I am trying to work out the moment generating function of a particular probability distribution $\rho(x)$ which is an even function. This means that all the odd moments are zero, so $$M(s)= \mathbb{E}[e^{sX}] = \sum_{n=0}^\infty \frac{s^{2n} \mu_{2n}}{(2n)!}$$ where $\mu_n$ is the nth moment. For my problem, it turns out that working out the above function is difficult and it is easier to work out $$\tilde{M}(s) = \mathbb{E}[e^{sX^2}] = \sum_{n=0}^\infty \frac{s^{n} \mu_{2n}}{n!}$$ which can be viewed as the generating function of the even moments. It seems to me that there ought to be a simple relationship between $M$ and $\tilde{M}$. If I have $\tilde{M}$ how can I recover $M$? The reason I would like to have $M$ is because it can (in principle) be inverse Laplace transformed to recover $\rho$, whereas I am not aware of any inversion formula for $\tilde{M}$.
Update:
Using the fact that the Fourier transform of a Gaussian is also a Gaussian I found that $$\tilde{M}(-s) = \frac{1}{\sqrt{4 \pi s}} \int_{-\infty}^\infty M(it) e^{-t^2/4s} dt$$ However this is not really what I want since I want to go from $\tilde{M}$ to $M$ not the other way round. Also it is too general since it doesn't rely on $M$ being even.