The two dimensional Fourier Transform of a function $f(x,y)$ is
$$\mathscr{F}\{f\}(k_x,k_y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x,y)e^{ik_xx}e^{ik_yy}dxdy$$
For $f(x,y)=\text{rect}_{L_x}(x)\text{rect}_{L_y}(y)$, the Fourier Transform is
$$\mathscr{F}\{f\}(k_x,k_y)=\int_{-L_y}^{L_y}\int_{-L_x}^{L_x}e^{ik_xx}e^{ik_yy}dxdy=2L_x\text{sinc}(k_xL_x)2L_y\text{sinc}(k_yL_y)$$
Define $I$ as the integral of interest
$$\begin{align}
I&=\int_{-\infty}^\infty \text{sinc}(8t)\text{sinc}^2(t)\cos(8\pi t)\,dt\\\\
&= \int_{-\infty}^\infty \text{sinc}(16t)\text{sinc}^2(t)\,dt
\end{align}$$
Now, the Fourier transforms of $f(t)=\text{sinc}(16 t)$ and $g(t)=\text{sinc}^2(t)$ are given respectively by
$$\begin{align}
\mathscr{F}\{f\}(\omega)&=\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt\\\\
&=\frac1{32}\left(\text{sgn}(\omega+16\pi)-\text{sgn}(\omega-16\pi)\right)
\end{align}$$
$$\begin{align}
\mathscr{F}\{g\}(\omega)&=\int_{-\infty}^\infty g(t)e^{i\omega t}\,dt\\\\
&=\frac{2\pi -\omega}{2\pi}\left(\text{sgn}(\omega+2\pi)-\text{sgn}(\omega-2\pi)\right)
\end{align}$$
Using the convolution theorem, we find that
$$\begin{align}
I&=\frac1{2\pi}\int_{-\infty}^\infty \frac{ \text{sgn}(\omega +16\pi)-\text{sgn}(\omega-16\pi)}{32} \left( \frac{2\pi -\omega}{2\pi}\left(\text{sgn}(\omega+2\pi)-\text{sgn}(\omega-2\pi)\right)\right)\,d\omega\\\\
&=\frac1{64\pi}\int_{-2\pi}^{2\pi}\frac{2\pi -\omega}{2\pi}\,d\omega\\\\
&=\frac1{16}
\end{align}$$
Note that WolframAlpha defines the sinc funtion as
$$\text{sinc}(x)=\frac{\sin(x)}{x}\ne \frac{\sin(\pi x)}{\pi x}$$
When accounting for this differnce in definition WA retunrs the correct answer (See Here) of $1/16$.
Best Answer
Rewriting the antiderivative in terms of $\sin$ and writing that function in its complex exponential form gives \begin{multline*} \int \frac{e^{-iz}}{z} \sin z \,dz = \int \frac{1}{t} e^{-iz} \cdot \frac{1}{2} \left(e^{iz} - e^{-iz}\right)\,dz = \frac{1}{2i}\int\frac{1}{z}(1 - e^{-2 i z})\,dz \\= -\frac{i}{2} \left[\int \frac{dz}{z} - \int \frac{e^{-2 i z}}{z} \,dt\right] = -\frac{i}{2} \left( \log z + \operatorname{Ei}_1(2iz)\right) + C\end{multline*}
For any $C$, this function has a removable singularity at $z = 0$, with limit $C + \frac{3\pi}{4} + i \cdot \frac{1}{2} (\log 2 + \gamma)$, where $\gamma$ is the Euler-Mascheroni constant. So, the function is an antiderivative everywhere except along the branch cut of $\log$ (we choose the branch cut so that $\arg z \in [-\pi, \pi)$). If we define $F(z)$ to be the antiderivative with $C = -\left[\frac{3\pi}{4} + i \cdot \frac{1}{2} (\log 2 + \gamma)\right]$, so that the limit of $F(0)$ as $t \to 0$ along the real axis, then the restriction $F\vert_\Bbb R$ satisfies $F\vert_\Bbb R(-t) = -\overline{F\vert_\Bbb R(t)}$.
For real $t$, $\lim_{t \to \pm \infty} \operatorname{Ei}_1(2 i t) = \pm \pi i$, which is finite. So, since $\log t$ is unbounded as $t \to \pm \infty$, $\int_{-\infty}^x \frac{e^{-it}}{t} \sin t \,dt$ does not converge for any $x$. This essentially occurs because $e^{-it} \sin t$ has a nonzero constant component.
Here $\operatorname{Ei}_1(z)$ is the analytic continuation of $$z \mapsto \int_1^\infty \frac{e^{-sz}}{s} \,ds$$ to $\Bbb C \setminus \{0\}$. This function is realized to the (complex-valued) one-argument exponential integral function, $\operatorname{Ei}(z)$, by $$\operatorname{Ei}(z) = -\operatorname{Ei}_1(-z) + \frac{1}{2} \log z - \frac{1}{2} \log \left(\frac{1}{z}\right) - \log(-z).$$