I have a question about a step in the construction of symmetric stable distributions. They were defined in the lecture as a distribution $\mu$ that has characteristic function
$f_{\mu}(t) = e^{-c|t|^{\alpha}}$ where $\alpha \in (0,2]$ and $c>0$.
Edit: I totally got something wrong stating the claim. The goal is to prove that a distribution $\mu$ exists, such that it has the before stated characteristic function. To do this a random variable X is defined via its density $g_X(x)=\frac{\alpha}{2}\frac{1}{|x|^{\alpha+1}}1_{\{|x|\geq 1\}}$ The following claim came up:
$\textbf{Claim}$: $1 - f_{X}(t) \sim K|t|^{\alpha}$ when $t\rightarrow 0$ (meaning for t small) for some $K>0$ where f denotes the characteristic function of X
First $t \geq 0$ is assumed. Using the identity $cos(tx)=\frac{e^{itx}+e^{-itx}}{2}$ and the substitution $u = tx$, one arrives at the equation
$1 - f_{X}(t) = \alpha t^{\alpha} \int_{t}^{\infty} \frac{1-cos(u)}{u^{\alpha+1}}du$
So far all the steps are clear to me, the confusing one comes now: Using Taylor, for $u\rightarrow 0$ we have $1-cos(u) \sim \frac{u^2}{2}$ , which implies that $\int_{0}^{\infty}\frac{1-cos(u)}{u^{\alpha+1}}du = K' < \infty$ and therefore $\alpha t^{\alpha} \int_{t}^{\infty} \frac{1-cos(u)}{u^{\alpha+1}}du \rightarrow K'\alpha|t|^{\alpha}$ for t small.
I'm not seeing why that integral should be finite, I see that $t \rightarrow 0$ implies $u \rightarrow 0$ due to the substitution, but we integrate u from 0 to $\infty$, so u is clearly not small. Replacing the 1-cos(u) term by $\frac{u^2}{2}$ doesn't give something finite either since
$\int_{0}^{\infty}\frac{u^2}{2u^{\alpha+1}}du = \int_{0}^{\infty}u^{1-\alpha}du = \frac{1}{2-\alpha}u^{2-\alpha}|^{\infty}_0=\infty$
If someone could point out my (for sure stupid) mistake, I would really appreciate it.
$f_\mu(t) = e^{-c|t|^\alpha} = 1 - c|t|^\alpha + o(|t|^\alpha) \Rightarrow 1- f_\mu(t) \sim c|t|^\alpha$.