A PDF (probability density function) f(x) is called a stable law if $f(y)=b\int_{-\infty}^{\infty}dx f(by-x)f(x)$ under appropriate values of b.
Rewrite this equation in terms of characteristic functions and show that the Levy distribution with characteristic function $f_X(k)=e^{-C|k|^{\alpha}}$ with $0<\alpha\le2 $ satisfies it.
Here's my approach to the problem:
Using the relation between characteristic functions and PDF: $f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-ikx}f_X(k)dk$ I was able to write the equation in terms of characteristic functions as
$ \int_{-\infty}^{\infty}e^{-ik_1y}f_Y(k_1)dk_1=\frac{b}{2\pi}\int_{-\infty}^{\infty}dx\Big(\int_{-\infty}^{\infty}e^{-ik_2(by-x)}f_X(k_2)dk_2\Big)\Big(\int_{-\infty}^{\infty}e^{-ik_3x}f_X(k_3)dk_3\Big)$
My idea was to substitute $f_X(k)=e^{-C|k|^{\alpha}}$ in this expression and find the value of $b$ so that it holds. The problem is that I am unable to compute the integrals required, that is,
$\int_{-\infty}^{\infty}e^{-ikx}e^{-C|k|^\alpha}dk$
My initial though was to split it in two integrals ranging from $(-\infty,0)$ and $(0,\infty)$, respectively but I am unable to continue from there.
Is there another way of doing it or can anyone guide me a bit on how to tackle this kind of integrals?
Let $F(f(x))=f(k)$ denote the Fourier transform of $f(x)$. Let $(f*g)(t)=\int_{-\infty}^{\infty}f(t-x)g(x)dx$ denote the convolution of f and g.
To rewrite the expression in terms of characteristic functions we apply Fourier transform on both sides. We have
$f(k)=bF((f*f)(by))=f^2(k/b) $
where we have used the properties
$F((f*g)(x))=f(k)g(k)$ and $F((f)(bx))=b^{-1}f(k/b)$
one can easily insert the Levy characteristic function in this expression and find the value b that makes it satisfy it.