I want to proof that $$\int_0^1 \exp(bx^{2a})\mathrm dx=\mathcal O(b^{1-2a}\exp(2b^a)).$$ I already found the similar bound ($a<1/2$), maybe I could use this by substitution: \begin{align} \int_1^m \exp \left(2 t^{2 a}\right) d t =\int_1^m\left(4 a t^{2 a-1} \exp \left(2 t^{2 a}\right)\right) \frac{d t}{4 a t^{2 a-1}} \\ \leqslant\int_1^m\left(4 a t^{2 a-1} \exp \left(2 t^{2 a}\right)\right) \frac{d t}{4 a m^{2 a-1}} \leqslant \frac{m^{1-2 a}}{4 a} \exp \left(2 m^{2 a}\right) \end{align} but I don't understand the last bound, i.e $$\int_1^m\left(4 a t^{2 a-1} \exp \left(2 t^{2 a}\right)\right)d t\le \exp \left(2 m^{2 a}\right).$$
Thanks
Use $2a < 1$ to have that $t^{2a - 1} \le 1$ and $t \le m$ to have $\exp\left({2t^2}\right) \le \exp \left(2m^2\right)$. You will have the inequality you are looking for.