Let $$\varphi(x):=\frac1{2\pi\sigma^2}e^{-\frac12\left(\frac{\|x\|}\sigma\right)^2}\;\;\;\text{for }x\in\mathbb R^2$$ for some $\sigma>0$. I want to numerically compute $$p(\omega):=\left|\hat\varphi(\omega)\right|^2\;\;\;\text{for }\omega\in[-1,1)^2.$$ Shouldn't be too complicated, I thought. For the numerical integration, I'm using $$[-a,a)=\bigcup_{i_1=-k_1}^{k_1-1}\bigcup_{i_2=-k_2}^{k_2-1}\left(\left[\frac{i_1}{k_1}a_1,\frac{i_1+1}{k_1}a_1\right)\times\left[\frac{i_2}{k_2}a_2,\frac{i_2+1}{k_2}a_2\right)\right)$$ for suitable chosen $a_1,a_2>0$ and $k_1,k_2\in\mathbb N$. This gives me the scheme $$\hat\varphi(\omega)\approx\frac{a_1a_2}{k_1k_2}\sum_{i_1=-k_1}^{k_1-1}\sum_{i_2=-k_2}^{k_2-1}e^{-{\rm i}2\pi\langle\omega,\left(\frac{i_1}{k_1}a_1,\:\frac{i_2}{k_2}a_2\right)}\varphi\left(\frac{i_1}{k_1}a_1,\frac{i_2}{k_2}a_2\right).$$ Plotting $p$ using this scheme and $a_1=a_2=k_1=k_2=100$, I obtain the following result:
Now, I already know that the analytical form of $\hat\varphi$ is $$\hat\varphi(\omega)=e^{-2\left(\pi\sigma\|\omega\|\right)^2}\;\;\;\text{for all }\omega\in\mathbb R^d,$$ where $d=2$ in our case. Plotting $p$ using this analytical form, I obtain this result:
They are obviously different. I think I should expect that the "spectrum is replicated" in the first result (can I somehow obtain the distance from the center until when the replication begins?). However, even if I cut out the centered spectrum, it doesn't match the one in the second result. So, is there anything wrong with the way I estimated $\hat\varphi$ in the numerical integration? All plots show $\sigma=1$.
