Self-reciprocal Fourier functions of the form $\frac{\cosh \alpha x}{\cosh x+c}$

There are some functions that are self reciprocal under cosine Fourier transform: $$\begin{equation} \frac{1}{\cosh x}, \frac{\cosh x}{\cosh 2x},\frac{1}{1+2\cosh x}. \end{equation}$$ In other words they satisfy the equation $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$. These three functions were known to Ramanujan and their detailed study can be found in the book "Ramanujan's lost notebook, part IV" by Andrews and Berndt, and also in Titchmarsh's book "Introduction to the theory of Fourier integrals".

One might ask a question whether it is possible to find all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ that are self-reciprocal? There are some functions which are similar to these three, but actually are sums of several functions, e.g. $\frac{\cosh x-(1+\sqrt{3})/2}{\cosh(3x/2)}$ (see the unpublished paper "On the transformation of infinite series" by Bryden Cais). The three functions found by Ramanujan are composed of a single term, and we are interested only in such functions.

In view of the following integral evaluation from Gradshteyn and Ryzhik, (p. 539, formula 3.983, no. 6) $$\int_0^\infty \frac{\cos ax\cosh \beta x}{\cosh \gamma x+\cos b}dx= \pi\frac{\cos\left[\frac{\beta}{\gamma}(\pi-b)\right]\cosh\left[\frac{a}{\gamma}(\pi+b)\right]- \cos\left[\frac{\beta}{\gamma}(\pi+b)\right]\cosh\left[\frac{a}{\gamma}(\pi-b)\right]} {\gamma\sin b\left(\cosh\frac{2\pi a}{\gamma}-\cos\frac{2\pi \beta}{\gamma}\right)},\\ |\text{Re} \beta|<\text{Re} \gamma,\quad 0< b <\pi,\quad a>0,$$ this question can be reduced to finding all $0< b <\pi, 0< \beta < 1$ such that $$\frac{\cos\beta(\pi-b)\cdot \cosh a(\pi+b)-\cos\beta(\pi+b)\cdot \cosh a(\pi-b)}{\cosh 2\pi a-\cos 2\pi \beta}=\frac{\cosh\frac{\pi-b}{1-\beta}\beta a\cdot \cos\beta(\pi-b)}{\cosh\frac{\pi-b}{1-\beta}a+\cos b}\qquad (1)$$ for all $a\ge 0$.

(When $\beta=0$ the equation is $2\cdot\frac{\cosh a(\pi+b)-\cosh a(\pi-b)}{\cosh 2\pi a-1}=\frac{1}{\cosh(\pi-b)a+\cos b}$ which after substitution $a=0$ becomes $2b(1+\cos b)=\pi$ and has a solution $b=\frac{\pi}{3}$. This leads to Ramanujan's third function.)

Substitution $a=0$ in equation $(1)$ gives: $$\sin\beta b\cdot(1+\cos b)=\cos\beta(\pi-b)\cdot\sin\pi\beta. \qquad (2)$$ Then we simplify equation $(1)$: $$\begin{align} &\cos\beta(\pi-b)\cosh\frac{2b-\pi\beta-b\beta}{1-\beta}a-\cos\beta(\pi+b)\cosh\frac{(\pi-b)(2-\beta)}{1-\beta}a+\\ &\cos(3\pi-b)\beta\cosh\frac{\pi-b}{1-\beta}\beta a+2\cos\beta(\pi-b)\cos b\cosh(\pi+b)a-\\ & 2\cos\beta(\pi+b)\cos b\cosh(\pi-b)a=\cos\beta(\pi-b)\cosh\frac{2\pi-3\pi\beta+b\beta}{1-\beta}a \end{align}$$ Here we have 6 different exponents: $$\frac{|2b-\pi\beta-b\beta|}{1-\beta},\quad\frac{(\pi-b)(2-\beta)}{1-\beta}, \quad\frac{\pi-b}{1-\beta}\beta,\quad\pi+b,\quad\pi-b,\quad\frac{|2\pi-3\pi\beta+b\beta|}{1-\beta}.$$ When $\beta<1/2$ the exponent $\frac{\pi-b}{1-\beta}\beta$ can be equal to only $\frac{|2b-\pi\beta-b\beta|}{1-\beta}$. When $\beta>1/2$ the exponent $\pi-b$ can equal either $\frac{|2b-\pi\beta-b\beta|}{1-\beta}$ or $\frac{|2\pi-3\pi\beta+b\beta|}{1-\beta}$. Together with $(2)$ we have two equations. All required $\beta$ and $b$ will be contained in the solutions of this system of equations. Solving these equations one can not only recover all of the Ramanujan's examples but find two more examples: $\beta=3/4,b=2\pi/3$ and $\beta=\sqrt{3}/2,b=(\sqrt{3}-1)\pi$ (note that in the last case $\cos\beta(\pi+b)=0$).

So there are five self-reciprocal functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$: $$\begin{equation} \frac{1}{\cosh x}, \frac{\cosh x}{\cosh 2x},\frac{1}{1+2\cosh x},\frac{\cosh\frac{3x}{4}}{2\cosh x-1},\frac{\cosh\frac{\sqrt{3}x}{2}}{\cosh x-\cos(\sqrt{3}\pi)}. \end{equation}$$