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)}}) $ ?