$\bf{1.}$ Consider the following self-obvious statement: If $f(x,y)=f(y,x)$ and $$\sqrt{\frac{2}{\pi}}\int_0^\infty f(x,y)\cos ax dx=g(a,y)=g(y,a),$$ (in other words, if partial Fourier transform of a symmetric function is symmetric) then $f(x,y)$ is a self-reciprocal Fourier function of two variables, i.e. $$ {\frac{2}{\pi}}\int\limits_0^\infty \int\limits_0^\infty f(x,y)\cos ax\cos by dxdy=f(a,b). $$ Using this simple observation one can find some non-trivial self-reciprocal functions. By non-trivial we mean non-separable functions, i.e functions that are not products $f_1(x)f_2(y)$ and also are not the functions of the form $f+f_c$.
Examples: Since $$ \int\limits_0^\infty\frac{\sin xy}{\sinh(\sqrt{\pi} x)}\cos ax dx=\frac{\sqrt{\pi}}{2}\frac{ \sinh(\sqrt{\pi}y)}{\cosh(\sqrt{\pi}y)+\cosh(\sqrt{\pi}a)}$$ we get two self-reciprocal Fourier transformations $$ \frac{2}{\pi}\int\limits_0^\infty \int\limits_0^\infty\frac{\cos ax\cos by}{\cosh\sqrt{\pi}x+\cosh\sqrt{\pi}y}dxdy=\frac{1}{\cosh\sqrt{\pi}a+\cosh\sqrt{\pi}a},\qquad (1) $$ $$ \frac{2}{\pi}\int\limits_0^\infty \int\limits_0^\infty\frac{\sin xy}{\sinh(\sqrt{\pi} x) \sinh(\sqrt{\pi}y)}\cos ax\cos bydxdy=\frac{\sin ab}{\sinh(\sqrt{\pi} a) \sinh(\sqrt{\pi}b)}.\qquad (2) $$ The integral from Gradshteyn and Ryzhik $$ \int_0^\infty K_0(\alpha\sqrt{x^2+\beta^2})\cos\gamma xdx=\frac{\pi}{2\sqrt{\alpha^2+\gamma^2}}e^{-\beta\sqrt{\alpha^2+\gamma^2}} $$ written in the form $$ \int_0^\infty K_0(\sqrt{\alpha^2+\beta^2}\sqrt{x^2+\beta^2})\cos\gamma xdx=\frac{\pi}{2\sqrt{\alpha^2+\gamma^2+\beta^2}}e^{-\beta\sqrt{\alpha^2+\gamma^2+\beta^2}} $$ leads to the following self-reciprocal functions $$ \frac{2}{\pi}\int\limits_0^\infty \int\limits_0^\infty \frac{e^{-\beta\sqrt{x^2+y^2+\beta^2}}}{\sqrt{x^2+y^2+\beta^2}}\cos ax\cos by\phantom{.}dxdy=\frac{e^{-\beta\sqrt{a^2+b^2+\beta^2}}}{\sqrt{a^2+b^2+\beta^2}},\qquad (3) $$ $$ \frac{2}{\pi}\int\limits_0^\infty \int\limits_0^\infty K_0(\sqrt{x^2+c^2}\sqrt{y^2+c^2}) \cos ax\cos by dxdy=K_0(\sqrt{a^2+c^2}\sqrt{b^2+c^2}).\qquad (4) $$ The functions $(1),(2),(3)$ can be used to generate many two dimensional analogs of identites similar to identities given by Hardy and Ramanujan (see "Ramanujan's lost notebook, part IV", by Andrews and Berndt), i.e. $$ \sqrt{\alpha\beta}\int\limits_0^\infty\int\limits_0^\infty\frac{e^{-x^2-y^2}}{\cosh\alpha x+\cosh \beta y}dxdy= \sqrt{\gamma\delta}\int\limits_0^\infty\int\limits_0^\infty\frac{e^{-x^2-y^2}}{\cosh\gamma x+\cosh \delta y}dxdy, $$ where $\alpha\gamma=\beta\delta=2\pi$. This identity also can be written as $$ \alpha\int\limits_0^\infty\int\limits_0^\infty\frac{e^{-x^2-y^2}\cos(2qxy)}{\cosh\alpha x\cosh \alpha y}dxdy=\beta\int\limits_0^\infty\int\limits_0^\infty\frac{e^{-x^2-y^2}\cos(2qxy)}{\cosh\beta x\cosh \beta y}dxdy,\qquad \alpha\beta=\pi\sqrt{1+q^2}. $$
Applying Poisson summation formula to the function $\frac{e^{-\beta\sqrt{x^2+y^2+\beta^2}}}{\sqrt{x^2+y^2+\beta^2}}$ one obtains Ramanujan's formula (see Hardy, "Twelve lectures", Lecture 5) $$ \sum_{n=0}^\infty\frac{r(n)}{\sqrt{n+a}}e^{-2\pi\sqrt{(n+a)b}}=\sum_{n=0}^\infty\frac{r(n)}{\sqrt{n+b}}e^{-2\pi\sqrt{(n+b)a}} $$ where $r(n)-$ is the sum of squares function, and also elliptic generalizations due to Hardy.
