Fourier Transforms: F[Rect] and F[sinc] Relationship Explanation

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Homework Statement



Explain how F[Rect] = sinc implies F[sinc] = REct +/- a few constants.


Homework Equations



2\pi f(-w) = \int^{\infty}_{-\infty} F(t) e^{-iwt} dt

The Attempt at a Solution



I have no idea!
 
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What's the formula for the inverse transform?
 
of which function exactly?
 
Of any function. What is the defining equation for getting the inverse transform?
 
I assume rect is short for "rectangle". The question asks you either to prove duality property of continuous Fourier transform or to use it. Search for the duality, you'll see what i mean.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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