Fourier Transform of image convoluting with kernel
The function
$$
\frac{\sin(x)}{x}
$$
has a well-known Fourier transform:
$$
\int_{-\infty}^{\infty}  \: \frac{\sin{x}}{x} e^{-j 2 \pi f x} dx = \begin{cases} \pi &amp; |f| &lt; \frac{1}{2 \pi} \\ 0 &amp; |f| &gt; \frac{1}{2 \pi} \\ \end{cases} 
$$
which is derived here.

Your function:

$$
\frac{\sin(x)\sin(y)}{xy}
$$

is a simple separable two dimensional version of it. The 2D Fourier transform will just be:

$$
\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}  \: \frac{\sin{x}}{x} \frac{\sin{y}}{y} e^{-j 2 \pi f_1 x} e^{-j 2 \pi f_2 y} dx dy = 
\int_{-\infty}^{\infty} \: \frac{\sin{x}}{x} e^{-j 2 \pi f_1 x}  dx \int_{-\infty}^{\infty}  \:\frac{\sin{y}}{y} e^{-j 2 \pi f_2 y}  dy =\begin{cases} \pi &amp; |f_1| \mbox{ and } |f_2|  &lt; \frac{1}{2 \pi} \\ 0 &amp; |f_1| \mbox{ or }  |f_2| &gt; \frac{1}{2 \pi} \\ \end{cases} 
$$



Below is a plot that takes a random image with a white square in the middle (top left), FFTs it, multiplies the FFT by a brick filter (top right), inverse FFTs it and plots the result (bottom right).





R Code Below

# 26383

img &lt;- runif(128*128, 0, 1)
dim(img) &lt;- c(128, 128)

img[32:96,32:96] &lt;- 1

N &lt;- 50

imgFFT &lt;- fft(img)
imgFFT[N:(128-N+2),] &lt;- 0.0000000000000
imgFFT[,N:(128-N+2)] &lt;- 0.0000000000000
img2 &lt;- fft(imgFFT, inverse = TRUE) / 128 / 128

img2[Mod(img2) &gt; 1] &lt;- 1

mx &lt;- max(log(Mod(imgFFT + 0.00001)))
mn &lt;- min(log(Mod(imgFFT + 0.00001)))

scaled_log_imgFFT &lt;- (log(Mod(imgFFT + 0.00001)) - mn)/(mx - mn)

par(pty="s") # make the plit square
plot(0:256, type='n')
rasterImage(as.raster(img),0,128,128,256)
rasterImage(as.raster(scaled_log_imgFFT),128,128,256,256)
rasterImage(as.raster(Mod(img2)),128,0,256,128)