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If we have a function that is an integral over the interval 0 to T enter image description here

Is is correct to say that its Fourier Transform is : enter image description here

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  • $\begingroup$ \delta there is the 2D Dirac delta? the asterisk denotes multiplication? $\endgroup$
    – Bob
    Mar 4, 2021 at 23:14
  • $\begingroup$ yes 2D delta, and the asterisk denotes convolution. $\endgroup$ Mar 5, 2021 at 1:38

1 Answer 1

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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

$$ \int \delta(x - a, y) f(x) dx = \delta(y) f(a, x)$$

$$\begin{eqnarray}g(x,y) &=& \int_{0}^{T} f(x,y) \delta(x - \nu_x t, y) dt \\ &=&\delta(y) f(\nu_x t, y) \end{eqnarray}$$

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.

$$\begin{eqnarray}G(u, v) &=& \int \left(\int \delta(y) f(\nu_x t, y) e^{-2i\pi i vy} dy \right) e^{-2i\pi i ux} dx \\ &=& \int \left( f(\nu_x t, 0) \right) e^{-2i\pi i ux} dx \\ &=& \delta(u) f(\nu_x t, 0) \\ &=& \delta(u) \int \left( \int F(u,v) e^{-2i\pi u v_x t} du \right) dv \end{eqnarray}$$

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