I would like to calculate the Fourier transform of Sinc[ b (ω1 - ω2)], but there are some problems as follows:
My target is
$\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\text{Sinc}(b (\text{$\omega_1$}-\text{$\omega_2$})) e^{-i \text{$ \omega_1 t_1$} } e^{-i \text{$ \omega_2 t_2 $} } d \text{$t_1$} d\text{$t_2$} =\frac{2 \pi ^2}{b} \Pi \left(\frac{\text{$t_1$}}{2 b}\right) \delta (\text{$t_1$}+\text{$t_2$})$
The problem is as follow:
I use the following code
Assuming[{b >0 }, FourierTransform[ Sinc[b (ω1 - ω2)], {ω1, ω2}, {t1, t2}, \!\(TraditionalForm\`FourierParameters -> {1, \(-1\)}\)]]
After runing the above code for a long time, the out put is
(I \[Pi] DiracDelta[t1+t2] (-Log[I b-I t1]+Log[-I b+I t1]+Log[-I (b+t1)]-Log[I (b+t1)]))/b
It can be further simplified to be 0, because
Log[-I (b - t1)] + Log[-I (b + t1)] - Log[I (b - t1)] - Log[I (b + t1)]
= Log[I (b - t1)*I (b + t1)] - Log[I (b - t1)*I (b + t1)]
= Log[-b^2 + t1^2] - Log[-b^2 + t1^2]
= 0
This means $\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\text{Sinc}(b (\text{$\omega_1$}-\text{$\omega_2$})) e^{-i \text{$ \omega_1 t_1$} } e^{-i \text{$ \omega_2 t_2 $} } d \text{$t_1$} d\text{$t_2$} =0$ ??
How to solve this porblem?
Clue 1: $Sinc[ω1 - ω2]$
FourierTransform[ Sinc[ω1 - ω2], {ω1, ω2}, {t1, t2}, \!\(TraditionalForm\`FourierParameters -> {1, \(-1\)}\)]
The result is $ π^2 DiracDelta[t1 + t2] (Sign[1 - t1] + Sign[1 + t1]) $.
But in Mathematica, (Sign[1 - t1] + Sign[1 + t1]) does not equal to 2 UnitBox[t1/2], because FullSimplify[Sign[1 - t1] + Sign[1 + t1] - 2 UnitBox[t1/2]] = Piecewise[{{-1, t1 == -1 || t1 == 1}}, 0]
Clue 2: $Sinc[b ω]$
FourierTransform[ Sinc[b ω], {ω}, {t}, \!\(TraditionalForm\`FourierParameters -> {1, \(-1\)}\)]
The result is $ \frac{\pi}{2 b} (Sign[b - t] + Sign[b + t])$.
Clue 3: $Sinc[3 (ω1 - ω2)]$
Assuming[{a > b }, FourierTransform[Sinc[3 (ω1 - ω2)], {ω1, ω2}, {t1, t2}, \!\(TraditionalForm\`FourierParameters -> {1, \(-1\)}\)]]
The result is 0.
Any comment or suggestion would be highly appreciated.
