Convergence to the stable law I am reading the book Kolmogorov A.N., Gnedenko B.V. Limit distributions for sums of independent random variables. 
From the general theory there it is known that if $X_i$ are symmetric i.i.d r.v such that  $P(|X_1|>x)=x^{-\alpha},\, x \geq 1$, then $(X_1+\ldots+X_n)n^{-1/\alpha}\to Y$, where c.f. of $Y$ equals $\varphi_Y(t)=e^{-c|t|^{\alpha}}, \alpha \in (0,2]$, so $Y$ has stable law of distribution. 
I want to check it without using that general theorems. So I start as the following, $X_1$ has density of distribution $f_X(x)=|x|^{-\alpha-1}\alpha/2, |x|>1$. Using Levy theorem one must prove that $\varphi^n_{X_1}(t/n^{1/\alpha})\to \varphi_Y(t),\, n \to \infty$ for all $t\in \mathbb R$. $$\varphi_{X_1}(t/n^{1/\alpha})=\int_{1}^{\infty}\cos(tx/n^{1/\alpha})\alpha x^{-\alpha-1}\,dx,$$ for all it is evident that $t$ $\varphi_{X_1}(t/n^{1/\alpha})\to 1, n \to \infty$ so we have indeterminate form $1^\infty$. 
So we are to find $n(\varphi_{X_1}(t/n^{1/\alpha})-1)$, but $\varphi_{X_1}(t/n^{1/\alpha})\sim 1+1/2(2txn^{-1/\alpha})^2$, and I can only say something about $\alpha=2$ and I got stuck here. Perhaps, I made a mistake somewhere.
Could you please help me? Thanks.

I yr integral make the change of variables $z = \frac y {n^{\frac 1 {\alpha}}}$.  This brings a factor $\frac 1n$ out  front.  The write $cos(tz) = 1 +  (cos(tz) -1)$.  Integrate the 1 explicitly, and the integral invlving $cos(tz)-1$ converges because it is nice at zero.
