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How can I prove that:

$$ \int_1^x e^{-\sqrt{\ln(t)}} \mathrm dt \sim_{x \rightarrow \infty} xe^{-\sqrt{\ln(x)}}$$

without using l'Hôpital's rule ?

Integration by parts:

$$ \int_1^x e^{-\sqrt{\ln(t)}} \mathrm dt= xe^{-\sqrt{\ln(x)}}-1+\frac{1}{2}\int_1^x \frac{e^{-\sqrt{\ln(t)}}}{\sqrt{\ln(t)}} \mathrm dt=xe^{-\sqrt{\ln(x)}}+o(xe^{-\sqrt{\ln(x)}})+\frac{1}{2}\int_1^x \frac{e^{-\sqrt{\ln(t)}}}{\sqrt{\ln(t)}} \mathrm dt $$

So how can I show that $$ \int_1^x \frac{e^{-\sqrt{\ln(t)}}}{\sqrt{\ln(t)}} \mathrm dt =o(xe^{-\sqrt{\ln(x)}}) $$ ?

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up vote 2 down vote accepted

Using the change of variable $t=\mathrm e^{s^2}$, one sees that one must show that $I(u)\ll K(u)$ when $u\to+\infty$, with $$ K(u)=\mathrm e^{u^2-u},\qquad I(u)=\int_0^uK(s)\mathrm ds. $$ Assume that $u\gt1$ and pick $v$ in $(1,u)$. Since $K\leqslant K(v)$ on $(0,v)$ and $K\leqslant K(u)$ on $(v,u)$, $$ I(u)\leqslant vK(v)+(u-v)K(u)\leqslant uK(v)+(u-v)K(u). $$ Assume that $v=u-w$ with $w\to0$ when $u\to+\infty$, then $(u-v)K(u)=wK(u)\ll K(u)$ and $$ K(v)=\mathrm e^{u^2-2uw+w^2-u+w}\leqslant\mathrm e^{u^2-u-uw}=K(u)\mathrm e^{-uw}, $$ for every $u$ large enough. Hence $uK(v)\leqslant u\mathrm e^{-uw}K(u)\ll K(u)$ for every choice of $w$ such that $w=o(1)$ and $u=o(\mathrm e^{uw})$, for example $w=1/\sqrt{u}$. QED.

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Thank you very much for this clear answer! – Chon May 28 '12 at 10:12

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