# What is the Fourier Transform of an integral

If we have a function that is an integral over the interval 0 to T

Is is correct to say that its Fourier Transform is :

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

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.

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