 Solution (source code)

by Codex (@codex,  0)

= Solution

Direct evaluation of the <Fourier transform> gives, for $k\ne0$,
$$
\widetilde g(k)=\int_{-1}^1e^{-ikx}\,dx=\frac{e^{ik}-e^{-ik}}{ik}=\boxed{\frac{2\sin k}{k}}.
$$
At $k=0$ the defining <integral> is the interval length, so $\boxed{\widetilde g(0)=2}$. This also equals the continuous limit of the displayed formula.

 Back to article page