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 ?
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Sign up to join this communityFor 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 ?
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 $\log x$.
I would estimate the integral as follows. If $$f(x):=xe^{-0.4\sqrt{\ln(x)}},$$ then \begin{align*} \int \limits_{2}^{x} \frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)} dt &=\int \limits_{2}^{f(x)} \frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)} dt +\int \limits_{f(x)}^{x} \frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)} dt\\ &\ll f(x)+x\sup_{f(x)\leq t\leq x}\frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)}. \end{align*} For $f(x)\leq t\leq x$ we have $$\ln(t)=\ln(x)+O\left(\sqrt{\ln(x)}\right)=\ln(x)\left(1+O\left(1/\sqrt{\ln(x)}\right)\right)=(1+o(1))\ln(x),$$ whence $$\sqrt{\ln(t)}=\sqrt{\ln(x)}\left(1+O\left(1/\sqrt{\ln(x)}\right)\right)=\sqrt{\ln(x)}+O(1),$$ and also, more trivially, $$\ln(t)^2=(1+o(1))\ln(x)^2.$$ This shows that $$\sup_{f(x)\leq t\leq x}\frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)}\ll\frac{e^{-0.3\sqrt{\ln(x)}}}{\ln^2(x)}.$$ Using this estimate in our initial bound, we conclude that $$ \int \limits_{2}^{x} \frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)} dt \ll f(x) + x\frac{e^{-0.3\sqrt{\ln(x)}}}{\ln^2(x)}\ll x\frac{e^{-0.3\sqrt{\ln(x)}}}{\ln^2(x)}.$$ That is, there is an absolute constant $C>0$ such that for any $x\geq 2$ we have $$ \int \limits_{2}^{x} \frac{e^{-0.3\sqrt{\ln(t)}}}{\ln^2(t)} dt < C x\frac{e^{-0.3\sqrt{\ln(x)}}}{\ln^2(x)}.$$ You can find an explicit value of $C>0$ by calculating all the $O$-constants in the above estimates, and I leave this calculation to you.