Similarly, applying Poisson summation formula to the function $K_0(\sqrt{x^2+c^2}\sqrt{y^2+c^2})$ one obtains the following curious identities $$ \sum_{n=0}^\infty (-1)^n \lambda(n) K_0(\pi\sqrt{n})=0 $$ $$ 2\sum_{n=0}^\infty\lambda(n)K_0(2\pi\sqrt{n})=\sum_{n=0}^\infty\frac{r(n)}{\sqrt{n+1}}e^{-2\pi\sqrt{n+1}} $$ where $\lambda(n)-$ is the number of integer solutions of the equation $(x^2+1)(y^2+1)=n$.
$\bf{2.}$ Consider now functions of the form $f(\sqrt{x^2+y^2})$. Then \begin{align} \frac{2}{\pi}\int\limits_0^\infty \int\limits_0^\infty f(\sqrt{x^2+y^2})\cos ax\cos by dxdy&=\frac{1}{2\pi}\int\limits_{-\infty}^\infty \int\limits_{-\infty}^\infty f(\sqrt{x^2+y^2})e^{i(ax+by)} dxdy\\ &=\frac{1}{2\pi}\int_0^\infty rdr\int_0^{2\pi}d\varphi f(r)e^{ir\sqrt{a^2+b^2}\cos\varphi}\\ &=\int\limits\limits_0^\infty rf(r)J_0(r\sqrt{a^2+b^2})dr. \end{align} This means that if $f(x)$ is its own Hankel transform of order 0, then $f(\sqrt{x^2+y^2})$ is a self reciprocal function of two variables.
$\it{Examples.}$ Functions $3.1$ and $3.3$ from this list with $\nu=0$ give $$ \frac{2}{\pi}\int\limits_0^\infty \int\limits_0^\infty J_0\left(\frac{x^2+y^2}{2}\right)\cos ax\cos bydxdy=J_0\left(\frac{a^2+b^2}{2}\right)\tag{5} $$ $$ \frac{2}{\pi}\iint\limits_{\substack{x\ge 0,y\ge 0\\x^2+y^2\ge z^2}}\frac{\cos\left(z\sqrt{x^2+y^2-z^2}\right)}{\sqrt{x^2+y^2-z^2}}\cos ax\cos bydxdy=\frac{\cos\left(z\sqrt{a^2+b^2-z^2}\right)}{\sqrt{a^2+b^2-z^2}},~a^2+b^2\ge z^2 \tag{6} $$
Partial Fourier transform of (6) using eq. 3.876.2 from Gradsteyn&Ryzhik gives another function $$ f(x)=\begin{cases} \displaystyle \sqrt{\frac2\pi}K_0\left(\sqrt{(x^2-z^2)(y^2-z^2)}\right),\qquad (x^2-z^2)(y^2-z^2)>0\\ \displaystyle -\sqrt{\frac\pi2}N_0\left(\sqrt{(z^2-x^2)(y^2-z^2)}\right),\qquad\ \ \ (x^2-z^2)(y^2-z^2)<0 \end{cases}\tag{7} $$
$\bf{3.}$ Another class of functions are constructed from self reciprocal functions of one variable as follows. Let $f_1(x)$ and $f_2(x)$ be such two functions and $\alpha\in\mathbb{R}$. Then the function $$ f(x,y)=f_1\left(x\cos\alpha+y\sin\alpha\right)f_2\left(y\cos\alpha-x\sin\alpha\right) $$ is a self reciprocal function of two variables. Note that (1) is of this form. Although these functions are non-trivial in a sense defined above, they are less interesting than the functions (2)-(7).
