A question about characteristic functions $1-\varphi(t) $ I am reading about stable laws on the book "Probability: Theory and Examples, by  Rick Durrett". II've been trying to understand this result for a long time and I can't.
Suppose the distribution of $X_{1}$ has
$$
P\left(X_{1}>x\right)=P\left(X_{1}<-x\right)=x^{-\alpha} / 2 \quad  \text { for } x \geq 1
$$
where $0<\alpha<2$.
If $\varphi(t)=E \exp \left(itX_{1}\right)$, then
$$
\begin{aligned}
1-\varphi(t) &=\int_{1}^{\infty}\left(1-e^{i t x}\right) \frac{\alpha}{2|x|^{\alpha+1}} d x+\int_{-\infty}^{-1}\left(1-e^{i t x}\right) \frac{\alpha}{2|x|^{\alpha+1}} d x \\
&=\alpha \int_{1}^{\infty} \frac{1-\cos (t x)}{x^{\alpha+1}} d x
\end{aligned}
$$
If $F(x)$  is the CDF of $X_1$, I know that:
$$\varphi(t)=E \exp \left(itX_{1}\right) = \int_{-\infty}^{\infty} e^{itx}dF(x)\quad \hbox{}\,\\\, 1-\varphi(t) =  \int_{-\infty}^{\infty} (1- e^{itx})dF(x) = \int_{-\infty}^{-1} (1- e^{itx})dF(x) + \int_{-1}^{1} (1- e^{itx})dF(x) + \int_{1}^{\infty} (1- e^{itx})dF(x) $$
But I don't know, from first equation, how to find exactly $F(x)$ for all $x$ and conclude that: $1-\varphi(t)=\alpha \int_{1}^{\infty} \frac{1-\cos (t x)}{x^{\alpha+1}} d x$?
Help
 A: You directly have that $\displaystyle P(X>x)=1-P(X\leq x)=1-F(x)\implies F(x)=1-\frac{x^{-\alpha}}{2}\,,x\geq 1$.
and $P(X\leq -x)=P(X\leq u)=F(u)=\frac{(-u)^{-\alpha}}{2}\,,u\leq -1$ where $u=-x$ .
This should also tell you that by right continuity and as $F$ is non-decreasing and as $F(-1)=F(1)=\frac{1}{2}$  you must have that $F(x)=\frac{1}{2}\,,-1\leq x< 1$
All in all we have $$F(x)=\begin{cases}1-\frac{x^{-\alpha}}{2}\,,x\geq 1\\\frac{1}{2}\,,-1\leq x< 1 \\ \frac{(-x)^{-\alpha}}{2}\,,x\leq -1\end{cases}$$
This gives you the density $\displaystyle f(x)=\mathbf{1}_{\{x\geq 1\}}\frac{\alpha x^{-(\alpha+1)}}{2}+\mathbf{1}_{\{x\leq -1\}}\frac{\alpha(-x)^{-(\alpha+1)}}{2}$ .
Now $1=\int_{\Bbb{R}}f(x)\,dx$.
This gives you
$$1-\varphi(t)=\int_{\Bbb{R}}f(x)dx-\int_{\Bbb{R}}f(x)e^{itx}\,dx$$
$$=\int_{1}^{\infty}(1-e^{itx}) \frac{\alpha}{2x^{\alpha+1}} \,d x +\int_{-\infty}^{-1}\left(1-e^{i t x}\right) \frac{\alpha}{2(-x)^{\alpha+1}} d x$$
Now in the 2nd integral you change variables $x=-u$ to get
$$=\int_{1}^{\infty}(1-e^{itx}) \frac{\alpha}{2x^{\alpha+1}} \,d x+\int_{1}^{\infty}\left(1-e^{-i t u}\right) \frac{\alpha}{2u^{\alpha+1}} d u$$
$$=\int_{1}^{\infty}\frac{1}{2}\cdot(2-(e^{itx}+e^{-itx}))\frac{\alpha}{2x^{\alpha+1}} d x=\int_{1}^{\infty}\alpha\frac{1-\cos (t x)}{x^{\alpha+1}} d x$$
