Since sinc is an entire function and decays with $1/\omega$, we can slightly shift the contour of integration in the inverse transform, and since there's no longer a singularity then, we can split the integral in two:
$$\begin{eqnarray}\int_{-\infty}^\infty e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega
&=&
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}-e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\\\
&=&
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{i\omega/2}}{i\omega}\mathrm{d}\omega -
\int_{-\infty+\epsilon i}^{\infty+\epsilon i} e^{i\omega x}\frac{e^{-i\omega/2}}{i\omega}\mathrm{d}\omega\;.
\end{eqnarray}$$
Now we can apply different substitutions to the two parts, $\omega'=\omega(x+\frac{1}{2})$ in the first part and $\omega'=\omega(x-\frac{1}{2})$ in the second part. That transforms the integrand into $e^{i\omega'}/(i\omega')$ in both cases. Now if $x$ lies outside the rectangle, the signs of the factors in the substitutions are the same, so the two integrals stay on the same side of the origin and go in the same direction, and hence yield the same value and cancel to $0$. But if $x$ lies inside the rectangle, then there's a sign change due to $x-\frac{1}{2}$ but not due to $x+\frac{1}{2}$, so we get
$$\int_{-\infty+\epsilon i}^{\infty+\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega' +
\int_{\infty-\epsilon i}^{-\infty-\epsilon i} \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'\;,$$
which is (again using the sufficient decay at infinity)
$$\oint \frac{e^{i\omega'}}{i\omega'}\mathrm{d}\omega'$$
on a contour that encloses the pole at the origin, and hence the value is $2\pi$.
The last line is wrong, it's supposed to be:
$$\frac{\sin(2\pi u)}{2\pi}\Big(\frac{1}{u}-\frac{1}{1-2u}-\frac{1}{1+2u}\Big)$$
I'll work out the first term, the rest are similar:
$$F(\omega)=\mathcal{F}_u\bigg[\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)\bigg]=\int_{-\infty}^{+\infty}\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)e^{-i\omega u}du$$
Keep in mind that $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ Hence:
$$F(\omega)=\int_{-\infty}^{+\infty}\frac 1{2\pi}\sin(2\pi u)\left(\frac 1u\right)e^{-i\omega u}du=\int_{-\infty}^{+\infty}\frac 1{4\pi i}\left(\frac 1u\right)e^{-i(\omega-2\pi) u}du-\int_{-\infty}^{+\infty}\frac 1{4\pi i}\left(\frac 1u\right)e^{-i(\omega+2\pi) u}du$$
Using two facts $$\mathcal{F}_t[e^{i\omega_0t}\cdot f(t)]=F(\omega-\omega_0)$$
where $G(\omega)=\mathcal{F}_t[f(t)]$
and $$\mathcal{F}_t\bigg[\frac 1t\bigg]=-i\pi\text{sgn} (\omega)$$
You get
$$F(\omega)=-\frac 1{4} \text{sgn}(\omega-2\pi)+\frac 1{4} \text{sgn}(2 \pi+\omega)$$
For the other parts, using the following property (very easy to prove):
$$\mathcal{F}_t(f(t-t_0))=e^{-it_0\omega}F(\omega)$$
$$F_1(\omega)=\frac 12 e^{-i\omega/2}F(\omega)$$
$$F_2(\omega)=\frac 12 e^{i\omega/2}F(\omega)$$
The transform $G_u(\omega)$ of the function $f(u)$ is:
\begin{align*}
G_u(\omega)&=F(\omega)+F_1(\omega)+F_2(\omega)\\
&=(1+\cos(w/2)F(\omega)\\
&=2\cos^2(\omega/4)F(\omega)
\end{align*}
So that $$\mathcal{F}_t(g(t))=G(\omega)=\frac T2 \cos^2(\frac{T\omega}
4) \left(\text{sgn}(2 \pi - T\omega) + \text{sgn}(2 \pi + T\omega)\right)$$
Best Answer
The Fourier transform is linear, so you can compute the $rect$ function that isn't multiplied by $cos$ by itself.
For the $rect$ * $cos$ term, remember that multiplication in the frequency domain is convolution in the time domain.
So $\mathscr F^{-1} (G(\omega)*F(\omega)) = \{g \ast f\}(t)$.
Consider $cos(pi \cdot f)$ and $rect(f/2)$ as $G(\omega)$ and $F(\omega)$ respectively. Take the inverse Fourier transforms of $G$ and $F$ separately. Then, try plugging $g(t)$ and $f(t)$ into the convolution definition.