# Fourier transform of a constant - Is it a function of distance only?

I have the following power spectrum $P(k)$, function of the modulus $k$ of the vector $\vec{k}$ in Fourier space:

$$P(k) = \begin{cases} P_0 \exp (-\frac{k^2 \sigma^2}{2}) & \text{for} \; k \leq k_0 \\ P_0 \exp (-\frac{k^2 \sigma^2}{2}) + \frac{P_0}{2} & \text{for}\; k > k_0 \end{cases}$$

where $P_0, \sigma$ and $k_0$ are known values. I was wondering if the correlation function (i.e. Fourier transform of this power spectrum), defined by

$$\xi (\vec{r}) = \int \frac{d^3 k}{(2 \pi)^3} \exp(- \text{i} \vec{k}\cdot \vec{r}) P(k)$$

is simply

$$\xi(r) = \begin{cases} P_0 \exp (\frac{-r^2}{2 \sigma^2}) & \text{for} \; r \leq r_0 \\ P_0 \exp (\frac{-r^2}{2 \sigma^2}) + \frac{P_0}{2}\delta (x) \delta(y) \delta(z) & \text{for}\; r > r_0 \end{cases}$$

In particular, my questions are the following:

1) Is this expression that I calculated correct?

2) If so, what is the value of $r_0$ (given $k_0$)?

3) can you help me visualize this correlation function as a function of the modulus $r$?

$\textbf{EDIT}$:

I came up with another solution, which replaces the one I provided before (which is wrong). Here it is:

$$\xi(\vec{r}) = \xi(r) = \exp{(-\text{i} k_0 r)} [(\frac{\delta(r)}{2} + \frac{1}{2 \pi \text{i} r})]$$

I was wondering:

1) Is it correct now?

2) How can I plot it as a function of the modulus $r$?