Smoothness of the density for McKean-Vlasov SDEs with measurable kernel

Consider the McKean-Vlasov SDE $$dX_t=\langle b(X_t-\cdot),\mu_t\rangle dt+dW_t,\quad \mu_t=\operatorname{Law}(X_t),$$ where $W$ is the $n$-dimensional Brownian motion and $b:\mathbb{R}^d\to\mathbb{R}^d$ is a measurable function. First assuming $b\in L^\infty$, we prove that the law $\mu_t$ of $X_t$ has a density $p_t$ with respect to the Lebesgue measure, which is continuously differentiable with gradient being $\gamma$-H\"older continuous for each $\gamma\in(0,1)$. Assume further that $b\in \mathcal{C}_b^1$, we prove that the density $p_t$ is infinitely differentiable. In the regularization by noise perspective, this shows McKean-Vlasov SDEs tend to have a smoother density function than SDEs without density dependence, under the same regularity assumption of the coefficients. We observe similar phenomenon for singular interaction kernels satisfying Krylov's integrability condition, for distributional kernels $b\in B_{\infty,\infty}^\alpha$, $\alpha\in(-1,0)$, and for processes driven by an $\alpha$-stable noise for $\alpha\in(1,2)$.