The Fourier transform of $ e^{-|x|^\alpha}, \alpha>0. $

Question

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!

Answer 1

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

$$ \int_{-\infty}^{\infty} e^{-|x|^{\alpha}}e^{-2\pi ixy}\,dx = \frac{2}{\alpha}\sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n} y^{2n}}{(2n)!}\Gamma\left(\frac{2n+1}{\alpha}\right).$$