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If we assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \end{align}\begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \tag{1} \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \tag{2} \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \tag{3} \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \tag{4} \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \end{align}\begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \tag{1'} \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \tag{2'} \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \tag{3'} \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \tag{4'} \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last expression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

If we assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last expression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

If we assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \tag{1} \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \tag{2} \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \tag{3} \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \tag{4} \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \tag{1'} \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \tag{2'} \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \tag{3'} \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \tag{4'} \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last expression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

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Nolord
Nolord
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If we assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last expression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

If assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last expression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

If we assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last expression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

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If assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last oneexpression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

If assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \end{align} Which of these four expressions make actual distributional sense? Of course, the last one does not, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

If assume that $$ \delta(z-1/2)e^{2\pi inz} = \delta(z-1/2)e^{\pi in} = \delta(z-1/2) (-1)^n, $$ where we used the Dirac delta. This implies $$ \delta(z-1/2) \sum_{n\in\mathbb Z} e^{2\pi inz} = \delta(z-1/2) \sum_{n\in\mathbb Z} (-1)^n. $$ On the left we have a Dirac times a Dirac comb $\mathrm{III}(z)$, and on the right we have a Dirac times a non convergent series. Something similar happens with the Fourier transform of $\text{sinc}$: \begin{align} \mathcal F[\text{sinc}(z-n)z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[\text{sinc}(z-n)] \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} e^{-2\pi inz}\text{rect}(z) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{e^{-2\pi inz}}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= n e^{-2\pi inz}\text{rect}(z) + \frac{(-1)^n}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)), \end{align} where $\text{rect}$ is the rectangular function. The last step is justified by the calculation \begin{align} e^{-2\pi inz} (\delta(z-1/2) - \delta(z+1/2)) &= e^{-2\pi inz} \delta(z-1/2) - e^{-2\pi inz}\delta(z+1/2) \\ = e^{-\pi in} \delta(z-1/2) - e^{\pi in} \delta(z-1/2) &= (-1)^n (\delta(z-1/2) - \delta(z+1/2)). \end{align} Since we have $$ \sum_{n=-\infty}^\infty \text{sinc}(z-n) = \lim_{N\to\infty} \sum_{n=-N}^N \text{sinc}(z-n) = 1, $$ by linearity \begin{align} \mathcal F[z] &= -\frac{1}{2\pi i}\frac{\text d}{\text dz}\mathcal F[1] = -\frac{1}{2\pi i}\frac{\text d}{\text dz} \delta(z) = -\frac{1}{2\pi i} \delta'(z) \\ &= -\frac{1}{2\pi i}\frac{\text d}{\text dz} \mathrm{III}(z)\text{rect}(z) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z) \text{rect}(z) - \frac{\mathrm{III}(z)}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)) \\ &= -\frac{1}{2\pi i}\mathrm{III}'(z)\text{rect}(z) - \left(\sum_{n=-\infty}^\infty (-1)^n \right) \frac{1}{2\pi i} (\delta(z-1/2) - \delta(z+1/2)). \end{align} Which of these four expressions make actual distributional sense? Or in other words, what step was invalid? Of course, the last expression does not make sense, but it followed from the identity $f(z) \delta(z-a) = f(a) \delta(z-a)$, so why did it yield a contradiction?

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