I would like to find the adjoint of parametrized integral operator with respect to $L^2$ norm \begin{align} &S: L^2[0,2\pi] \to L^2[0,2\pi]\\ &(S\varphi)(t)=\int_{0}^{2\pi} \Phi(p(t),z(\tau))\varphi(\tau)|z'(\tau)|d\tau \end{align} where $z(t):=(z_1(t),z_2(t)), t\in [0,2\pi]$ and $p(\tau):=(p_1(\tau),p_2(t)), \tau\in [0,2\pi]$,
also $\Phi(p(t),z(\tau))=K_0(|(p(t)-z(\tau)|)$ which is modified bessel function and it is real
By using $<S\varphi, \psi>=<\varphi,S^*\psi>$, the obtained adjoint of $S$ equals \begin{align} \int_{0}^{2\pi} \Phi(z(t),p(\tau))\psi(\tau)|p'(t)|dt \end{align}
Is this result true ? Thank you in advance.
\begin{align} \langle S\varphi,\psi \rangle&=\int_0^{2\pi}\int_0^{2\pi}\psi(t)\Phi(p(t),z(\tau))\varphi(\tau)|z'(\tau)|\,\mathrm{d}\tau\mathrm{d}t \\ &=\int_0^{2\pi}\int_0^{2\pi}\psi(\tau)\Phi(p(\tau),z(t))\varphi(t)|z'(t)|\,\mathrm{d}\tau\mathrm{d}t \\ &=\langle \varphi,S^*\psi\rangle, \end{align} with $$S^*\psi(t)=\int_0^{2\pi}\Phi(p(\tau),z(t))\psi(\tau)|z'(t)|\,\mathrm{d}\tau.$$ Of course, this is the long way by interchanging the integrating variables. One could just recognize that Hilbert-Schmidt operators of the form $$ \varphi \mapsto \int_a^bK(x,y)\varphi(y)\,\mathrm{d}y $$ have adjoint $$ \psi\mapsto \int_a^b \overline{K(y,x)}\psi(y)\,\mathrm{d}y. $$ In this case, it appears everything is real, so conjugates are irrelevant.