I have a time domain function$[f(t)=\cos(wt).e^{-t^{2}}]$
I want to find out the laplace transform of the above function.
The convolution property says that a product in time domain can be obtained as a convolution in frequency domain.
So, I can obtain the laplace transform of $f(t)$ as $F(s)= \left[\frac{s}{s^{2} + w^{2}}\right] *\left[\frac{5.7s^{2} -18.2s + 92.416}{s^{5}+8.3s^{4}+33s^{3}+74.8s^{2}+94.5s + 52.3}\right] $
where $\left[\frac{s}{s^{2} + w^{2}}\right]$ is the laplace transform of $\cos(wt)$ and $\left[\frac{5.7s^{2} -18.2s + 92.416}{s^{5}+8.3s^{4}+33s^{3}+74.8s^{2}+94.5s + 52.3}\right]$ is the laplace transform of $e^{-t^{2}}$ (which is obtained by rational approximation) and $*$ stands for convolution
How can the final expression for $F(s)$ be obtained?
I assume that your definition of $f(t)$ only holds for $t>0$, and that $f(t)=0$ for $t<0$. In this case
$$f(t)=\cos(wt)e^{-t^2}u(t)$$
where $u(t)$ is the Heaviside step function. The function
$$g(t)=e^{-t^2}u(t)$$
has a closed-form Laplace transform:
$$G(s)=\frac{\sqrt{\pi}}{2}e^{s^2/4}\text{erfc}(s/2)$$
Note that $f(t)$ can be written as
$$f(t)=\cos(wt)g(t)=\frac12\left(e^{iwt}g(t)+e^{-iwt}g(t)\right)$$
Now you can use the shifting property of the Laplace transform to obtain
$$F(s)=\frac12\left(G(s-iw)+G(s+iw)\right)$$