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I define the Fourier transform of a function $g$ to be $ \hat{g}(\lambda) = \int_{-\infty}^\infty g(x)e^{-i\lambda x} \, \mathrm{d}x $ with an associated inverse formula

$$ g(x)= \frac{1}{2\pi} \int_{-\infty}^\infty \hat{g}(\lambda) e^{-i\lambda x} \, \mathrm{d}\lambda. $$

Fix $c>0$. I wish to verify this latter formula for the rectangular function $f(x) = \mathbf{1}_{|x| < \frac{c}{2}}$. Computing the Fourier transform is straightforward: after some manipulation, I find $ \hat{f}(\lambda) = \frac{2}{\lambda} \sin\left(\frac{\lambda c}{2} \right) $ so my task must involve evaluating $$ \int_{-\infty}^\infty \frac{2}{\lambda} \sin\left(\frac{\lambda c}{2} \right ) e^{i\lambda x}\, \mathrm{d}\lambda. $$ It seems like I can do this using contour integration. It seems like I should be using the following contour (image taken from this question):

Putative contour

and sending $\epsilon \to 0$ and $R \to \infty$ to evaluate the integral. But I'm not sure how I should handle the boundary terms: for instance, parametrising the outer semicircle as $\gamma(t) = Re^{i \pi t}$ for $t\in [0,1]$ leaves the contribution $$ \lim_{R \to \infty} \int_0^1 \frac{2}{Re^{i \pi t}} \sin\left( \frac{cRe^{i\pi t}}{2} \right) e^{ix Re^{i\pi t}} \cdot \left( Ri\pi e^{i\pi t} \right) \, \mathrm{d}t = 2\pi i \lim_{R \to \infty} \int_0^1 \sin\left( \frac{cRe^{i\pi t}}{2} \right) e^{ix Re^{i\pi t}} \, \mathrm{d}t $$ which seems intractable, even after assuming I can pass the limit through the integral. Even if I can find the resulting value, it's not clear to me where the dependence on $c$ arises, since we want to attain $1$ if $|x|<c/2$ and $0$ otherwise. Am I missing something obvious?

RDL
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1 Answers1

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Let $F(\lambda)= \frac{2\sin(\lambda c/2)}{\lambda}$. Then, the inverse Fourier Transfom of $f$ is

$$\begin{align} \mathscr{F^{-1}}\{F\}(x)&=\frac1{2\pi}\int_{-\infty}^\infty F(\lambda)e^{i\lambda x}\,d\lambda\\\\ &\frac1{2\pi}\int_{-\infty}^\infty \frac{2\sin(\lambda c/2)}{\lambda} e^{i\lambda x} \,d\lambda\\\\ &=\frac1{2\pi i} \int_{-\infty}^\infty \frac{e^{i\lambda (x+c/2)}}{\lambda}\,d\lambda-\frac1{2\pi i} \int_{-\infty}^\infty \frac{e^{i\lambda (x-c/2)}}{\lambda}\,d\lambda\tag1 \end{align}$$

The integral $\int_{-\infty}^\infty \frac{e^{i\lambda t}}{\lambda}\,d\lambda$ can be evaluated applying a host of different methodologies. Here, we will use contour integration. When $t>0$ ($t<0$), we close the contour in the upper-half (lower-half) plane. The contour also deforms around the origin with a semi-circular arc. The Proceeding, we find

$$\frac1{2\pi i}\int_{-\infty}^\infty \frac{e^{i\lambda t}}{\lambda}\,d\lambda=\frac12 \text{sgn}(t)\tag2$$

Using $(2)$ in $(1)$ yields the coveted result

$$\mathscr{F^{-1}}\{F\}(x)=\frac12\left(\text{sgn}(x+c/2)-\text{sgn}(x-c/2)\right)$$

J.G.
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Mark Viola
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