What is the Fourier Transform of an integral
When integrating functions with dirac delta you have to remember that
$$\int \delta(x - a) f(x) dx = f(a)$$ so, it works as  change of variables
and for the 2D dirac, integrating over an axis gives the 1D delta
$$ \int \delta(x - a, y) f(x) dx = \delta(y) f(a, x)$$
$$\begin{eqnarray}g(x,y) 
  &amp;=&amp; \int_{0}^{T} f(x,y) \delta(x - \nu_x t, y) dt \\
  &amp;=&amp;\delta(y) f(\nu_x t, y)
\end{eqnarray}$$
This does not depend on $x$ or $u$, it is expressed in terms of $f(\nu_x t, 0)$, if you want to express in terms of $F(u, v)$ you must calculate the inverse fourier tranform for these for this point.
$$\begin{eqnarray}G(u, v) 
&amp;=&amp; \int \left(\int \delta(y) f(\nu_x t, y) e^{-2i\pi i vy} dy \right) e^{-2i\pi i ux} dx \\
&amp;=&amp; \int \left( f(\nu_x t, 0) \right) e^{-2i\pi i ux} dx \\
&amp;=&amp; \delta(u) f(\nu_x t, 0) \\
&amp;=&amp; \delta(u) \int \left( \int F(u,v) e^{-2i\pi u v_x t} du \right) dv
\end{eqnarray}$$