} \left(\dfrac{1}{2}\right)^{2n} \left[ \sum_{k=-n}^{n} (-1)^{k+n}{2n \choose k+n}\delta\left( s+\dfrac{k}{\pi}\right)\right] \\ \\ &= \sum_{n=0}^{\infty} \left(\dfrac{1}{2}\right)^{2n} \left[ \sum_{k=-n}^{n} (-1)^{k+n}{2n \choose k+n}\delta\left( s+\dfrac{k}{\pi}\right)\right] \\ \mathscr{F}\left\{\dfrac{1}{1+\sin^2(x)}\right\} &= \sum_{m=-\infty}^{\infty} c_m \delta\left( s+\dfrac{m}{\pi}\right) \end{align*}$$

With

$$\begin{align*}c_m &= \sum_{n=|m|}^{\infty} \left(\dfrac{1}{2}\right)^{2n} (-1)^{m+n}{2n \choose m+n}\\ \\ &= c_{-m}\\ \\ &= \sum_{n=|m|}^{\infty} \left(\dfrac{1}{2}\right)^{2n} (-1)^{|m|+n}{2n \choose |m|+n}\\ \end{align*}$$

So putting the above results together

$$\begin{align*}\mathscr{F}\left\{\dfrac{x^3}{\sinh(x)\left(1+\sin^2(x)\right)}\right\} &= \mathscr{F}\left\{\dfrac{x^3}{\sinh(x)}\right\} * \mathscr{F}\left\{\dfrac{1}{1+\sin^2(x)}\right\}\\ \\ &= \left(-\dfrac{\pi^4}{4} \mathrm{sech}^4\left(\pi^2 s\right)\left[2\sinh^2(\pi^2 s)-1\right]\right) * \sum_{m=-\infty}^{\infty} c_m \delta\left( s+\dfrac{m}{\pi}\right)\\ \\ &= \sum_{m=-\infty}^{\infty} -c_m\dfrac{\pi^4}{4} \mathrm{sech}^4\left(\pi^2 \left[s+\dfrac{m}{\pi}\right]\right)\left[2\sinh^2\left(\pi^2 \left[s+\dfrac{m}{\pi}\right]\right)-1\right]\\ \\ \mbox{with}\\ \\ c_m = c_{-m} &= \sum_{n=|m|}^{\infty} \left(\dfrac{1}{2}\right)^{2n} (-1)^{|m|+n}{2n \choose |m|+n}\\ \end{align*}$$

So your Fourier Transform ends up being the summation of scaled, shifted copies of the third derivative of $\tanh(\pi^2 s)$ with the scaled copies centered at frequencies $s = \dfrac{m}{\pi}$

See this question for a possible closed from for the $c_m$: Closed form for $c_m = \sum_{n=|m|}^{\infty} \left(\dfrac{1}{2}\right)^{2n} (-1)^{m+n}{2n \choose m+n}$, $m$ integer