The Fourier transform of $ e^{-|x|^\alpha}, \alpha>0. $ Do you know the Fourier transform of
$$ 
e^{-|x|^\alpha},  \alpha>0.
$$
Does it have an implicit formula.
In the spacial cases
$$
(e^{-|x|})^\hat{\,}(\xi)=\frac{2}{1+4\pi^2\xi^2},\  (e^{-\pi|x|^2})^\hat{\,}(\xi)=e^{-\pi|\xi|^2},\ \xi\in R.
$$
Here the Fourier transform for $f\in L^1(R)$ takes this form
$$
\hat{f}(\xi)=\int_{R}f(x)e^{-2\pi ix\xi}dx,\ \xi\in R.
$$
Thanks for any hints!
 A: We wish to evaluate
$$\int_{-\infty}^{\infty} e^{-|x|^{\alpha}}e^{-2\pi ixy}\,dx.$$
Making use of evenness, we get
$$2\int_0^{\infty} e^{-x^{\alpha}} \cos(\pi xy)\,dx.$$
Expanding $\cos$ in a power series, we have
$$2\sum_{n=0}^{\infty}\frac{(-1)^n\pi^{2n} y^{2n}}{(2n)!}\int_0^{\infty} e^{-x^{\alpha}} x^{2n}\,dx.$$
So we need only to evaluate the integral now. Making a change of variable $y = x^{\alpha}$, we have $dy = \alpha x^{\alpha-1} = \alpha y^{1-\frac{1}{\alpha}}\,dx$ and so
$$\int_0^{\infty} e^{-x^{\alpha}} x^{2n}\,dx = \int_0^{\infty} e^{-y}\left(y^{\frac{1}{\alpha}}\right)^{2n}\frac{1}{\alpha y^{1-\frac{1}{\alpha}}}\,dy = \frac{1}{\alpha}\int_0^{\infty} e^{-y} y^{\frac{2n+1}{\alpha}-1}\,dy.$$
From the definition of the Gamma function
$$\Gamma(z) = \int_0^{\infty} e^{-t}t^{z-1}\,dt,$$
we get that
$$\int_0^{\infty} e^{-x^{\alpha}} x^n\,dx = \frac{1}{\alpha}\Gamma\left(\frac{2n+1}{\alpha}\right).$$
Plugging this into the series gives
$$ \int_{-\infty}^{\infty} e^{-|x|^{\alpha}}e^{-2\pi ixy}\,dx = \frac{2}{\alpha}\sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n} y^{2n}}{(2n)!}\Gamma\left(\frac{2n+1}{\alpha}\right).$$
