Let $(R,\mathfrak m)$ be a Noetherian local ring of positive dimension. For an ideal $J$ of $R$, let $\bar J$ denote the integral closure of $J$ (https://en.m.wikipedia.org/wiki/Integral_closure_of_an_ideal) .
Is it true that $\cap_{n\ge 1} \overline{\mathfrak m^n}=0$ ? If this is not true in general, then under what additional assumption on $R$ , is it true (like regular local, minimal multiplicity) ?
It is known (Krull intersection Theorem) that $\cap_{n\ge 1} \mathfrak m^n=0$ and trivially, $\mathfrak m^n \subseteq \overline{\mathfrak m^n} \subseteq \mathfrak m$ but unfortunately that doesn't say much. One special case I know is that if $R$ is regular local of dimension $2$ then $\mathfrak m^n=\overline{\mathfrak m^n}, \forall n$ , so in that case we're good. More generally, It is known by a result of Ratliff that if $R$ is analytically unramified (i.e. the completion of $R$ is reduced https://en.m.wikipedia.org/wiki/Analytically_unramified_ring )
then for some $s>0$, we have $\overline{\mathfrak m^{ns}}=({\overline{\mathfrak m^s}})^n\subseteq \mathfrak m^n,\forall n$, so even in that case, our required intersection is $0$.
The answer is no in general. Let $R=k[\![x,y]\!]/(x^2,xy)$. Then, for $n>1$, $\mathfrak{m}^n=(y^n)$. But we notice $x^2+0x+0=0$ for any $n$, so $x$ is integral over $(y^n)$ for any $n$. In particular, this means $\bigcap^{\infty}_{i=1} \overline{\mathfrak{m}^n}$ is not zero in this case.
More generally, the nilradical is contained in the integral closure of any ideal, and so, in order for your desired intersection to be $0$, $R$ must at least be reduced. As you already mention, it is true if $R$ is analytically unramified, which is relatively mild and not far away from this.