= Solution
The <Translation property of the Fourier transform> gives
$$
\begin{aligned}
\widetilde r(k)&=\widetilde u(k)\sum_{n=1}^4e^{ik(4n-10)}\\
&=\frac{4\sin^2k}{k^2}\left(e^{-6ik}+e^{-2ik}+e^{2ik}+e^{6ik}\right)\\
&=\boxed{\frac{16\sin^2k}{k^2}\cos(4k)\cos(2k)},\qquad \boxed{\widetilde r(0)=16}.
\end{aligned}
$$
For the remaining <integral>, the same <Translation property of the Fourier transform> identifies $e^{20ik'}\widetilde g(k')$ as the <Fourier transform> of $g(x+20)$. By the product form of the <convolution theorem>, with inverse normalization $1/(2\pi)$,
$$
\int_{\mathbb R}e^{20ik'}\widetilde g(k')\widetilde r(k-k')\,dk'=2\pi\,\mathcal F\{g(x+20)r(x)\}(k).
$$
But $g(x+20)$ is supported in $[-21,-19]$, whereas the four translated <triangular functions> making up $r$ are all supported in $[-8,8]$. Their product is identically zero. Therefore the requested <integral> is $\boxed{0}$ for every real $k$.
Back to article page
