Consider a Borel-measurable function $a:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{C}$, and let $T_a$ be the linear operator on $L^2(\mathbb{R})$ with domain \begin{equation} \mathcal{D}(T_a)=\left\{f\in L^2(\mathbb{R}):\int\left|\int a(x,y)f(y)\;\mathrm{d}y\,\right|^2\mathrm{d}x<\infty\right\} \end{equation} acting as \begin{equation} \left(T_af\right)(x)=\int a(x,y)f(y)\;\mathrm{d}y, \end{equation} that is, the integral operator with kernel $a(x,y)$. We assume $\mathcal{D}(T_a)$ to be dense in $L^2(\mathbb{R})$.
It is fairly immediate to prove that, if the kernel is assumed to be square-integrable with respect to both entries: \begin{equation} \int\left|a(x,y)\right|^2\;\mathrm{d}x\,\mathrm{d}y<\infty, \end{equation} then $T_a$ is a bounded operator (in fact, it is Hilbert-Schmidt) whose adjoint $T_a^*$ is equal to the integral operator associated with the function $a^*(x,y)=\overline{a(y,x)}$, that is, $T_a^*=T_{a^*}$. In particular, if $\overline{a(y,x)}=a(x,y)$, then $T_a$ is self-adjoint.
My question is whether, or under what conditions, the same holds if the condition above is relaxed, i.e. if the equality $T_a^*=T_{a^*}$ also holds for unbounded integral operators. In general, it is easy to prove that $T_{a^*}$ and $T_a$ have the same domain and \begin{equation} T_{a^*}\subseteq T_a^*, \end{equation} since, given $f,g\in\mathcal{D}(T_a)$, \begin{equation} \left\langle T_{a^*}g,f\right\rangle=\left\langle g,T_af\right\rangle. \end{equation} However, we are left with proving the converse inequality $\mathcal{D}(T_a^*)\subseteq\mathcal{D}(T_a)$. I was trying to do that by using the very definition of adjoint: given $g\in\mathcal{D}(T_a^*)$, there must exist some $\tilde{g}\in L^2(\mathbb{R})$ such that, for every $f\in\mathcal{D}(T_a)$, \begin{equation} \left\langle\tilde{g},f\right\rangle=\left\langle g,T_af\right\rangle, \end{equation} that is, writing the integrals explicitly and taking complex conjugates, for every $f\in\mathcal{D}(T_a)$, \begin{equation} \int\left[\tilde{g}(y)-\int \overline{a(y,x)}(x)\,\mathrm{d}x\right]\overline{f(y)}\:\mathrm{d}y=0. \end{equation} The idea is to use the arbitrariety of $f$ above to show that the quantity between square brackets must vanish. EDIT: More precisely, I thought about that: given $f\in L^2(\mathbb{R})$, by assumption there is a sequence of functions $\{f_n\}_{n\in\mathbb{N}}\subset\mathcal{D}(T_a)$ with $f_n\to f$; for such a sequence, we have \begin{equation} \int\left[\tilde{g}(y)-\int \overline{a(y,x)}g(x)\,\mathrm{d}x\right]\overline{f_n(y)}\:\mathrm{d}y=0. \end{equation} Formally, taking the limit $n\to\infty$ and using the continuity of the $L^2(\mathbb{R})$ scalar product, one would show that indeed \begin{equation} \int\left[\tilde{g}(y)-\int \overline{a(y,x)}g(x)\,\mathrm{d}x\right]\overline{f(y)}\:\mathrm{d}y=0, \end{equation} which would prove the desired property, since $f$ is arbitrary. However, to do that, the assumption $g\in\mathcal{D}(T_a^*)$ alone must imply that, indeed, the function $y\mapsto\int \overline{a(y,x)}g(x)\,\mathrm{d}x$ be square-integrable.
Notice, as a final consideration, that a similar procedure is usually carried out for multiplication operators $(M_Af)(x)=A(x)f(x)$ (which, roughly speaking, may be interpreted as integral operators with a degenerate kernel $a(x,y)=A(x)\delta(x-y)$, the latter being the Dirac distribution). In such a case, a procedure analogous to the one above allows one to prove that indeed $M_{A}^*=M_{A^*}$.