Computing $\int_{-\infty}^{+\infty} e^{-iky} e^{-c|k|^{\alpha}} |k|^{\beta} dk$

Question

I want to solve the above integral, where $c >0$, $\beta = 0,2$, $0<\alpha<2 $, $y \in \mathbb{R}$, $k \in \mathbb{R}$. I tried to use the Residue Theorem adding the semi-circle contour of infinite radius in the half-plane with negative immaginary numbers.
I don't see singularity inside this half-plane to use for the residue computation; what am I missing?
Any comment is welcome
[EDIT]
As pointed by metamorphy I have a branch cut due to $|k|^{\alpha}$ at exponent, so I can't apply the residue theorem. How should I proceed to solve the integral?

Answer 1

The best you can get is a Fox-Wright-type power series.

Substitute $k=c^{-1/\alpha}t$ (essentially reducing to $c=1$) and split $\int_{-\infty}^\infty=\int_{-\infty}^0+\int_0^\infty$ (only $\alpha>0$, $\alpha\neq 1$ and $\Re\beta>-1$ are assumed below). The problem is then reduced to $$f(\alpha,\beta,z)=\int_0^\infty t^\beta\exp(-zt-t^\alpha)\,dt,$$ where $z\in\mathbb{C}$, and $\Re z\geqslant 0$ if $\alpha<1$ (to have a convergent integral). Now, if $\alpha>1$, then $$f(\alpha,\beta,z)=\sum_{n=0}^\infty\frac{(-z)^n}{n!}\int_0^\infty t^{n+\beta}e^{-t^\alpha}dt=\frac1\alpha\sum_{n=0}^\infty\frac{(-z)^n}{n!}\Gamma\left(\frac{n+\beta+1}{\alpha}\right).$$ If $0<\alpha<1$ and $\Re z>0$, we expand $e^{-t^\alpha}$ this time, and get $\int_0^\infty t^{\eta-1}e^{-zt}\,dt$, which is equal to $\Gamma(\eta)/z^\eta$ even in our case of complex $z$ (this is where contour integration comes into play): $$f(\alpha,\beta,z)=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^\infty t^{\alpha n+\beta}e^{-zt}\,dt=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\frac{\Gamma(\alpha n+\beta+1)}{z^{\alpha n+\beta+1}}.$$ (The case $\Re z=0$ is covered by taking a limit. The case $\alpha=1$ is trivial.)