Since$$\prod\_{p \leq n} \left(1-\frac{1}{p}\right) =\frac{ e^{-\gamma}+o(1)}{ \log n},$$ by Mertens theorem, the density of integers in $$(X^{\theta},X],$$ which aren't divisible by primes $$p \leq X^{\theta}$$
is $$\rho(X)\sim \frac{ e^{-\gamma}}{\theta \log X}.$$


How small a subinterval in this interval can inherit this density?



Local density of numbers not divisible by small primes
Actually, the density you indicate is false, because $X^\theta$ is too large for the primes $p\leq X^\theta$ to behave sufficiently independently. For example, for $\theta=1/2$ the sifted set consists precisely of the primes in $(X^{1/2},X]$ whose density is $1/\log X$ (instead of $2e^{-\gamma}/\log X$) by the Prime Number Theorem.


Let $\Phi(x,y)$ denote the number of integers up to $x$ which are not divisible by any prime up to $y$. Let $u:=\log x/\log y$. Then Theorem 3 in Section of III.6.2 of Tenenbaum: Introduction to analytic and probabilistic number theory (Cambridge University Press, 1995) gives the asymptotic formula
$$ \Phi(x,y)=\frac{x\omega(u)-y}{\log y}+O\left(\frac{x}{(\log y)^2}\right),\qquad x\geq y\geq 2,$$
where $\omega:[1,\infty)\to[1/2,1]$ is the [Buchstab function](https://en.wikipedia.org/wiki/Buchstab_function) satisfying (cf. Corollary 3.1 after the theorem)
$$ \omega(u)=e^{-\gamma}+O(u^{-u/2}),\qquad u\geq 1.$$
Better bounds and various refinements are also available: see Chapter III.6 of the book in more detail.


