How can I show that a function defined by an integral is entire 
It is in the form of a Fourier transform, and the question asks to show that this is  entire function of $z$, but I don't know the way to prove that it is entire.
I have tried once using contour integration. Its integral is zero on any circle, but I cannot link it to entireness.
 A: Morera's Theorem and the related Cauchy's integral theorem and may be what you're seeking.
$$\oint_\gamma f(z) dz = 0$$
If $f$ is continuous on an open set that contains $\gamma$, and satisfies the above integral, then $f$ is holomorphic on that set (analytic). By extension, if the above integral vanishes for every possible $\gamma$, then $f$ is analytic everywhere.
A: Hint.
Apply the complex form of differentiation under the integral sign theorem.
If you define $$f_\alpha(z,t)=e^{-\vert t \vert^\alpha} e^{2\pi i z t},$$ you have $$F_\alpha(z)=\int_{-\infty}^{+\infty} f_\alpha(z,t) dt.$$ The partial derivative $$\frac{\partial f_\alpha}{\partial z}(z,t)=2\pi i t e^{-\vert t \vert^\alpha} e^{2\pi i z t}$$ is continuous on $\mathbb C \times \mathbb R$ and we have: $$\left\vert \frac{\partial f_\alpha}{\partial z}(z,t) \right\vert \le 2 \pi \vert t \vert e^{-\vert t \vert^\alpha - 2 \pi \mathfrak{I}(z)t}$$ where $\mathfrak{I}(z)$ is the imaginary part of $z$. As $\alpha > 1$, one can find $1 < \beta < \alpha$ such that $$-\vert t \vert^\alpha - 2 \pi \mathfrak{I}(z)t \le -\vert t \vert^\beta$$ for $t$ in the neighborhood of $\pm \infty$. As $\displaystyle \int_{-\infty}^{+\infty} 2 \pi \vert t \vert e^{-\vert t \vert^\beta} dt$ converges, $\displaystyle \int_{-\infty}^{+\infty} \frac{\partial f_\alpha}{\partial z}(z,t) dt$ also.
