It helps with that integral, I know that it does not converge, it will be oscillating since they are complex exponentials. If there is any way to get to known functions or solve it in some way, thank you. I did the following, with Ramanujan's master theorem, I do not know if it is correct as I did:
Ramanujan's master theorem says: If a complex-valued function $f(x)$ has an expansion of the form $$f(x) = \sum_{k=0}^{\infty} \frac{(-x)^k}{k!} \phi(k)$$ then the Mellin transform of $f(x)$ is given by $$\int_{0}^{\infty} x^{s-1} f(x) dx = \Gamma(s) \phi(-s),$$ where $\Gamma(s)$ is the gamma function.
For the case that I have, I did: $$(k^2 +\alpha)^{1/2} \equiv \omega_{k}\quad\text{and}\quad e^{-i 2\omega_{k} t - i k x} = \sum_{j=0}^{\infty} \frac{(-i2\omega_{k}t)^{j}}{(j!)} \cdot \sum_{j=0}^{\infty} \frac{(-i k x)^{j}}{(j!)},$$ where $\sum_{j=0}^{\infty} \frac{(-i k x)^{j}}{(j!)} \equiv \phi(j)$, so that
$$\int_{0}^{\infty} k^{s-1} f(k) dk = \Gamma(s) \phi(-s),$$
and similarly for the other term but negative sign. Then:
$$\int_{0}^{\infty} e^{-i2\omega_{k}t} (e^{ikx} + e^{-ikx}) dk \rightarrow \int_{0}^{\infty} k^{s-1}f(k)dk + \int_{0}^{\infty} (k')^{s-1} f(k') dk' = 2\Gamma(s) \phi(-s)$$
Is it correct how I take $\phi$ and what happens to the other half of the integral, the one that goes from $-\infty$, to $0$?