Rectangular Width Fourier Function

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Given the rectangular width of a function (as defined in the image), prove that the product of the rectangular width of f and its fourier transformation is 2pi

Working on #7, I've tried writing out the Fourier transformation and plugging it into the formula and multiplying it with Wf, but I'm getting mixed up about how I'm allowed to combine integrals and mix variables. I also think there might be a trick or some property of Fourier series that I don't know that makes it easy to solve.

Thanks in advance for any tips to solve.

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You have the relationship $$ \mathcal{F}[f](0) = \int_{-\infty}^{\infty} f(x) e^{-i0x} \, dx = \int_{-\infty}^{\infty} f(x) \, dx $$ and, for the inverse transform, the Fourier Inversion Theorem gives $$ f(0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \mathcal{F}[f](k) \, dk, $$ putting the $2\pi$ all on the inverse transform (the result is different if you use the $e^{2\pi i k x}$ convention, which changes the size of the Fourier transform, but makes it unitary).

Then it should be fairly obvious what happens to $W_f \cdot W_{\mathcal{F}[f]}$.