Let $\xi>0$ and be real, $0<\alpha < 1$ and be real and $m\in\mathbb{N}$. Consider the integral on the complex plane z,
$$\mathcal{F}(u) = \int_{\mathcal{C}}\exp(-(z\xi)^{\alpha})(u-z)^mdz$$
Where $u \in \mathbb{C}$. What would be the choice of the contour $\mathcal{C}$ such that the inverse Laplace transform of $\mathcal{F}(u)$, $\mathcal{L}^{-1}(\mathcal{F})(t)$ converges? My problem is that the exponential term is not analytical and I am not sure how to consider the branch cut.
P.S. I want to calculate the Laplace transform of the multiplication of two functions, let say $g(t)$ and $h(t)$. Thus, $$\mathcal{L}(h(t)g(t))(u) = \frac{1}{2\pi i}\int_{\mathcal{C}}\mathcal{H}(z)\mathcal{G}(u-z)dz$$ where $\mathcal{H}(u)$ and $\mathcal{G}(u)$ are the Laplace transforms of $h(t)$ and $g(t)$ respectively. After simplification I got $\mathcal{F}(u)$.
Thanks in advance.