For start, is $\int \limits\_{2}^{x} \frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)} dt \leq x\ln(\ln(x)) \frac{e^{-0.3\sqrt{\ln(x)}}}{\ln^2(x)}$ ?


If the above is true, what is a better bound for the integral ?



What is the upper bound for $\int \limits_{2}^{x} \frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)} dt$?
Write $f(x)=\frac{e^{-0.3\sqrt{\log t}}}{\log^2 t}$. On the interval $[2, xf(x)]$ bound the integral trvially by $xf(x)$. On $[xf(x), x]$ the integrand is close to constant. More precisely, we have
$$
\frac{f(xf(x))}{f(x)} \leq e^{-0.3\left(\sqrt{\log xe^{-0.4\sqrt{\log x}})}-\sqrt{\log x)}\right)}\frac{\log^2 x}{\log^2(xe^{-0.4\sqrt{\log x}})}\leq e^{0.25},
$$
provided that $x$ is sufficiently large. Hence the integral in question satisfies
$$
x\frac{e^{-0.3\sqrt{\log x}}}{\log^2 x} \leq\int\_2^x\frac{e^{-0.3\sqrt{\log t}}}{\log^2 t}\;dt \leq e^{0.25} x\frac{e^{-0.3\sqrt{\log x}}}{\log^2 x}.
$$
If you need higher precision, you can get them by choosing the subdivision more carefully. With some work you should be able to get an asymptotic formula with relative error term some negative power of $\log x$.


