Editing Projection-slice theorem (section)

== Proof in two dimensions ==
[[Image:ProjectionSlice.png|frame|center|A graphical illustration of the projection slice theorem in two dimensions. ''f''('''r''') and ''F''('''k''') are 2-dimensional Fourier transform pairs. The projection of ''f''('''r''') onto the ''x''-axis is the integral of ''f''('''r''') along lines of sight parallel to the ''y''-axis and is labelled ''p''(''x''). The slice through ''F''('''k''') is on the ''k''<sub>''x''</sub> axis, which is parallel to the ''x'' axis and labelled ''s''(''k''<sub>''x''</sub>). The projection-slice theorem states that ''p''(''x'') and ''s''(''k''<sub>''x''</sub>) are 1-dimensional Fourier transform pairs.]]

The projection-slice theorem is easily proven for the case of two dimensions.
Without loss of generality, we can take the projection line to be the ''x''-axis.
There is no loss of generality because if we use a shifted and rotated line, the law still applies. Using a shifted line (in y) gives the same projection and therefore the same 1D Fourier transform results. The rotated function is the Fourier pair of the rotated Fourier transform, for which the theorem again holds.

If ''f''(''x'',&nbsp;''y'') is a two-dimensional function, then the projection of ''f''(''x'',&nbsp;''y'') onto the ''x'' axis is ''p''(''x'') where

:<math>p(x)=\int_{-\infty}^\infty f(x,y)\,dy.</math>

The Fourier transform of <math>f(x,y)</math> is

:<math>
F(k_x,k_y)=\int_{-\infty}^\infty \int_{-\infty}^\infty
f(x,y)\,e^{-2\pi i(xk_x+yk_y)}\,dxdy.
</math>

The slice is then <math>s(k_x)</math>

:<math>s(k_x)=F(k_x,0)
=\int_{-\infty}^\infty \int_{-\infty}^\infty f(x,y)\,e^{-2\pi ixk_x}\,dxdy
</math>
:::<math>=\int_{-\infty}^\infty
\left[\int_{-\infty}^\infty f(x,y)\,dy\right]\,e^{-2\pi ixk_x} dx
</math>
:::<math>=\int_{-\infty}^\infty p(x)\,e^{-2\pi ixk_x} dx
</math>

which is just the Fourier transform of ''p''(''x''). The proof for higher dimensions is easily generalized from the above example.