# What are all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ self-reciprocal under Fourier transform?

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}

It seems that they have been discovered by Ramanujan. Detailed study of these three functions 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".

Recently I have found that \begin{equation} \frac{\cosh\frac{\sqrt{3}x}{2}}{\cosh x-\cos(\sqrt{3}\pi)} \qquad (1) \end{equation} is self-reciprocal under Fourier transform. Although I doubt that this hasn't been derived before, I can't find any information about this particular function in the literature. Can someone point me to the literature where this function is studied in more detail?

Is it 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. $\frac{\cosh x-(1+\sqrt{3})/2}{\cosh(3x/2)}$. The three functions found by Ramanujan are composed of a single term, and we are interested only in such functions.

Below I show how self-reciprocity of (1) can be established, which is pretty straightforward: Let's substitute $\gamma=1,\beta=\frac{\sqrt{3}}{2}, b=(\sqrt{3}-1)\pi$ in the formula p. 539, formula 3.983, no. 6, from Gradshteyn and Ryzhik, \begin{align} \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 \end{align} We see that since $\frac{\sqrt{3}}{2}<1, 0<\sqrt{3}-1<1$ all the conditions for the validity of this integral evaluation are satisfied, and we get $$\int_0^\infty \frac{\cos ax \cosh\frac{\sqrt{3}x}{2}}{\cosh x-\cos(\sqrt{3}\pi)}d x=\pi\frac{\cosh(a\sqrt{3}\pi)}{\cosh 2\pi a-\cos(\sqrt{3}\pi)}$$ It can be written in the symmetrical form $$\sqrt{\frac{2}{\pi}}\int_0^\infty\frac{\cosh(\sqrt{\frac{3\pi}{2}}x)}{\cosh(\sqrt{2\pi}x) -\cos(\sqrt{3}\pi)}\cos ax dx=\frac{\cosh(\sqrt{\frac{3\pi}{2}}a)}{\cosh(\sqrt{2\pi}a) -\cos(\sqrt{3}\pi)}$$

Also the following identity can be derived from the integral mentioned above \begin{align} \frac{\Gamma(\frac{1}{2}+\frac{s}{2})}{\pi^{\frac{s}{2}}}\sum_{n=-\infty}^\infty\text{sgn}(n)\frac{\sin(\sqrt{3}\pi p n)}{{\bigl\lvert n+\frac{\sqrt{3}}{2p}\bigr\rvert}^s}= \frac{\Gamma(1-\frac{s}{2})}{\pi^{\frac{1-s}{2}}}\sum_{n=-\infty}^\infty\text{sgn}(n)\frac{\sin(\sqrt{3}\pi q n)}{{\bigl\lvert n+\frac{\sqrt{3}}{2q}\bigr\rvert}^{1-s}},\quad pq=1,\quad\frac{\sqrt{3}}{2}<p<\frac{2}{\sqrt{3}}. \end{align} This identity is an 8 term functional equation for Lerch zeta function and in fact it can be shown that it is a consequence of Lerch's functional equation. I have searched the literature about Lerch zeta function, but nobody mentions that the Lerch Zeta function has any symmetry at the point $\sqrt{3}/2$.

$\bf{Update:}$ The question can be reduced to finding all $0<b<\pi, \beta<1$ such that $$\frac{\cos\beta(\pi-b)\cdot \cosh a(\pi+b)-\cos\beta(\pi+b)\cdot \cosh a(\pi-b)}{\cosh 2\pi a-\cos 2\pi \beta}=\frac{\cosh\frac{\pi-b}{1-\beta}\beta a\cdot \cos\beta(\pi-b)}{\cosh\frac{\pi-b}{1-\beta}a+\cos b} \quad(2)$$ for all $a\ge 0$.

The case $\beta=0$ doesn't lead to anything new. When $\beta\neq 0$ we can substitute $a=0$ and obtain: $$\sin\beta b\cdot(1+\cos b)=\cos\beta(\pi-b)\cdot\sin\pi\beta. \qquad (3)$$

Then we simplify equation (2): \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 (3) we have two equations. I have solved this equations and besides recovering all of the Ramanujan's examples and the case $\beta=\sqrt{3}/2,b=(\sqrt{3}-1)\pi$ I found one more example: $\beta=3/4,b=2\pi/3$.

So the answer is there are 5 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}

• Can you please clarify what, exactly, do you mean by self-reciprocal? Do you want $\mathcal F f=f$? Or is it sufficient to have this with some scaling? – mickep Oct 17 '15 at 20:01
• I have used residue calculus to calculate the Fourier transform of $$\dfrac{\cosh(\beta x)}{\cosh(x)-\cos(\delta \pi)}.$$ When I applied that to your third and fourth function I got the transformations $$\dfrac{1}{3}\dfrac{\pi\sqrt{3}(\cosh(\frac{4\pi a}{3})-\cosh(\frac{2\pi a}{3}))}{\cosh(2\pi a)-1}$$ respectively $$\dfrac{\pi}{\sqrt{6}}\dfrac{\cosh(\frac{5\pi a}{3})+\cosh(\frac{\pi a}{3})}{\cosh(2\pi a)}.$$ For me it does not looks like self-reciprocal functions. – JanG Oct 19 '15 at 9:38
• Why are you editing your question in a way to make the existing answer obsolete (as well as changing it to a state where it doesn't even contain a question in the body of the text)? – mrf Oct 26 '15 at 10:18
• @mrf Existing "answer" has nothing to do with the question I asked. Also I want to delete this question, but stubborn moderators say that question with active bounty can't be deleted. – user278906 Oct 26 '15 at 10:27
• Regarding the comment made by @JanG above: that functions can be brought to the desired form to make the self-reciprocity explicit by trivial arithmetics. I leave the excercise for you. In fact in the body of the question I did exactly that in more general setting. I will not be able to respond to anybody, because tomorrow my account will be deleted. Bye. – user278906 Oct 27 '15 at 20:31

