I am looking for an equivalent of the following integral : $$ \int_{0}^{A}{e^{t^\alpha}}dt $$ with $0<\alpha$ when $A$ tends to infinity.
I call an equivalent a simpler function $f$ of $A$ such that $$ \int_{0}^{A}{e^{t^\alpha}}dt = f(A)(1 + o(1)) $$ when $A$ tends to infinity with the notation of Landau.
I have already tried an integration by part and a change of variable but I cannot control the last integral to show it is negligible.
If you have any idea. Thanks.
By the L'Hospital rule $$ \mathop {\lim }\limits_{A \to + \infty } \frac{{\int_0^A {e^{t^\alpha } dt} }}{{\frac{1}{\alpha }A^{1 - \alpha } e^{A^\alpha } }} = \mathop {\lim }\limits_{A \to + \infty } \frac{1}{{1 + \frac{{1 - \alpha }}{{\alpha A^\alpha }}}} = 1, $$ hence $$ \int_0^A {e^{t^\alpha } dt} = \frac{1}{\alpha }A^{1 - \alpha } e^{A^\alpha } (1 + o(1)) $$ as $A\to +\infty$.