Suppose we are given a fourier transform $$ F(\omega) = \frac{1}{\omega^2+4} $$ Can we use inverse laplace tranform by taking $i\omega = p$ to find the inverse fourier transform?
I did this and got the answer $$ \frac{1}{4} \cdot (e^{-2t} - e^{2t}) $$ This term should not contain the $e^2$ part, if we compare the solution with the inverse fourier transform.
Note that the inverse Fourier transform of $F(\omega)$ is
$$f(t)=\frac14 e^{-2|t|}\tag{1}$$
You can obtain this result via the inverse Laplace transform by using $s=j\omega$:
$$F(s)=\frac{1}{4-s^2}=\frac14\frac{1}{2-s}+\frac14\frac{1}{2+s}\tag{2}$$
Note that you also need to specify the region of convergence (ROC) in order to find the correct time function. Since we know that the Fourier transform exists, the region of convergence must include the complex axis (i.e. $\text{Re}\{s\}=0$). For the first term of (2) there is a pole at $s=2$, so the ROC must be the left half-plane $\text{Re}\{s\}<2$, which corresponds to a left-sided time function $f_1(t)=\frac14 e^{2t}u(-t)$, where $u(t)$ is the unit step function. The second term in (2) has a pole at $s=-2$, so its ROC is the right half-plane $\text{Re}\{s\}>-2$ with the corresponding time function $f_2(t)=\frac14 e^{-2t}u(t)$. So the complete time function is
$$f(t)=\frac14(f_1(t)+f_2(t))=\frac14 (e^{2t}u(-t)+e^{-2t}u(t))\tag{3}$$
which equals (1).