I would  like to calculate the Fourier transform of  Sinc[ b (ω1 - ω2)], but there are some problems as follows:

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**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$})$

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**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?



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***Clue 1***


        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]`

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***Clue 2***

        

        FourierTransform[ Sinc[b ω], {ω}, {t}, \!\(TraditionalForm\`FourierParameters -> {1, \(-1\)}\)]


The result is  $ \frac{\pi}{2 b} (Sign[b - t] + Sign[b + t])$

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Any comment or suggestion would be highly appreciated.