Let $\alpha > 1$. The function $$f(z) = \int_0^{\infty} \exp(-t^{\alpha}) \cos(zt) dt$$ is entire. Proof: write $g(z,t) = \exp(-t^{\alpha}) \cos(zt)$. Given $a \in \mathbb C$, $\partial_a g$ exists and $\partial_a g(z,t) = -t \exp(-t^{\alpha}) \sin(at)$ and this is integrable with respect to $t$. Hence, $f$ is differentiable at $a$ and $f'(a) = \int_0^{\infty} -t\exp(-t^{\alpha}) \sin(at) dt$.
I am wondering how to find the order of $f$. I wrote $f$ as a power series but the coefficients are just too much for me to handle. Is there a clever way to do this? If not, how to compute the order? I am okay if you can just handle a specific case (say $\alpha = 2$).
$$\left|\,f^{(n)}(0)\right|\leq \int_{0}^{+\infty}t^n \exp\left(-t^{\alpha}\right)\,dt =\frac{1}{\alpha}\cdot\Gamma\left(\frac{n+1}{\alpha}\right)$$ implies that the order of $f$ is bounded by $\frac{\alpha}{\alpha-1}$.