## 2 Answers

In Erdelyi et al., Table of integral transforms volume 1 on the Fourier cosine transform, the relation 1.9(17) page 31 gives (using $\beta=1$) $${\mathcal F}_{\cos}\left[\frac{\cosh ax}{\cosh x+\cos c}\right](z)= -2\pi\frac{\cos(ac)\sinh(\pi)}{\sin c}\; \frac{\sinh(cz)} {\cosh(2\pi z)-\cos(2\pi a)}$$ where $0<c<\pi$ and $|\operatorname{Re}(a)|<1.$

Therefore, your claim will be true only for tuned parameters $a$ and $c$, as in the examples you treated above.

$$\newcommand{\res}{\operatorname{Res}} \newcommand{\re}{\operatorname{Re}}$$

I cannot give you the references you are looking for but I will calculate the Fourier transform of your fifth function.

Put $$I = \dfrac{1}{2}\int_{-\infty}^{\infty}\dfrac{e^{iax}\cosh\frac{\sqrt{3}x}{2}}{\cosh x -\cos(\sqrt{3}\pi)}\, dx.$$ Since $I$ is an even function of $a$ it is no essential restriction to assume that $a>0$.

We split $I$ into two terms $I = \frac{1}{2}I_1+ \frac{1}{2}I_2$ where

$$I_1 = \int_{-\infty}^{\infty}\dfrac{e^{iax}e^{\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\, dx$$ and $$I_2 = \int_{-\infty}^{\infty}\dfrac{e^{iax}e^{-\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\, dx = \int_{-\infty}^{\infty}\dfrac{e^{-iax}e^{\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\, dx = \overline{I_1}.$$ Consequently $\displaystyle I = \re(I_1).$

To be able to calculate $I_1$ we integrate in positive direction around a rectangle with corners in $(R,0), (R,i2\pi), (-R,i2\pi)$ and $(-R,0)$. Since $a>0$ and $\frac{\sqrt{3}}{2}<1$ there will be no contributions from the vertical edges as $R \to \infty$. There are two singularities $i\sqrt{3}\pi$ and $i(2-\sqrt{3})\pi$ inside the curve. The residue theorem then gives that $$\begin{gathered} I_1-\int_{-\infty}^{\infty}\dfrac{e^{ia(x+i2\pi)}e^{\frac{\sqrt{3}(x+i2\pi)}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\, dx = (1-e^{-2a\pi+i\sqrt{3}\pi})I_1 \\[2ex] = 2{\pi}i\left(\underset{x=i\sqrt{3}\pi}{\res}\dfrac{e^{iax}e^{\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}+\underset{x=i(2-\sqrt{3})\pi}{\res}\dfrac{e^{iax}e^{\frac{\sqrt{3}x}{2}}}{e^{x}+e^{-x}-2\cos(\sqrt{3}\pi)}\right)\\[2ex] = 2{\pi}i\left(\dfrac{e^{-a\sqrt{3}\pi}e^{\frac{i3\pi}{2}}}{2i\sin(\sqrt{3}\pi)}+ \dfrac{e^{-a(2-\sqrt{3})\pi}e^{\frac{i\sqrt{3}(2-\sqrt{3})\pi}{2}}}{2i\sin((2-\sqrt{3})\pi)}\right) = \dfrac{-{\pi}i}{\sin(\sqrt{3}\pi)}(e^{-a\sqrt{3}\pi}+e^{-2a\pi +a\sqrt{3}\pi +i\sqrt{3}\pi}). \end{gathered}$$ We now make some preparations for the calculation of $\re(I_1).$ $$\begin{gathered} I_1 = \dfrac{-{\pi}i}{\sin(\sqrt{3}\pi)}(e^{-a\sqrt{3}\pi}+e^{-2a\pi +a\sqrt{3}\pi +i\sqrt{3}\pi})\dfrac{1-e^{-2a\pi-i\sqrt{3}\pi}}{(1-e^{-2a\pi+i\sqrt{3}\pi})(1-e^{-2a\pi-i\sqrt{3}\pi})}\\[2ex] = \dfrac{-{\pi}i}{\sin(\sqrt{3}\pi)}\dfrac{e^{-a\sqrt{3}\pi}-e^{-2a\pi-a\sqrt{3}\pi -i\sqrt{3}\pi}+ e^{-2a\pi +a\sqrt{3}\pi +i\sqrt{3}\pi} - e^{-4a\pi+a\sqrt{3}\pi}}{1+e^{-4a\pi}-2e^{-2a\pi}\cos(\sqrt{3}\pi)}. \end{gathered}$$ Finally we have that $$\begin{gathered} I =\re(I_1) = \dfrac{{\pi}}{\sin(\sqrt{3}\pi)}\dfrac{e^{-2a\pi +a\sqrt{3}\pi}\sin(\sqrt{3}\pi)+e^{-2a\pi-a\sqrt{3}\pi}\sin(\sqrt{3}\pi)}{1+e^{-4a\pi}-2e^{-2a\pi}\cos(\sqrt{3}\pi)}\\[2ex] = \pi\dfrac{e^{a\sqrt{3}\pi}+e^{-a\sqrt{3}\pi}}{e^{2a\pi}+e^{-2a\pi}-2\cos(\sqrt{3}\pi)} = \pi\dfrac{\cosh(a\sqrt{3}\pi)}{\cosh(2a\pi)-\cos(\sqrt{3}\pi)}. \end{gathered}$$