Please can some one help me on the following integration.
$$ G(\nu)=\frac{1}{\Delta t}\int_{t_a - \frac{\Delta t}{2}}^{t_a + \frac{\Delta t}{2}} f(t_a -t)e^{-2\pi\nu it}dt $$
where $f(x)=\mbox{sinc}(x)$;
I can see that the is some relationship with the convolution, but I am able to solve only for $f(t_a-t)=1$
Simplifying a bit the expression we have: $$ G(\nu) = \frac{e^{-2\pi \nu i t_a}}{\Delta t}\int_{-\Delta t/2}^{\Delta t/2}f(-t)e^{-2\pi\nu i t}\,dt$$ and since $\operatorname{sinc}(x)$ is an even function we have: $$ G(\nu) = \frac{2 e^{-2\pi \nu i t_a}}{\Delta t}\int_{0}^{\Delta t/2}f(-t)\cos(2\pi\nu t)\,dt$$ and since in a right neighbourhood of zero both $f(-t)$ and $\cos(2\pi\nu t)$ behave like $1+o(t)$, by letting $\Delta t\to 0^+$ we simply have $$ G(\nu) = e^{-2\pi\nu i t_a} $$ due to the integral version of the mean value theorem.