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*** 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*** FourierTransform[ Sinc[b ω], {ω}, {t}, \!\(TraditionalForm\`FourierParameters -> {1, \(-1\)}\)] The result is $ \frac{\pi}{2 b} (Sign[b - t] + Sign[b + t])$ --------------------------------------------------------------------------- Any comment or suggestion would be highly appreciated.